221 free, high-quality past-paper-style questions on Kinematics for IB Physics SL, each with a full mark scheme. Covers: Distance, displacement & averages; Motion graphs; Equations of motion; and more. For when you have finished the past papers, or want to practise one topic at a time.A.1 at Higher Level →Read the A.1 notes →My revision plan →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.1,029 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1A-1A-01Distance, displacement & averagesPaper 1AEasy1 mark
An athlete runs once around a circular track of circumference 400 m, returning to the starting point after 50 s.
What are the magnitude of the average velocity and the average speed of the athlete?
Average velocity / m s−1Average speed / m s−1
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Step 1Displacement is the change in position. After one complete lap the athlete is back at the start, so the displacement is 0 and the average velocity is 0.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Distance is the total path length, 400 m, so the average speed is 400 / 50 = 8.0 m s−1.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the displacement after one complete lap is zero, so the average velocity is zero; the distance travelled is 400 m, so the average speed is 8.0 m s⁻¹.
BThe average speed cannot be zero: the athlete covers a distance of 400 m, and average speed is distance divided by time.
CThis reverses the two quantities. Velocity uses displacement, which is zero here; speed uses distance, which is not.
DThis treats displacement as equal to distance. They are equal only for motion in a straight line without reversal.
The graph shows how the velocity v of a tram varies with time t as it travels along a straight track between two stops.
Which graph shows how the displacement s of the tram from the first stop varies with t?
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Step 1The velocity is the gradient of the displacement–time graph. From 0 to 10 s the velocity increases from zero, so the graph starts with zero gradient and curves upwards; from 10 s to 40 s the velocity is constant at 12 m s−1, so the graph is a straight line.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2From 40 s to 55 s the velocity decreases to zero but is never negative, so s keeps increasing while the gradient decreases, and the graph becomes horizontal at 55 s: Graph D.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis copies the shape of the velocity–time graph. The displacement is the area under the velocity–time graph, so it keeps increasing while the velocity is positive; it does not fall back to zero.
BThe last stage curves the wrong way. The tram is slowing down, so the gradient of the displacement–time graph, which is the velocity, must decrease, not increase.
CThe curvatures are swapped. At the start the tram speeds up from rest, so the gradient must start at zero and increase; at the end it slows down, so the gradient must decrease.
DCorrect: the gradient (the velocity) rises from zero, stays constant, then falls to zero, so the graph curves upwards, is straight, then curves over to become horizontal.
6A-1A-47Distance, displacement & averagesPaper 1AEasy1 mark
A cyclist starts from rest at a traffic light and moves along a straight road. The graph shows how her displacement s varies with time t. The dashed line is the tangent to the curve at t = 4.0 s.
What are her average velocity from t = 0 to t = 4.0 s and her instantaneous velocity at t = 4.0 s?
Graph drawn to scale
Average velocity / m s−1Instantaneous velocity / m s−1
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Step 1Average velocity = change in displacement ÷ time = (12 − 0) m ÷ 4.0 s = 3.0 m s−1 (the gradient of the chord).
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Instantaneous velocity = gradient of the tangent at 4.0 s = (24 − 0) m ÷ (6.0 − 2.0) s = 6.0 m s−1.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis uses the chord for both values. The instantaneous velocity at 4.0 s is the gradient of the tangent at that point, which is steeper than the chord.
BCorrect: the chord from the origin to (4.0 s, 12 m) gives 3.0 m s−1; the tangent at 4.0 s gives 6.0 m s−1.
CThis swaps the two quantities: the chord gives the average velocity and the tangent gives the instantaneous velocity.
DThis uses the tangent for both values. The average velocity over the interval is the total displacement divided by the total time, which is the gradient of the chord.
A ball is projected from level ground at an angle above the horizontal. Air resistance is negligible.
Which statement about the ball at the highest point of its path is correct, whatever the angle of projection?
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Step 1At the highest point the vertical component of velocity is zero, while the horizontal component is unchanged throughout the flight because no horizontal force acts. So the speed is the least there and equals that horizontal component.
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Answer C
Answer: C · 1 stage of work, one mark
Every option, and why
AOnly the vertical component of the velocity is zero. The horizontal component is unchanged, so the ball is still moving.
BThe only force is the weight, which is vertical at every point. This option assumes that a moving body needs a force in its direction of motion.
CCorrect: the vertical component is zero and the horizontal component is constant, so the speed is least and equal to the horizontal component of the launch velocity.
DThe fraction of the kinetic energy left at the top is cos² of the launch angle, which equals ½ only for a 45° launch; it is not true in general.
8A-1A-67Distance, displacement & averagesPaper 1AEasy1 mark
A runner uses a GPS app to record a run along winding paths through a park. The run finishes at a different point from where it started.
Which statement about the whole run must be correct?
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Step 1Average speed = distance travelled ÷ time; magnitude of average velocity = magnitude of displacement ÷ the same time.
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Step 2The distance along any path is at least as long as the straight line from start to finish, so the average speed can never be less than the magnitude of the average velocity.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThe mean of speeds recorded at equal time intervals is the average speed (distance ÷ time), not the magnitude of the average velocity, which uses the straight-line displacement.
BA constant speed does not make the route straight. Along winding paths the distance is greater than the magnitude of the displacement, whatever the speed.
CThis reverses the inequality. The magnitude of the displacement can never exceed the distance travelled, so the magnitude of the average velocity can never exceed the average speed.
DCorrect: distance ≥ magnitude of displacement for any path, and both are divided by the same time.
A lifeguard standing at the edge of a pier throws a rescue bag to a swimmer in the sea below. She throws bag X horizontally. She then throws an identical bag, Y, from the same point with the same speed but at 25° below the horizontal. Air resistance is negligible.
Which row compares the motion of Y with the motion of X?
Time taken by Y to reach the seaHorizontal distance travelled by Y
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Step 1Vertically, both bags fall through the same height with the same acceleration g, but Y starts with a downward component of velocity, u sin 25°, and X with none, so Y reaches the sea sooner.
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Step 2Horizontally, Y moves at u cos 25°, which is less than u, and for a shorter time, so it travels a smaller horizontal distance than X.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the downward component of the launch velocity shortens the fall, and a smaller horizontal component acting for less time gives a smaller horizontal distance.
BThis assumes that throwing the bag partly downwards makes it travel further. Its horizontal component, u cos 25°, is smaller than u, and it is in the air for less time, so it lands closer to the pier.
CThis treats the time of fall as depending only on the height. That is true only when the vertical component of the launch velocity is the same; Y starts with a downward component, so it falls faster.
DThis assumes that the same speed from the same height gives the same motion. The velocity must be resolved: Y has a larger downward and a smaller horizontal component than X.
13A-1A-72Velocity & accelerationPaper 1AEasy1 mark
A juggler throws a ball vertically upwards and catches it at the height from which it was thrown. Air resistance is negligible. Upwards is positive.
Which graph shows how the acceleration a of the ball varies with time t while the ball is in the air?
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Step 1The only force on the ball is its weight, so its acceleration is g downwards, −9.81 m s−2, at every instant — including the highest point, where the velocity is zero but still changing.
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Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: a constant negative acceleration of −g from release to catch.
BThis reverses the sign of the acceleration when the ball starts to fall. The weight, and so the acceleration, is downwards on the way up and on the way down.
CThis is the shape of the velocity–time graph (velocity decreasing uniformly through zero), not the acceleration–time graph.
DThis makes the acceleration zero at the highest point. The velocity is zero there, but it is still changing at the rate g.
14IBP-A1-N1-01Motion graphs & definitionsPaper 1AEasy1 mark
A data logger records the acceleration of a trolley on a laboratory track and plots an acceleration–time graph. What does the area between this graph and the time axis represent?
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Acceleration is the rate of change of velocity, a = Δv/Δt, so Δv = aΔt. The area under an a–t graph therefore gives the change in velocity. (Displacement is the area under a v–t graph; rate of change of velocity is the value read off the a–t graph itself.)
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Answer B
Answer: B
Common error: Students often apply the rule "area = displacement" to every graph. Check the axes first: area under v–t gives displacement, area under a–t gives change in velocity.
15IBP-A1-N1-02Motion graphs & definitionsPaper 1AEasy1 mark
A cyclist rides exactly half a lap of a circular track of radius 35 m. What is the magnitude of the cyclist's displacement?
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After half a lap the cyclist is diametrically opposite the starting point. Displacement is the straight-line distance from start to finish: 2r = 2 × 35 = 70 m. The distance travelled along the track is πr ≈ 110 m, which is not the displacement.
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Answer B
Answer: B
Exam tip: Displacement depends only on the start and end positions, not on the path. Sketch the start and finish points and join them with a straight line.
16IBP-A1-N1-03Motion graphs & definitionsPaper 1AEasy1 mark
The displacement–time graph for a train on a straight track is a straight line. At t = 0 the displacement is 120 m and at t = 30 s the displacement is 480 m. What is the velocity of the train?
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Velocity = gradient of the s–t graph = Δs/Δt = (480 − 120)/30 = 360/30 = 12 m s−1.
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Answer B
Answer: B
Common error: A gradient uses the change in displacement. Dividing a single reading (480 m) by the time ignores that the graph does not start at the origin.
17IBP-A1-N2-01Equations of motionPaper 1AEasy1 mark
A cyclist moving at 2.0 m s−1 along a straight road accelerates uniformly at 1.5 m s−2 for 6.0 s. What is the speed of the cyclist at the end of the 6.0 s?
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v = u + at = 2.0 + 1.5 × 6.0 = 11 m s−1.
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Answer C
Answer: C
Common error: The initial speed is easy to forget. at is only the change in velocity; add it to u to get the final velocity.
18IBP-A1-N2-02Equations of motionPaper 1AEasy1 mark
An aircraft touches down on a straight, level runway at 70 m s−1. It then decelerates uniformly at 3.5 m s−2 until it stops. What distance does the aircraft travel along the runway while decelerating?
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v2 = u2 + 2as with v = 0 and a = −3.5 m s−2: s = u2 / (2 × 3.5) = 4900 / 7.0 = 7.0 × 102 m.
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Answer B
Answer: B
Exam tip: When the final velocity is zero and no time is given, v² = u² + 2as gets you there in one step. Put the deceleration in as a negative acceleration.
A smartphone slips out of a person's hand and falls from rest. Air resistance is negligible and g = 9.8 m s−2. What is the speed of the phone 1.5 s after it is released?
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From rest, v = u + at = 0 + 9.8 × 1.5 = 14.7 m s−1 ≈ 15 m s−1.
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Answer C
Answer: C
Common error: Using s = ½gt² gives the distance fallen (11 m), not the speed. Check which quantity the question asks for before choosing an equation.
A juggler throws a ball vertically upwards. Air resistance is negligible. Which row describes the velocity and the acceleration of the ball at the instant it reaches its highest point?
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At the highest point the ball is momentarily at rest, so its velocity is zero. The only influence on its motion is gravity, so its acceleration is 9.8 m s−2 downwards throughout the flight, including at the top.
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Answer B
Answer: B
Common error: Zero velocity does not mean zero acceleration. In free fall the acceleration is the same at every instant, whether the object is rising, at the top or falling.
21IBP-A1-N5-01Kinematics data & measurementPaper 1AEasy1 mark
In a road-safety class, a student tests her reaction time. A partner holds a vertical ruler with its zero mark level with her open fingers and releases it without warning. She catches it after it has fallen 0.15 m. Take g = 9.8 m s−2 and assume air resistance is negligible. What is her reaction time?
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The ruler starts from rest, so d = ½gt2 and t = √(2d/g) = √(2 × 0.15 / 9.8) = √0.0306 = 0.17 s (0.175 s).
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Answer D
Answer: D
Common error: Forgetting the square root in t = √(2d/g) gives a time with the wrong size by a large factor. A human reaction time of a few tenths of a second is a useful sense check.
22IBP-A1-N5-02Kinematics data & measurementPaper 1AEasy1 mark
A model railway wagon carries a card of length 4.0 cm. As the wagon passes a light gate, the card blocks the beam for 16 ms. What is the average speed of the wagon while the card passes through the gate?
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Convert to SI units first: length = 0.040 m, time = 0.016 s. v = 0.040 / 0.016 = 2.5 m s−1.
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Answer C
Answer: C
Exam tip: Convert every quantity to SI units (cm to m, ms to s) before substituting. A quick check that the answer is a sensible speed for the object catches most conversion slips.
23IBP2S-001-05-01Velocity & accelerationPaper 1AEasy1 mark
A skydiver jumps from a stationary hot-air balloon. Before she opens her parachute, her speed increases, but the gradient of her velocity–time graph steadily decreases. Which describes her motion during this time?
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The gradient of a velocity–time graph is the acceleration. A decreasing gradient means the acceleration is getting smaller. The velocity values are still rising, so her speed is increasing. Air resistance grows as she speeds up, so her acceleration falls from g towards zero.
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Answer B
Answer: B
Common error: A falling gradient does not mean the object is slowing down. Read the values on the graph for speed, and read the gradient for acceleration.
24IBP2S-001-05-02Motion graphs & definitionsPaper 1AEasy1 mark
A delivery drone starts from rest and rises vertically. Its acceleration increases uniformly from 0 to 4.0 m s−2 over 3.0 s. What is the speed of the drone at 3.0 s?
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The change in velocity equals the area under the acceleration–time graph. Here that area is a triangle: ½ × 3.0 × 4.0 = 6.0 m s−1. The drone starts from rest, so its speed at 3.0 s is 6.0 m s−1. The option 12 m s−1 treats the final acceleration as if it acted for the whole 3.0 s. The option 1.3 m s−1 divides by the time instead of multiplying.
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Answer C
Answer: C
Exam tip: When acceleration is not constant, v = u + at cannot be used. Find the area under the a–t graph instead.
25IBP2S-001-05-07Fluid resistance & terminal speedPaper 1AEasy1 mark
A raindrop is falling vertically at its terminal speed. Which statement about the raindrop is correct?
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At terminal speed, the air resistance is equal in size to the weight. The resultant force is zero, so the acceleration is zero and the velocity stays constant.
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Answer A
Answer: A
Common error: 'Terminal' does not mean the drop has stopped. It means the drop has stopped speeding up.
A decorative fountain launches a jet of water at an angle above the horizontal. Air resistance is negligible. Which describes the velocity and the acceleration of a water droplet at the highest point of its path?
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With no air resistance the horizontal velocity component is constant and non-zero, so at the top the velocity is horizontal. Only the vertical component is zero there. The only force is gravity, so the acceleration is g downwards at every point, including the top. Options with zero velocity treat the motion as purely vertical. Options with zero acceleration confuse zero vertical velocity with zero acceleration.
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Answer C
Answer: C
Common error: Zero vertical velocity does not mean zero acceleration. Gravity never switches off during the flight.
27IBP2S-001-00-02Motion graphs & definitionsPaper 1AEasy1 mark
A track cyclist completes exactly two laps of a 250 m velodrome in 36 s. She finishes at her starting point. What are her average speed and the magnitude of her average velocity?
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Average speed = total distance ÷ time = 500 m ÷ 36 s = 13.9 m s−1. Average velocity = displacement ÷ time. She returns to her start, so the displacement is zero and the average velocity is zero. Option A treats speed and velocity as the same. Option C swaps the two definitions. Option D uses only one lap for the distance.
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Answer B
Answer: B
Exam tip: Use distance for speed and displacement for velocity. A closed loop always gives zero displacement.
28IBP2S-001-00-06Velocity & accelerationPaper 1AEasy1 mark
A tennis ball travelling at 20 m s−1 hits a racket. It leaves at 30 m s−1 in the opposite direction. The contact time is 5.0 ms. What is the magnitude of the average acceleration of the ball during contact?
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Velocity is a vector. Take the outgoing direction as positive. The change in velocity is +30 − (−20) = 50 m s−1. Then a = 50 ÷ 5.0×10−3 = 1.0×104 m s−2. Option A subtracts the two speeds and ignores the reversal of direction. Options B and C use only one of the two speeds.
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Answer D
Answer: D
Common error: When an object rebounds, add the speeds to find the change in velocity. Do not subtract them.
29IBP2S-001-00-10Motion graphs & definitionsPaper 1AEasy1 mark
A car's speedometer reads a steady 30 km h−1 as the car drives once around a circular roundabout. Which statement is correct?
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Velocity has both magnitude and direction. The speed is constant, but the direction of motion changes continuously, so the velocity changes. Acceleration is the rate of change of velocity, so it is non-zero. Option A treats velocity as speed. Options C and D are internally inconsistent: velocity changes if and only if acceleration is non-zero.
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Answer B
Answer: B
Common error: Constant speed does not mean constant velocity. A change in direction alone is an acceleration.
30IBP2S-001-04-01Fluid resistance & terminal speedPaper 1AEasy1 mark
A delivery drone releases a small medical package. The package falls vertically and soon reaches terminal speed. Buoyancy is negligible. Which statement about the package while it falls at terminal speed is correct?
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At terminal speed the velocity is constant, so the acceleration and the resultant force are zero. The only two forces are the weight, acting down, and the drag, acting up. They must therefore be equal in magnitude. The weight still acts, so A is wrong. B treats the package as if it were in free fall. D confuses constant speed with a force imbalance.
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Answer C
Answer: C
Common error: Constant velocity means zero resultant force, not zero force. The weight never disappears; it is balanced by the drag.
31IBP2S-001-04-02Fluid resistance & terminal speedPaper 1AEasy1 mark
A ball-throwing machine for dogs launches a ball, without spin, at an angle above the horizontal across level ground. Air resistance on the ball is not negligible. Compared with the same launch in a vacuum, which statement is correct?
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Drag reduces the horizontal velocity continuously. On the way down the ball moves more slowly horizontally than it did on the way up, so the descent is steeper than the ascent and the path is not symmetrical. B ignores the horizontal component of the drag. C would only be true for a symmetrical path with no air resistance. A is wrong because drag also lowers the maximum height, which generally shortens the flight.
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Answer D
Answer: D
Exam tip: With air resistance, look for three changes: a lower maximum height, a shorter range and an asymmetric path with a steeper descent.
32IBP2S-001-04-07Fluid resistance & terminal speedPaper 1AEasy1 mark
A sediment sampler is released from rest just below the surface of the sea and sinks vertically. The water exerts a drag force that increases with speed. Which describes the velocity–time graph for the sampler?
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At release the speed is zero, so there is no drag and the acceleration (the gradient) is at its largest. As the speed rises, the drag increases, the resultant force falls and the gradient decreases. The velocity approaches the terminal value, where the gradient is zero. A ignores drag. B reverses the trend in acceleration. D suggests that drag can reverse the motion, but drag only reduces the resultant force towards zero.
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Answer C
Answer: C
Exam tip: The gradient of a velocity–time graph is the acceleration. Link the gradient at each stage to the size of the resultant force.
33IBP2S-001-04-09Equations of motionPaper 1AEasy1 mark
An engineer models the drag on a racing cyclist as F = kv2, where v is the cyclist's speed. What are the SI base units of k?
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k = F/v2 = (kg m s−2)/(m2 s−2) = kg m−1. Option A is the unit of k for a linear drag law, F = kv. Option C is just the newton. Option D multiplies by v2 instead of dividing by it.
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Answer B
Answer: B
Exam tip: Rearrange the equation for the constant first. Then substitute base units, writing the newton as kg m s⁻².
34IBP2S-001-03-01Motion graphs & definitionsPaper 1AEasy1 mark
A velocity–time graph is plotted for a hospital lift as it travels between floors. What quantity is represented by the area between the graph line and the time axis?
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Area under a velocity–time graph is velocity × time, which is displacement. Acceleration is the gradient of the graph, not the area. Average velocity is the area divided by the time interval. Rate of change of acceleration would come from the gradient of an acceleration–time graph.
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Answer B
Answer: B
Common error: Mixing up gradient and area. For a v–t graph, gradient gives acceleration and area gives displacement. Check this by multiplying or dividing the axis units.
35IBP2S-001-03-02Motion graphs & definitionsPaper 1AEasy1 mark
A delivery drone's displacement–time graph is a horizontal straight line at a displacement of 12 m for a time interval of 5.0 s. What is the drone doing during this interval?
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The gradient of a displacement–time graph is the velocity. A horizontal line has zero gradient, so the velocity is zero and the drone stays at 12 m. Reading 12 off the axis as a velocity, or dividing 12 m by 5.0 s, wrongly treats the displacement value as a change in displacement.
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Answer A
Answer: A
Exam tip: On a displacement–time graph the height of the line gives position. Only the gradient tells you about velocity.
A juggler throws a ball vertically upwards and catches it at the same height. Upwards is taken as positive and air resistance is negligible. Which describes the velocity–time graph from release to catch?
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The only force on the ball is gravity, so the acceleration is −9.8 m s−2 throughout, including at the top. The v–t graph is therefore one straight line of gradient −9.8 m s−2. The velocity passes through zero at the highest point and becomes negative on the way down. The V-shape is a speed–time graph. A horizontal section would mean zero acceleration at the top. A curve would need air resistance.
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Answer B
Answer: B
Common error: At the top of the flight the velocity is zero but the acceleration is still g downwards.
37IBP2S-001-03-09Motion graphs & definitionsPaper 1AEasy1 mark
A cyclist's displacement–time graph rises from 0 m at t = 0 to 120 m at t = 40 s. It then falls to 40 m at t = 60 s. What is the cyclist's average velocity over the full 60 s?
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Average velocity is the overall change in displacement divided by the time: (40 − 0)/60 = 0.67 m s−1. The value 3.3 m s−1 is the average speed, 200 m/60 s. The value 2.0 m s−1 uses the maximum displacement over the full time. The value 3.0 m s−1 is the velocity during the first 40 s only.
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Answer A
Answer: A
Common error: Using total distance instead of net displacement when the motion reverses direction.
38IBP2S-001-02-01Motion graphs & definitionsPaper 1AEasy1 mark
A swimmer completes four lengths of a straight 50 m pool in 160 s. She finishes at the same end of the pool where she started. What are her average speed and the magnitude of her average velocity?
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Average speed = total distance ÷ time = 200 m ÷ 160 s = 1.25 m s−1. She ends where she started, so her displacement is zero and her average velocity is zero. Option A treats speed and velocity as the same quantity. Option B swaps the two quantities. Option D uses only one 50 m length for the distance.
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Answer C
Answer: C
Common error: Average velocity depends only on the start and end positions. Average speed depends on the whole path travelled.
39IBP2S-001-02-03Kinematics data & measurementPaper 1AEasy1 mark
An autonomous warehouse trolley travels a distance of (2.00 ± 0.02) m in a time of (4.0 ± 0.2) s. What is the percentage uncertainty in its average speed?
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The percentage uncertainty in distance is 0.02/2.00 = 1 %. The percentage uncertainty in time is 0.2/4.0 = 5 %. For a quotient, the percentage uncertainties add: 1 % + 5 % = 6 %. Option A subtracts the uncertainties because the quantities are divided. Option B keeps only the larger uncertainty. Option C averages the two.
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Answer D
Answer: D
Common error: Percentage uncertainties add for both multiplication and division. They never subtract.
40IBP2S-001-02-09Distance, displacement & averagesPaper 1AEasy1 mark
Which statement about the motion of an object is always true?
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The straight line between the start and end points is the shortest possible path. So |displacement| ≤ distance, with equality only for straight-line motion in one direction. Option A reverses this inequality. Option B is false: displacement is a vector and its sign depends on the chosen direction. Option C is false: distance is accumulated path length and can only increase or stay constant.
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Answer D
Answer: D
Exam tip: Distance is a scalar that never decreases. Displacement is a vector that can increase, decrease or change sign.
41IBP2S-001-01-01Non-uniform accelerationPaper 1AEasy1 mark
A maglev test vehicle starts from rest and accelerates uniformly along a straight guideway. It travels 2.0 m in the first second. What distance does it travel during the third second of its motion?
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Starting from rest, s = ½at2, so s ∝ t2. The total distances after 1 s, 2 s and 3 s are in the ratio 1 : 4 : 9, giving 2.0 m, 8.0 m and 18 m. The distance covered during the third second is 18 − 8.0 = 10 m. Option A assumes distance grows linearly with time. Option B is the total distance after 2 s. Option D is the total distance after 3 s, not the distance in the third second.
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Answer C
Answer: C
Common error: 'During the third second' means between t = 2 s and t = 3 s. It does not mean the total distance up to t = 3 s.
A robot vacuum cleaner moves along a straight line in the negative x-direction. Its speed is increasing. Which describes the signs of its velocity and its acceleration?
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The robot moves in the negative direction, so its velocity is negative. Its speed is increasing, so the acceleration acts in the same direction as the velocity. The acceleration is therefore also negative. Option A comes from thinking that 'speeding up' always means a positive acceleration. Options C and D give the velocity the wrong sign.
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Answer B
Answer: B
Exam tip: An object speeds up when velocity and acceleration have the same sign. It slows down when they have opposite signs.
43IBP2S-001-01-07Distance, displacement & averagesPaper 1AEasy1 mark
The equation s = ut + ½at2 is used to analyse the motion of an object. Which condition must be satisfied for the equation to be valid?
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The suvat equations are derived by assuming that the acceleration is uniform, so constant acceleration is the condition required. Option A is a special case (a = 0), not a requirement. Option C is also a special case (u = 0). Option D is not required: the equation applies to any straight-line motion with constant acceleration.
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Answer B
Answer: B
Common error: Do not apply suvat equations to motion where the acceleration changes, such as a falling object approaching terminal speed.
A stone is thrown vertically upwards with speed u from the edge of a cliff. It later lands in the sea, a vertical distance h below the point of release. Air resistance is negligible.
What is the speed of the stone as it reaches the sea?
Diagram not to scale
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Step 1Take upwards as positive. Over the whole flight the displacement is s = −h and the acceleration is a = −g; the path up and back down does not need to be split.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2v² = u² + 2as = u² + 2(−g)(−h) = u² + 2gh, so the speed is √(u² + 2gh).
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis is the speed of a stone dropped from rest. The stone returns to the cliff edge moving downwards at u, so its initial speed cannot be ignored.
BThis takes the displacement as +h while keeping a = −g: a sign error. The stone ends up below the point of release, so s and a have the same sign.
CThis adds the two speeds as numbers. The contributions combine through v² = u² + 2as, not by addition.
DCorrect: with s = −h and a = −g, v² = u² + 2gh. The same result follows from energy conservation, ½mv² = ½mu² + mgh.
A driver sees a hazard and, after a fixed reaction time, brakes with a constant deceleration until the car stops. The reaction time and the deceleration do not depend on the speed of the car.
The initial speed of the car is doubled. What happens to the distance travelled during the reaction time and to the braking distance?
Distance during reaction timeBraking distance
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Step 1During the reaction time the speed is constant, so that distance is ut, proportional to u: × 2.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2While braking, 0 = u² − 2as gives s = u²/2a, proportional to u²: × 4.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the reaction distance ut doubles, and the braking distance u²/2a increases by 2² = 4.
BThis assumes the braking time stays the same. At twice the speed the car needs twice as long to stop (t = u/a) at twice the average speed, so the braking distance increases four times.
CThis applies the square law to both distances. During the reaction time there is no acceleration, so that distance is simply ut.
DThis swaps the two dependences: it is the braking distance that goes as u² and the reaction distance as u.
A ball is projected horizontally from the top of a vertical cliff and lands on the level ground below. Air resistance is negligible.
The ball is projected horizontally again from the same point with twice the speed. What happens to the time of flight and to the horizontal distance travelled?
Time of flightHorizontal distance
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Step 1The vertical motion starts from zero vertical velocity in both cases and covers the same height, so the time of flight is unchanged: × 1.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The horizontal velocity is constant, so the horizontal distance = uxt doubles: × 2.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis assumes a faster ball reaches the ground sooner, so that the two effects cancel. The horizontal velocity has no effect on the vertical motion.
BThis assumes the faster ball also stays in the air longer. The time of flight depends only on the height and on g.
CThe time is right, but the distance is taken to depend on the square of the launch speed. With the time fixed, the horizontal distance is simply proportional to ux.
DCorrect: the time of flight is set by the vertical motion alone, √(2h/g), so it is unchanged, and the horizontal distance uxt doubles.
48A-1A-21Non-uniform accelerationPaper 1AMedium1 mark
A body starts from rest and moves in a straight line. Its acceleration increases uniformly with time from zero, as shown on the graph. The speed of the body at time T is v.
What is the speed of the body at time 2T?
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Step 1The change in speed is the area under the acceleration–time graph. Up to time T the area is a triangle: v = ½ × T × kT = ½kT², where a = kt.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Up to time 2T the triangle has twice the base and twice the height: area = ½ × 2T × 2kT = 2kT² = 4v.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis assumes the speed is proportional to time, which is true only for uniform acceleration. Here the acceleration itself grows.
BThis is the area between T and 2T only, ½(kT + 2kT)T = 3v: the increase in speed. The speed v already gained at T must be added.
CCorrect: the area under the graph grows as t², so doubling the time multiplies the speed by 4.
DThis cubes the time ratio. A quantity proportional to t³ here is the displacement, not the speed.
49A-1A-36Fluid resistance & terminal speedPaper 1AMedium1 mark
A ball is projected from level ground. In each graph the dashed line shows how the height h of the ball varies with horizontal distance x when air resistance is negligible.
Which solid line shows the trajectory of the same ball, launched with the same velocity, when air resistance is not negligible?
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Step 1Air resistance opposes the velocity throughout, so the ball rises less high and travels less far, and the horizontal velocity keeps falling; the path is therefore no longer symmetric: the ball comes down more steeply than it went up, and the highest point lies beyond the middle of the range.
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Answer B
Answer: B · 1 stage of work, one mark
Every option, and why
AThe height and range are reduced correctly, but the path is still a symmetric parabola. With air resistance the horizontal velocity falls during the flight, so the descent is steeper than the ascent.
BCorrect: lower maximum height, shorter range, and an asymmetric path whose descent is steeper than its ascent, because the horizontal velocity keeps decreasing.
CThis path is asymmetric the wrong way round: it rises more steeply than it falls. The ball is moving horizontally more slowly on the way down, so the descent is the steeper part (and the launch direction is the same as without air resistance).
DThis assumes air resistance acts only on the vertical motion, so that the range is unchanged. It also opposes the horizontal motion, so the range is reduced.
50A-1A-37Velocity & accelerationPaper 1AMedium1 mark
A body moves along a straight line.
Which statement about the motion of the body must be correct?
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Step 1Acceleration is the rate of change of velocity, so zero acceleration means the velocity, and therefore the speed, does not change; each of the other statements has a counter-example (a ball at the top of its flight; a body speeding up in the negative direction; a body that goes out and comes back).
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Answer B
Answer: B · 1 stage of work, one mark
Every option, and why
AA ball thrown upwards has zero velocity at the top of its flight, yet its acceleration is still g downwards. Zero velocity does not mean zero rate of change of velocity.
BCorrect: zero acceleration means zero rate of change of velocity, so the velocity, and with it the speed, stays constant.
CA negative acceleration slows the body only if the velocity is positive. A body moving in the negative direction with negative acceleration is speeding up.
DThis confuses an average value with instantaneous values. A body that moves out and returns to its start has zero average velocity but was moving.
An electric bus starts from rest and moves along a straight road with uniform acceleration.
What is (distance travelled during the third second) ÷ (distance travelled during the first second)?
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Step 1From rest, s = ½at², so the distances travelled by t = 1 s, 2 s and 3 s are in the ratio 1 : 4 : 9.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Distance in the third second = 9 − 4 = 5 units; distance in the first second = 1 unit; ratio = 5.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis is s(3 s) ÷ s(1 s), the ratio of the total distances travelled from the start, not the distance covered during the third second alone.
BThis is the distance covered from t = 1 s to t = 3 s (9 − 1 = 8 units), which spans two seconds rather than the third second only.
CCorrect: the distances from rest grow as t², so the third second covers 9 − 4 = 5 units against 1 unit in the first second.
DThis is the ratio of the speeds at the ends of the third and first seconds (3 : 1). The distance covered in a second depends on the average speed during that second: 2.5a in the third second against 0.5a in the first, a ratio of 5.
A stone is thrown from a bridge with speed u at an angle θ below the horizontal. It reaches the river a time t later. Air resistance is negligible.
What is the height of the bridge above the river?
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Step 1Take downwards as positive. The vertical component of the initial velocity is u sin θ, downwards, and the vertical acceleration is g, downwards.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Vertical displacement = (u sin θ)t + ½gt².
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: the stone already moves downwards at u sin θ when released, and then accelerates downwards at g.
BThis uses u cos θ, which is the horizontal component of the initial velocity. The angle θ is measured from the horizontal, so the vertical component is u sin θ.
CThis treats the stone as thrown upwards at θ above the horizontal, so the vertical component of the initial velocity has the wrong sign.
DThis assumes that the initial velocity is horizontal, so it ignores the downward component u sin θ.
53A-1A-51Fluid resistance & terminal speedPaper 1AMedium1 mark
A light foam ball is thrown vertically upwards. Air resistance on the ball is not negligible.
Which statement about the motion of the ball is correct?
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Step 1While the ball rises, air resistance acts downwards, in the same direction as the weight, so the resultant downward force is greater than mg and the deceleration is greater than g.
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Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: on the way up the drag and the weight both act downwards, so the magnitude of the acceleration exceeds g.
BEnergy is transferred to the air on the way up and on the way down, so the ball returns with less kinetic energy and a smaller speed.
CThis assumes a symmetric flight. The ball is slower at each height on the way down, so it takes longer to fall than to rise.
DThis has the direction of the drag wrong for the fall. While the ball falls, air resistance acts upwards, opposing the weight, so the acceleration is less than g.
A remote-controlled boat moves in a straight line back towards a jetty, slowing down as it approaches. The graph shows how its displacement s from the jetty varies with time t. Displacement away from the jetty is positive.
Which row gives the signs of the velocity and of the acceleration of the boat at point X?
Graph NOT drawn to scale
Velocity at XAcceleration at X
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Step 1The gradient of the s–t graph at X is negative, so the velocity is negative (towards the jetty).
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The gradient is becoming less negative, so the velocity is increasing algebraically: the acceleration is positive (away from the jetty), opposite to the velocity, which is why the boat slows down.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis takes the sign of the velocity from the displacement, which is positive, and assumes that slowing down always means a negative acceleration.
BThe velocity is negative, but slowing down means the acceleration is opposite to the velocity, so here the acceleration is positive.
CThis takes both signs from the displacement, which is positive. The velocity is the gradient, which is negative at X.
DCorrect: negative gradient → negative velocity; gradient becoming less steep → velocity changing in the positive direction → positive acceleration.
During the push-start of a bobsleigh, the sled moves along a straight section of track with uniform acceleration. Two timing gates a distance d apart record its speed as v1 at the first gate and v2 at the second.
What is the acceleration of the sled?
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Step 1v² = u² + 2as with u = v1, v = v2 and s = d gives v2² = v1² + 2ad, so a = (v2² − v1²)/2d.
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Answer A
Answer: A · 1 stage of work, one mark
Every option, and why
ACorrect: rearranging v2² = v1² + 2ad.
BThis squares the difference of the speeds. The equation involves the difference of the squares, v2² − v1², which is not equal to (v2 − v1)².
CThis drops the factor of 2 from 2as, giving twice the true acceleration.
DThis subtracts the squares the wrong way round, giving a negative acceleration for a sled that is speeding up.
A basketball is thrown towards a hoop that is higher than the point of release. The ball rises and then falls into the hoop. Air resistance is negligible and upwards is positive.
In each graph the solid line shows the vertical component of the velocity and the dashed line shows the horizontal component. Which graph is correct for the flight from release to the hoop?
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Step 1The horizontal component is constant because no horizontal force acts; the vertical component decreases uniformly at the rate g, passing through zero at the highest point.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2The hoop is higher than the release point, so the ball is caught on the way down before it falls back to its release height: the final vertical component is negative but smaller in magnitude than the initial one.
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThis line ends at a vertical component equal and opposite to the initial one, which happens only when the ball returns to its release height. The hoop is higher, so the flight ends sooner.
BThe horizontal component is drawn decreasing. With no air resistance there is no horizontal force, so it stays constant.
CThis plots the magnitude of the vertical component, which is always positive. With upwards positive the vertical component becomes negative once the ball is falling.
DCorrect: constant horizontal component; vertical component decreasing linearly from positive to a smaller negative value.
Water leaves a horizontal outlet pipe in the wall of a storm drain at a speed of 8.0 m s−1 and then falls freely. Air resistance is negligible.
At what angle below the horizontal is the water moving 0.60 s after it leaves the pipe?
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Step 1Vertical component after 0.60 s: vy = gt = 9.81 × 0.60 = 5.89 m s−1; the horizontal component stays 8.0 m s−1.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2tan θ = 5.89/8.0 = 0.736, so θ = 36°.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis is the direction of the displacement from the pipe (tan θ = 1.77 m / 4.8 m), not the direction of the velocity. The vertical displacement is ½gt², but the vertical velocity is gt.
BCorrect: tan θ = vy/vx = 5.89/8.0, so θ = 36°.
CThis uses sin θ = 5.89/8.0, treating the horizontal component 8.0 m s−1 as the resultant speed.
DThis is the angle between the velocity and the vertical, tan θ = 8.0/5.89, not the angle below the horizontal.
59A-1A-78Fluid resistance & terminal speedPaper 1AMedium1 mark
A table-tennis practice robot fires a ball horizontally from a fixed height above the floor. Air resistance on the ball is not negligible.
Compared with the motion the ball would have without air resistance, which statement is correct?
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Step 1Air resistance acts opposite to the velocity. Its horizontal component slows the horizontal motion; once the ball is moving downwards, its vertical component acts upwards.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2So the resultant downward force is less than the weight: the vertical acceleration is less than g, the ball takes longer to reach the floor, and it lands closer to the robot.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThe horizontal component of the air resistance acts backwards throughout the flight, so the horizontal component of the velocity decreases.
BCorrect: the upward component of the air resistance on the falling ball partly opposes its weight, so the downward acceleration is less than g.
CThe vertical acceleration is less than g, so the ball takes longer, not less time, to fall the same height.
DThe horizontal velocity decreases and the time of flight changes, so the horizontal distance is not the same; it is shorter.
60A-1A-79Fluid resistance & terminal speedPaper 1AMedium1 mark
A small steel ball is fired vertically downwards into a deep tank of oil. It enters the oil at a speed greater than its terminal speed in the oil.
In each graph the dashed line shows the terminal speed. Which graph shows how the speed v of the ball varies with time t after it enters the oil?
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Step 1Above the terminal speed the drag is greater than the weight minus the buoyancy force, so the resultant force is upwards and the ball slows down.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2As it slows the drag decreases, so the deceleration decreases: the speed falls ever more slowly towards the terminal speed, at which the resultant force is zero.
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Answer A
Answer: A · 2 stages of work, one mark
Every option, and why
ACorrect: speed decreasing at a decreasing rate, approaching the terminal speed from above.
BThis shows a constant deceleration followed by a sudden change to constant speed. The drag, and so the deceleration, decreases gradually as the ball slows.
CThis shows the ball slowing towards rest, as if the drag were the only force. The weight keeps the ball moving at the terminal speed.
DThis is the graph for a ball released from rest in the oil. It ignores the initial speed, which is greater than the terminal speed.
61A-1A-80Distance, displacement & averagesPaper 1AMedium1 mark
An indoor positioning system gives the position of a warehouse drone as coordinates (x, y) in a horizontal plane, measured from a fixed reference point. At t = 0 the drone is at (2.0 m, 6.0 m). At t = 4.0 s it is at (8.0 m, −2.0 m).
What are the magnitude of the displacement of the drone and the magnitude of its average velocity for this interval?
Diagram not to scale
Displacement / mAverage velocity / m s−1
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Step 1Displacement = change in position = (8.0 − 2.0, −2.0 − 6.0) = (6.0 m, −8.0 m); magnitude = √(6.0² + 8.0²) = 10 m.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Magnitude of average velocity = 10 m ÷ 4.0 s = 2.5 m s−1.
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Answer C
Answer: C · 2 stages of work, one mark
Every option, and why
AThis subtracts the two distances from the reference point, 8.2 m − 6.3 m. The displacement is the vector change in position, not the change in distance from the reference point.
BThis uses the final position, 8.2 m from the reference point, as the displacement. The drone did not start at the reference point.
CCorrect: |Δr| = √(6.0² + 8.0²) = 10 m and 10/4.0 = 2.5 m s−1.
DThis adds the sizes of the components, 6.0 m + 8.0 m. Perpendicular components combine by Pythagoras.
62IBP-A1-N1-04Motion graphs & definitionsPaper 1AMedium1 mark
A lift in an office building starts from rest. Its speed increases uniformly to 2.4 m s−1 in 3.0 s, stays at 2.4 m s−1 for 8.0 s, then decreases uniformly to zero in 2.0 s. What is the average speed of the lift for the whole journey?
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Distance = area under the v–t graph = ½(3.0)(2.4) + (8.0)(2.4) + ½(2.0)(2.4) = 3.6 + 19.2 + 2.4 = 25.2 m. Total time = 3.0 + 8.0 + 2.0 = 13.0 s. Average speed = 25.2/13.0 = 1.9 m s−1 (1.94 m s−1).
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Answer C
Answer: C
Exam tip: Average speed is always total distance ÷ total time. Only for a single stage of uniform acceleration can you average the initial and final speeds.
63IBP-A1-N1-05Motion graphs & definitionsPaper 1AMedium1 mark
A delivery drone moves along a straight horizontal line. At t = 0 its velocity is +2.0 m s−1. Its acceleration–time graph shows +1.5 m s−2 from t = 0 to t = 4.0 s, then −0.50 m s−2 from t = 4.0 s to t = 10.0 s. What is the velocity of the drone at t = 10.0 s?
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Change in velocity = signed area under the a–t graph = (1.5)(4.0) + (−0.50)(6.0) = 6.0 − 3.0 = +3.0 m s−1. Final velocity = 2.0 + 3.0 = +5.0 m s−1.
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Answer B
Answer: B
Common error: The area under an a–t graph gives only the change in velocity. Add it to the initial velocity, and treat area below the time axis as negative.
64IBP-A1-N2-03Equations of motionPaper 1AMedium1 mark
A rider on an electric scooter is travelling in a straight line at 6.0 m s−1 when she sees an obstacle ahead. Her reaction time is 0.80 s, during which the speed does not change. The scooter then decelerates uniformly at 2.5 m s−2 until it stops. What is the total distance travelled from the moment she sees the obstacle until the scooter stops?
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Reaction distance = 6.0 × 0.80 = 4.8 m. Braking distance = u2/(2a) = 36 / 5.0 = 7.2 m. Total = 4.8 + 7.2 = 12 m.
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Answer C
Answer: C
Common error: Stopping distance has two stages. During the reaction time the vehicle moves at constant speed, so the distance for that stage is simply speed × time.
65IBP-A1-N2-04Equations of motionPaper 1AMedium1 mark
A lift in a tall building starts from rest and moves vertically upwards. It accelerates uniformly at 1.2 m s−2 for 2.5 s and then moves at constant speed for a further 8.0 s. What is the total distance travelled by the lift in the 10.5 s?
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Speed after acceleration: v = 1.2 × 2.5 = 3.0 m s−1. Stage 1: s = ½ × 1.2 × 2.52 = 3.75 m. Stage 2: s = 3.0 × 8.0 = 24 m. Total = 28 m (27.75 m).
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Answer B
Answer: B
Exam tip: In two-stage motion, the final velocity of the first stage is the initial velocity of the second. Treat each stage with its own equation, then add the displacements.
A water jet in a decorative fountain leaves its nozzle vertically upwards at 6.0 m s−1. Air resistance is negligible and g = 9.8 m s−2. What is the maximum height reached by the water above the nozzle?
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At the top v = 0. Using v2 = u2 − 2gs: s = u2/(2g) = 36/19.6 = 1.84 m ≈ 1.8 m.
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Answer C
Answer: C
Exam tip: When only the launch speed is given, v² = u² + 2as reaches the maximum height in one step without finding the time first.
Before serving, a tennis player tosses a ball vertically upwards and lets it fall back to the height from which it was released. Upwards is taken as positive and air resistance is negligible. Which describes the velocity–time graph for the ball from release until it returns to the release height?
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The acceleration is constant (−g), so the graph is a straight line with gradient −g. The velocity starts positive, passes through zero at the top and becomes negative. With no air resistance the motion is symmetric, so the ball returns with the same speed in the opposite direction: a straight line ending at a negative value of equal magnitude.
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Answer C
Answer: C
Exam tip: On a velocity–time graph with a sign convention, a change of direction appears as the line crossing the time axis, not as a bounce back to positive values. A V-shape is what a speed–time graph looks like.
68IBP-A1-N5-03Kinematics data & measurementPaper 1AMedium1 mark
On a test track, the deceleration of a braking car is found from a = v2 / (2s), where v is the speed when the brakes are applied and s is the stopping distance. The percentage uncertainty in v is 3% and the percentage uncertainty in s is 2%. What is the percentage uncertainty in a?
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For a product or quotient, percentage uncertainties add, and a power multiplies its percentage uncertainty. %a = 2 × 3% + 2% = 8%. The factor 2 in the denominator is exact and adds no uncertainty.
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Answer D
Answer: D
Common error: Uncertainties never subtract, even when a quantity is in the denominator. Square terms double their percentage uncertainty.
69IBP-A1-N5-04Kinematics data & measurementPaper 1AMedium1 mark
A sledge is released from rest at the top of a snowy slope. A motion sensor records its displacement s along the slope at time t. A graph of s against t2 is a straight line through the origin that passes through the point (t2 = 4.0 s2, s = 2.4 m). What is the acceleration of the sledge?
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From rest, s = ½at2, so the gradient of s against t2 is a/2. Gradient = 2.4 / 4.0 = 0.60 m s−2, so a = 2 × 0.60 = 1.2 m s−2.
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Answer C
Answer: C
Exam tip: When a relationship is linearised, write the equation in the form y = mx and match each term: here y = s, x = t², and the gradient m = a/2.
A tennis ball is hit vertically upwards. Air resistance on the ball is not negligible. What is the magnitude of the acceleration of the ball at the instant it reaches its highest point?
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At the highest point the ball is momentarily at rest. Air resistance depends on speed, so it is zero at that instant. The only force acting is the weight, so the acceleration is exactly g, directed downwards. The ball's velocity is zero there, but its acceleration is not.
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Answer B
Answer: B
Common error: Zero velocity does not mean zero acceleration. If the acceleration were zero at the top, the ball would stay there.
71IBP2S-001-05-04Fluid resistance & terminal speedPaper 1AMedium1 mark
A cricket ball is thrown vertically upwards and caught at the same height from which it was released. Air resistance is significant. tup is the time taken to rise and tdown is the time taken to fall. Which is correct?
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On the way up, drag and weight both act downwards, so the deceleration is greater than g and the ball stops quickly. On the way down, drag opposes weight, so the acceleration is less than g and the fall takes longer over the same height. Drag removes energy throughout the flight, so the return speed is less than the launch speed.
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Answer B
Answer: B
Exam tip: Treat the up and down journeys separately. Decide the direction of drag on each leg before you compare the accelerations.
72IBP2S-001-05-05Motion graphs & definitionsPaper 1AMedium1 mark
The displacement–time graph for a hospital lift has three sections: • 0 to 4 s: a straight line of positive gradient; • 4 s to 6 s: a curve whose gradient decreases smoothly to zero; • after 6 s: a horizontal line. Which describes the acceleration of the lift in these three sections, in order?
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The gradient of a displacement–time graph is the velocity. • A straight line means constant velocity, so the acceleration is zero. • A decreasing gradient means the lift is slowing down, so the acceleration is opposite to the velocity. • A horizontal line means the lift is at rest, so the acceleration is zero.
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Answer A
Answer: A
Common error: A straight line on a displacement–time graph means constant velocity, not constant acceleration. Do not read an s–t graph as if it were a v–t graph.
73IBP2S-001-05-06Motion graphs & definitionsPaper 1AMedium1 mark
A probe in deep space starts from rest and fires its engine. As fuel is used, its acceleration increases linearly with time. Which pair best describes the acceleration–time graph and the velocity–time graph of the probe?
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Acceleration increasing linearly with time gives a sloping straight line on the a–t graph. The gradient of the v–t graph equals the acceleration. Because the acceleration keeps growing, the v–t graph is a curve that gets steeper with time. A straight v–t line would mean constant acceleration.
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Answer B
Answer: B
Exam tip: Each graph is the gradient of the one above it: s–t → v–t → a–t. Check that the shapes you choose agree with this.
74IBP2S-001-05-08Fluid resistance & terminal speedPaper 1AMedium1 mark
A football is kicked from level ground and lands back on level ground. Air resistance is significant. Which statement about its flight is correct?
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Drag continually reduces the horizontal component of velocity. By the end of the flight the horizontal speed is small compared with the vertical speed, so the ball comes down more steeply than it went up. The path is therefore not symmetrical. • The peak occurs beyond the halfway point. • The horizontal velocity is not constant. • The best launch angle is less than 45°.
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Answer C
Answer: C
Exam tip: With drag, describe the trajectory as asymmetric: lower peak, shorter range, and a steeper descent than ascent.
75IBP2S-001-05-09Motion graphs & definitionsPaper 1AMedium1 mark
A tram travels along a straight track. At t = 0 its velocity is 2.0 m s−1. Its acceleration is then: • +0.50 m s−2 for 20 s; • zero for 60 s; • −0.25 m s−2 for 30 s. What is the velocity of the tram at the end of this motion?
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The change in velocity is the signed area under the a–t graph: (0.50 × 20) + 0 + (−0.25 × 30) = 10 − 7.5 = +2.5 m s−1. Adding the initial velocity gives 2.0 + 2.5 = 4.5 m s−1. • 2.5 m s−1 forgets the initial velocity. • 12 m s−1 ignores the deceleration. • 19.5 m s−1 adds the negative area as if it were positive.
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Answer B
Answer: B
Common error: Area below the time axis on an a–t graph counts as negative. Then add the change in velocity to the initial velocity.
76IBP2S-001-00-03Projectile motionPaper 1AMedium1 mark
A delivery drone flies horizontally at a constant velocity of 8 m s−1. It releases a package of medical supplies. Air resistance is negligible. What path does the package follow as seen by an observer on the drone, and as seen by an observer standing on the ground?
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After release, the package keeps the drone's horizontal velocity of 8 m s−1. The drone also keeps moving at 8 m s−1. Relative to the drone there is therefore no horizontal motion, and the package falls straight down below it. Relative to the ground there is constant horizontal velocity plus uniform vertical acceleration, which gives a parabola. Option D assumes the package loses its horizontal velocity on release.
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Answer B
Answer: B
Common error: An object released from a moving vehicle keeps the vehicle's velocity at the instant of release. It does not drop straight down relative to the ground.
77IBP2S-001-00-04Projectile motionPaper 1AMedium1 mark
Two irrigation nozzles fire water at the same speed from ground level. One is set at 30° and the other at 60° above the horizontal. The ground is level and air resistance is negligible. Which statement is correct?
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Range R = u2 sin2θ / g. Since sin60° = sin120°, the two ranges are equal. Time of flight t = 2u sinθ / g. Because sin60° > sin30°, the 60° jet stays in the air longer, by a factor of √3. Option A assumes equal range means equal time. Options C and D assume that a steeper or a shallower angle always travels further.
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Answer B
Answer: B
Exam tip: On level ground, complementary launch angles (θ and 90° − θ) give equal ranges but different flight times and maximum heights.
78IBP2S-001-00-05Fluid resistance & terminal speedPaper 1AMedium1 mark
A badminton shuttlecock experiences significant air resistance. It is hit upwards at an angle to the horizontal. Compare its motion with that of a projectile launched with the same initial velocity but with no air resistance. Which statement about the shuttlecock is correct?
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On the way up, drag and gravity both act downwards, so the shuttlecock decelerates quickly. On the way down, drag acts upwards and opposes gravity, so the downward acceleration is less than g. It therefore takes longer to fall than to rise. Option B is wrong because drag reduces the horizontal component. Option C is wrong because drag also reduces the maximum height. Option D is wrong because at the top, horizontal drag adds to gravity, so the acceleration magnitude exceeds g.
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Answer A
Answer: A
Exam tip: With air resistance the path becomes asymmetric: a lower peak, a shorter range, a steeper descent and a longer descent time.
79IBP2S-001-00-07Projectile motionPaper 1AMedium1 mark
A lifeguard on a bridge throws three identical buoyancy aids, each with the same speed. One is thrown vertically upwards, one vertically downwards and one horizontally. All land in the river below. Air resistance is negligible. Which correctly compares their speeds as they reach the water?
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For each throw, the horizontal component of velocity is unchanged. The vertical component satisfies vy2 = uy2 + 2gh. The final speed squared is therefore v2 = ux2 + uy2 + 2gh = u2 + 2gh, whatever the direction of launch. Because the speed and the drop height are the same, all three hit the water at the same speed. They take different times to arrive and land in different directions. The other options confuse a different flight time or direction with a different final speed.
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Answer D
Answer: D
Exam tip: Without drag, the impact speed depends only on the launch speed and the height fallen, not on the launch direction.
80IBP2S-001-00-09Projectile motionPaper 1AMedium1 mark
An astronaut on the Moon throws a rock with the same initial velocity as a rock thrown on Earth. The gravitational field strength on the Moon is one-sixth of that on Earth. Air resistance is negligible in both cases. By what factors do the time of flight, the maximum height and the range on level ground change?
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The results are t = 2u sinθ / g, H = u2sin2θ / 2g and R = u2 sin2θ / g. Each one is inversely proportional to g, so dividing g by 6 multiplies all three by 6. Option B wrongly uses square-root relations. Options C and D assume the height or the range depends on 1/g2.
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Answer A
Answer: A
Common error: Write out each formula before scaling. Do not guess the power of g.
81IBP2S-001-04-03Fluid resistance & terminal speedPaper 1AMedium1 mark
A pollen grain falls from rest through still air. The drag force on it is proportional to its speed, and buoyancy is negligible. Its terminal speed is vt. What is the magnitude of its acceleration when its speed is ½vt?
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At terminal speed the drag is kvt = mg. At half that speed the drag is mg/2. The resultant force is then mg − mg/2 = mg/2, so a = 0.50g. Option C (0.75g) comes from using a drag proportional to v2. Option A (0.25g) gives the drag as a fraction of the weight for a v2 law, not the acceleration. Option D ignores drag altogether.
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Answer B
Answer: B
Exam tip: Write the drag as a fraction of the weight first, using the drag law given. Then subtract that fraction from 1 to get a/g.
82IBP2S-001-04-04Fluid resistance & terminal speedPaper 1AMedium1 mark
A humanitarian supply crate falls under a parachute. The drag force on the crate and parachute is proportional to the square of the speed. The same parachute is then used with a crate of double the total mass. What happens to the terminal speed?
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At terminal speed kvt2 = mg, so vt = √(mg/k). The parachute is the same, so k is unchanged. Doubling m therefore multiplies vt by √2. A wrongly applies the idea that all objects fall at the same rate, which holds only in a vacuum. C would be correct for drag proportional to v. D squares the factor instead of taking the square root.
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Answer B
Answer: B
Common error: "Everything falls at the same rate" applies only when there is no air resistance. With drag, a heavier object with the same shape has a higher terminal speed.
83IBP2S-001-04-05Fluid resistance & terminal speedPaper 1AMedium1 mark
Glass microspheres used in a medical diagnostic test sink through a liquid and quickly reach terminal speed. The drag on each sphere is given by Stokes' law, F = 6πηrv. A second batch of spheres is made of the same glass but with twice the radius. These spheres are released in the same liquid. What is the ratio of the terminal speed of the new spheres to that of the original spheres?
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At terminal speed, weight − upthrust = drag. Weight minus upthrust is (4/3)πr3(ρglass − ρliquid)g, so it is proportional to r3. The drag is proportional to rv. Equating gives r3 ∝ rvt, so vt ∝ r2. Doubling the radius multiplies the terminal speed by 22 = 4, which is option C. Option D (8) uses only the r3 dependence of the weight and ignores the r in the drag. Option B assumes the speed is simply proportional to r. Option A uses the r in the drag formula but ignores the growth in weight.
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Answer C
Answer: C
Exam tip: For proportionality questions, write how each force depends on r, equate the forces at terminal speed, then cancel. Both weight and upthrust scale with volume, so with r³.
84IBP2S-001-04-06Fluid resistance & terminal speedPaper 1AMedium1 mark
A volleyball is thrown vertically upwards. Air resistance is not negligible. Point P is part-way up the path. Which describes the magnitude of the ball's acceleration as it passes P on the way up and again on the way down?
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Drag always opposes the velocity. On the way up, the weight and the drag both act downwards, so |a| = g + D/m, which is greater than g. On the way down, the drag acts upwards, so |a| = g − D/m, which is less than g. B ignores drag. C reverses the direction of the drag. D assumes the drag always acts downwards.
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Answer A
Answer: A
Common error: Drag does not have a fixed direction. It reverses whenever the velocity reverses, so it adds to the weight going up and opposes the weight coming down.
85IBP2S-001-04-08Fluid resistance & terminal speedPaper 1AMedium1 mark
A skydiver falls at terminal speed and then opens her parachute. Immediately after it opens, her speed is greater than her new terminal speed. Which describes her motion over the next few seconds?
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Once the parachute opens, the drag exceeds her weight, so the resultant force and acceleration point upwards. Her velocity is still downwards, so she slows down. As her speed falls, the drag falls, and the upward acceleration shrinks towards zero at the new terminal speed. A confuses an upward acceleration with upward motion. C ignores the fact that drag depends on speed. D gives the wrong direction for the acceleration while the speed is decreasing.
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Answer B
Answer: B
Common error: An upward acceleration does not mean upward motion. Here the velocity stays downward the whole time; only its size decreases.
86IBP2S-001-03-03Motion graphs & definitionsPaper 1AMedium1 mark
A sprinter starts from rest. Her speed increases uniformly to 10 m s−1 in 4.0 s. She then runs at a constant 10 m s−1 for a further 6.0 s. What distance does she cover in the 10 s?
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The distance is the area under the v–t graph. The triangle gives ½ × 4.0 × 10 = 20 m and the rectangle gives 6.0 × 10 = 60 m, so the total is 80 m. The answer 100 m assumes 10 m s−1 throughout. The answer 60 m ignores the acceleration phase. The answer 120 m treats the triangle as a full rectangle and adds an extra 20 m.
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Answer B
Answer: B
Common error: Using speed × total time when part of the motion is an acceleration. Split the area into simple shapes.
87IBP2S-001-03-04Motion graphs & definitionsPaper 1AMedium1 mark
A rocket sled starts from rest on a test track. Its acceleration–time graph shows a constant acceleration of 15 m s−2 from t = 0 to t = 2.0 s, then zero acceleration. What is the speed of the sled at t = 3.0 s?
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The area under an acceleration–time graph is the change in velocity: 15 × 2.0 = 30 m s−1. After 2.0 s the acceleration is zero, so the sled keeps moving at 30 m s−1. The answer 15 m s−1 reads the graph value as a velocity. The answer 7.5 m s−1 divides instead of multiplying. The answer 60 m s−1 uses at2.
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Answer C
Answer: C
Exam tip: Zero acceleration means constant velocity, not zero velocity.
88IBP2S-001-03-06Motion graphs & definitionsPaper 1AMedium1 mark
A train leaves a station. Its displacement–time graph has the shape x = kt2, where k is a positive constant. Which pair correctly describes the train's velocity–time and acceleration–time graphs?
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The gradient of x = kt2 is v = 2kt, which is a straight line through the origin. The gradient of that line is a = 2k, which is constant, so the a–t graph is a horizontal line. Option A copies the shape of the x–t graph. Option C confuses a curving graph with a uniform velocity. Option D differentiates only once and assumes the acceleration graph has the same shape as the velocity graph.
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Answer B
Answer: B
Exam tip: Going from x–t to v–t to a–t, take the gradient each time. A parabola becomes a sloping line, which becomes a constant.
89IBP2S-001-03-07Fluid resistance & terminal speedPaper 1AMedium1 mark
A skydiver steps out of a hovering helicopter and falls until she reaches terminal velocity. Air resistance is not negligible. Which statement about her velocity–time graph, taking downwards as positive, is correct?
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At the instant of release her speed is zero, so there is no air resistance and her acceleration equals g = 9.8 m s−2. As her speed rises, the drag force increases and the resultant force falls. The gradient therefore decreases smoothly and reaches zero at terminal velocity. Option A ignores the gradual growth of drag. Options C and D confuse increasing speed with increasing acceleration.
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Answer B
Answer: B
Common error: An increasing speed does not mean an increasing acceleration. The v–t graph can rise while its gradient falls.
90IBP2S-001-03-08Motion graphs & definitionsPaper 1AMedium1 mark
A robot vacuum cleaner moves in a straight line. Its velocity–time graph shows +0.30 m s−1 for 10 s and then −0.20 m s−1 for 6.0 s. What are its final displacement from the start and the total distance travelled?
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The area above the axis is 0.30 × 10 = 3.0 m and the area below the axis is 0.20 × 6.0 = 1.2 m. Displacement takes direction into account: 3.0 − 1.2 = 1.8 m. Distance adds the magnitudes: 3.0 + 1.2 = 4.2 m. The other options either add the areas for displacement or subtract them for distance.
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Answer A
Answer: A
Exam tip: Area below the time axis counts as negative displacement but positive distance.
91IBP2S-001-02-02Distance, displacement & averagesPaper 1AMedium1 mark
A capsule on a Ferris wheel of radius r moves at constant speed through exactly half a revolution. What is the ratio \(\frac{\text{magnitude of average velocity}}{\text{average speed}}\) for this half revolution?
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After half a revolution the capsule is diametrically opposite its starting point. The displacement is therefore 2r. The distance travelled is half the circumference, πr. Both quantities are divided by the same time, so the ratio is 2r/(πr) = 2/π ≈ 0.64. Option D inverts the ratio. Option C assumes that constant speed makes velocity and speed equal. Option A uses r rather than 2r for the displacement.
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Answer B
Answer: B
Exam tip: The ratio of the magnitude of average velocity to average speed can never exceed 1. Any option greater than 1 can be rejected immediately.
92IBP2S-001-02-04Kinematics data & measurementPaper 1AMedium1 mark
The braking distance of a train is calculated using d = v2/(2a). The speed v is known to within 3 % and the deceleration a is known to within 2 %. What is the percentage uncertainty in d?
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Because v is squared, its percentage uncertainty is doubled: 2 × 3 % = 6 %. The 2 % uncertainty in a is then added, giving 8 %. The factor of 2 in the denominator is an exact number and contributes no uncertainty. Option A ignores the power. Option C doubles the wrong quantity. Option D squares the percentage (32 = 9) instead of multiplying it by the power.
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Answer B
Answer: B
Exam tip: For a quantity raised to the power n, multiply its percentage uncertainty by |n|.
93IBP2S-001-02-05Motion graphs & definitionsPaper 1AMedium1 mark
A lift starts from rest and makes the following journey: • it accelerates uniformly to 3.0 m s−1 in 2.0 s; • it then moves at 3.0 m s−1 for 6.0 s; • it then decelerates uniformly to rest in 2.0 s. What is the average speed of the lift for the whole journey, in m s−1?
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The distance in each stage is the area under the velocity–time graph: • acceleration: ½ × 3.0 × 2.0 = 3.0 m; • constant speed: 3.0 × 6.0 = 18 m; • deceleration: 3.0 m. The total distance is 24 m in 10 s, so the average speed is 2.4 m s−1. Option D uses only the maximum speed. Option A averages the start and maximum speeds. Option B ignores the distance covered while accelerating and decelerating.
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Answer C
Answer: C
Exam tip: Average speed is always total distance ÷ total time. Find the distance from the area under the velocity–time graph.
94IBP2S-001-02-06Distance, displacement & averagesPaper 1AMedium1 mark
A cyclist rides up a hill at a constant 6.0 m s−1. He then returns down the same road at a constant 18 m s−1. What is his average speed for the round trip, in m s−1?
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Let each leg have length L. The total time is L/6 + L/18 = 4L/18. The average speed is then 2L ÷ (4L/18) = 9.0 m s−1. More time is spent at the lower speed, so the average lies below the arithmetic mean. Option D is the arithmetic mean of the two speeds. Option C is the geometric mean. Option A is the average velocity, not the average speed.
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Answer B
Answer: B
Common error: Speeds can be averaged directly only when equal times are spent at each speed, not equal distances.
95IBP2S-001-02-07Motion graphs & definitionsPaper 1AMedium1 mark
A sled is released from rest on a uniform slope. Its displacement s is measured at several times t. A graph of s against t2 is a straight line through the origin with gradient (2.45 ± 0.10) m s−2. What is the acceleration of the sled, in m s−2?
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For motion from rest, s = ½at2, so the gradient equals a/2. This gives a = 2 × gradient = 4.90 m s−2. Multiplying by an exact factor of 2 also doubles the absolute uncertainty, to 0.20 m s−2. Option B takes the gradient itself as a. Option A halves the gradient instead of doubling it. Option C fails to scale the uncertainty.
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Answer D
Answer: D
Exam tip: Always compare the graph equation with y = mx + c to find what the gradient represents physically.
96IBP2S-001-01-03Motion graphs & definitionsPaper 1AMedium1 mark
In a warehouse, delivery robot P moves along a straight aisle at a constant speed of 1.2 m s−1. As P passes robot Q, Q starts from rest and accelerates uniformly in the same direction. What is the speed of Q at the instant it draws level with P again?
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When the robots are level again, they have covered the same distance in the same time, so their average velocities are equal. Q accelerates uniformly from rest, so its average velocity is half its final speed. Setting ½vQ = 1.2 m s−1 gives vQ = 2.4 m s−1. Option A is the speed Q has when the gap between them is largest, not when they meet. Option B wrongly applies a square-root relationship. Option D doubles twice.
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Answer C
Answer: C
Exam tip: Equal displacement in equal time means equal average velocity. This often gives the answer without solving a quadratic.
97IBP2S-001-01-04Equations of motionPaper 1AMedium1 mark
An e-scooter rider travelling at speed v has a reaction distance of 3.0 m and a braking distance of 4.0 m. The reaction time and the braking deceleration stay the same. What is the total stopping distance when the rider travels at 1.5v?
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Reaction distance = speed × reaction time, so it scales with v: 3.0 × 1.5 = 4.5 m. Braking distance = v2/2a, so it scales with v2: 4.0 × 2.25 = 9.0 m. The total stopping distance is 4.5 + 9.0 = 13.5 m. Option A gives only the new braking distance. Option B scales both distances linearly. Option D scales both distances with v2.
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Answer C
Answer: C
Common error: Reaction distance and braking distance scale differently with speed. Treat the two parts of the stopping distance separately.
98IBP2S-001-01-05Velocity & accelerationPaper 1AMedium1 mark
Water drips from a leaking pipe high above a laboratory floor. One drop leaves the pipe from rest, and the next drop leaves from rest 0.50 s later. Air resistance is negligible. While both drops are falling, what happens to the vertical distance between them?
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Both drops have the same acceleration, g. The velocity of the first drop relative to the second is therefore constant, equal to g × 0.50 s = 4.9 m s−1. A constant relative velocity means the separation increases at a constant rate. Option A confuses equal accelerations with equal velocities. Option C applies the acceleration to the separation itself. Option D gets the direction of the relative motion wrong.
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Answer B
Answer: B
Exam tip: For two bodies with the same acceleration, work in terms of relative motion. Their relative velocity is constant.
99IBP2S-001-01-08Projectile motionPaper 1AMedium1 mark
A lunar lander descends vertically at 3.0 m s−1. Its engines cut out when it is 5.0 m above the Moon's surface. The gravitational field strength on the Moon is 1.6 N kg−1. What is the speed of the lander as it reaches the surface?
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Use v2 = u2 + 2as with the downward direction positive: v2 = 3.02 + 2 × 1.6 × 5.0 = 9.0 + 16 = 25, so v = 5.0 m s−1. Option B ignores the initial velocity. Option D adds the speeds 3.0 and 4.0 directly instead of adding the squares. Option A subtracts the speeds instead of adding the squares.
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Answer C
Answer: C
Common error: In v² = u² + 2as, it is the squares of the speeds that combine, not the speeds themselves.
100IBP2S-001-01-09Motion graphs & definitionsPaper 1AMedium1 mark
In a simplified model of blood flow, a red blood cell accelerates uniformly from 0.10 m s−1 to 0.50 m s−1 while travelling 0.060 m along a narrowing vessel. How long does this take?
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For uniform acceleration, s = ½(u + v)t. Rearranging gives t = 2s/(u + v) = 0.12/0.60 = 0.20 s. Option A divides the distance by the final speed. Option B divides the distance by the change in speed. Option D uses (v − u) in place of (u + v).
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Answer C
Answer: C
Exam tip: If the acceleration is not given, s = ½(u + v)t avoids calculating it first.
Two identical stones are projected from level ground with the same speed. Stone P is projected at 30° above the horizontal and stone Q at 60° above the horizontal. Air resistance is negligible.
What are the ratios of the range and of the maximum height of Q to those of P?
Range of Q ÷ range of PMaximum height of Q ÷ maximum height of P
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Step 1Range = (u cos θ)(2u sin θ/g) ∝ sin θ cos θ, and sin 60° cos 60° = sin 30° cos 30°, so the ranges are equal: ratio 1.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Maximum height = (u sin θ)²/2g ∝ sin²θ, so the ratio is (sin 60°/sin 30°)² = (√3)² = 3.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThe range is right, but the height ratio uses sin 60°/sin 30° without squaring. The height depends on the square of the vertical component, (u sin θ)²/2g.
BCorrect: Q stays in the air √3 times longer but moves horizontally √3 times more slowly, so the ranges are equal; the heights are in the ratio sin²60°/sin²30° = 3.
CThe height ratio is right, but the range is taken to depend only on the time of flight. Q's horizontal component, u cos 60°, is smaller by the same factor √3.
DThis scales both quantities with the vertical component of velocity alone, forgetting the smaller horizontal component for the range and the square for the height.
102A-1A-17Distance, displacement & averagesPaper 1AHard1 mark
A cyclist travels a distance d along a straight road at a constant speed v, and then a further distance d at a constant speed 2v.
What is the average speed of the cyclist for the whole journey?
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Step 1Total time = d/v + d/2v = 3d/2v.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Average speed = total distance ÷ total time = 2d ÷ (3d/2v) = 4v/3.
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Answer B
Answer: B · 2 stages of work, one mark
Every option, and why
AThis divides only one of the two distances, d, by the total time. The total distance is 2d.
BCorrect: the journey takes 3d/2v, so the average speed is 2d ÷ 3d/2v = 4v/3.
CThis is the mean of the two speeds, which would be right for equal times at each speed. The cyclist spends twice as long at the lower speed, so the average is nearer to v.
DThis is the instantaneous speed during the second half, not the average speed over the journey.
The graph shows how the velocity of a body along a straight line varies with time.
What is the displacement of the body over the 10 s shown?
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Step 1Displacement is the area under a velocity–time graph, counted as negative below the time axis.
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All 4 steps must be completed — there is no mark for a part-answer.
Step 2From 0 to 6.0 s the area is ½ × 6.0 × 12 = +36 m.
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Step 3The line falls from +12 m s⁻¹ to −8.0 m s⁻¹ over 4.0 s, so it crosses zero at 6.0 + 4.0 × 12/20 = 8.4 s. From 6.0 to 8.4 s the area is ½ × 2.4 × 12 = +14.4 m.
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Step 4From 8.4 to 10 s the area is ½ × 1.6 × 8.0 = 6.4 m below the axis, so the displacement is 36 + 14.4 − 6.4 = 44 m.
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Answer B
Answer: B · 4 stages of work, one mark
Every option, and why
AThis adds all three areas as positive, which gives the total distance travelled (56.8 m). Displacement counts area below the axis as negative.
BCorrect: the areas are +36 m (0 to 6.0 s), +14.4 m (6.0 to 8.4 s) and −6.4 m (8.4 to 10 s), giving 36 + 14.4 − 6.4 = 44 m.
CThis is the area of the first triangle only, up to 6.0 s. The motion continues for another 4.0 s.
DThis treats the whole section from 6.0 s to 10 s as a single triangle below the axis. Part of that section is still above the axis.
105A-1A-50Velocity & accelerationPaper 1AHard1 mark
Water drips from a leaking gutter at regular time intervals. Air resistance is negligible.
Two successive drops are both falling. Which row describes how the distance between the drops and the speed of the lower drop relative to the upper drop change as they fall?
Distance between the dropsSpeed of lower drop relative to upper drop
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Step 1The lower drop has fallen for Δt longer, so it is always faster by gΔt: both drops gain speed at the same rate g, so this difference stays constant.
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All 2 steps must be completed — there is no mark for a part-answer.
Step 2Because the lower drop is always faster, the distance between the drops grows steadily (by gΔt every second).
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Answer D
Answer: D · 2 stages of work, one mark
Every option, and why
AThe drops do have the same acceleration, but the lower one started earlier and is faster, so the gap between them keeps growing.
BThe lower drop is faster, but both gain speed at the same rate g, so the difference in their speeds does not grow.
CEqual accelerations do not mean equal velocities: the lower drop has been accelerating for longer, so it is faster by gΔt.
DCorrect: the relative speed is constant at gΔt, so the distance between the drops increases at a steady rate.
A cargo e-bike starts from rest at traffic lights and accelerates uniformly along a straight road. In time T it travels a distance D.
What is (speed of the bike when it has travelled D/2) ÷ (speed of the bike at time T/2)?
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Step 1From rest v = at, so at T/2 the speed is half the final speed v.
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All 3 steps must be completed — there is no mark for a part-answer.
Step 2v² = 2as, so v ∝ √s: at D/2 the speed is v/√2.
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Step 3Ratio = (v/√2) ÷ (v/2) = 2/√2 = √2.
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Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AThis takes the time to travel D/2 as T/4, applying the square law the wrong way round. Since s ∝ t², D/2 is reached at T/√2, not T/4.
BThis is the inverse ratio: (speed at T/2) ÷ (speed at D/2).
CThis assumes that the bike reaches half the distance at half the time, which is true only at constant speed. The bike covers less distance in the first half of the time.
DCorrect: speed at D/2 = v/√2 and speed at T/2 = v/2, so the ratio is √2.
110IBP-A1-N1-06Motion graphs & definitionsPaper 1AHard1 mark
A remote-controlled toy car moves along a straight track. Its velocity decreases uniformly from +4.0 m s−1 at t = 0 to −2.0 m s−1 at t = 6.0 s. What is the average speed of the car between t = 0 and t = 6.0 s?
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The velocity changes by 6.0 m s−1 in 6.0 s, so it falls by 1.0 m s−1 each second and reaches zero at t = 4.0 s. Forward distance = ½(4.0)(4.0) = 8.0 m; backward distance = ½(2.0)(2.0) = 2.0 m. Total distance = 10.0 m, so average speed = 10.0/6.0 = 1.7 m s−1 (1.67 m s−1). (The displacement is 8.0 − 2.0 = 6.0 m, giving an average velocity of 1.0 m s−1.)
✓ 1
Answer C
Answer: C
Exam tip: When a v–t graph crosses the time axis, find the crossing time first. For distance, add the areas above and below the axis as positive values; for displacement, subtract the area below the axis.
111IBP-A1-N2-05Equations of motionPaper 1AHard1 mark
A fish swims in a straight line at a constant speed of 3.0 m s−1. When the fish is 8.0 m ahead of a sea lion, the sea lion starts from rest and swims after it along the same line with a uniform acceleration of 2.0 m s−2. How long after starting does the sea lion reach the fish?
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Measure positions from the sea lion's start. Sea lion: x = ½ × 2.0 × t2 = t2. Fish: x = 8.0 + 3.0t. Equal positions: t2 − 3.0t − 8.0 = 0, so t = (3.0 + √41)/2 = 4.7 s (4.70 s). The negative root is rejected.
✓ 1
Answer D
Answer: D
Common error: Catching up means equal positions, not equal speeds. When the speeds are equal the gap is at its largest, not zero.
A hot-air balloon is rising vertically at a constant speed of 4.0 m s−1. When the basket is 30 m above the ground, a sandbag is released from it. Air resistance on the sandbag is negligible and g = 9.8 m s−2. How long after release does the sandbag reach the ground?
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At release the sandbag shares the balloon's velocity, 4.0 m s−1 upwards. Taking upwards as positive: −30 = 4.0t − 4.9t2, so 4.9t2 − 4.0t − 30 = 0. t = (4.0 + √(16 + 588))/9.8 = 2.9 s (2.92 s). The negative root is rejected.
✓ 1
Answer D
Answer: D
Common error: An object released from a moving balloon or lift is not released from rest: it starts with the velocity of whatever was carrying it. Choose a positive direction and give every vector its sign.
113IBP2S-001-05-10Motion graphs & definitionsPaper 1AHard1 mark
An electric scooter accelerates from rest to 12 m s−1 in 6.0 s. Its velocity–time graph is a smooth curve whose gradient decreases continuously throughout the 6.0 s. What is the distance travelled in the 6.0 s?
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Uniform acceleration would give a straight line from (0, 0) to (6.0 s, 12 m s−1), with area ½ × 6.0 × 12 = 36 m. A curve with decreasing gradient that joins the same two points lies above this straight line, so its area is greater than 36 m. The speed never exceeds 12 m s−1, so the area is less than 12 × 6.0 = 72 m.
✓ 1
Answer C
Answer: C
Exam tip: Bound the area under a curved v–t graph by comparing it with simple shapes. Use a triangle below the curve and a rectangle above it.
114IBP2S-001-00-08Projectile motionPaper 1AHard1 mark
Parcels leave the end of a horizontal conveyor belt at speed v, from height h above the floor. They land a horizontal distance d from the end of the belt. The belt is then raised to height 9h and its speed is reduced to v/2. Air resistance is negligible. What is the new horizontal landing distance?
Diagram not to scale. Left: original set-up. Right: belt raised to 9h and slowed to v/2.
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The fall time comes from h = ½gt2, so t ∝ √h. Multiplying the height by 9 multiplies the fall time by 3. The horizontal distance is vt, so the new distance = (v/2)(3t) = 1.5vt = 1.5d. Option B ignores the change in speed. Option C assumes the time is proportional to h. Option D divides by the speed factor instead of multiplying by it.
✓ 1
Answer A
Answer: A
Exam tip: For horizontal launches, find the time from the vertical drop first. Then multiply by the constant horizontal speed.
115IBP2S-001-04-10Projectile motionPaper 1AHard1 mark
A space probe descends through the atmosphere of a distant moon. At the surface, the gravitational field strength is half of the Earth's value and the atmospheric density is 1/50 of the Earth's value. The drag force is F = ½ρCdAv2, where Cd and A are the same on both bodies. What is the probe's terminal speed near the moon's surface compared with near the Earth's surface?
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At terminal speed mg = ½ρCdAvt2, so vt ∝ √(g/ρ). The ratio is √(0.5 × 50) = √25 = 5. Option C forgets the square root. Option D also inverts the effect of g (50 × 2 = 100). Option A inverts the whole ratio, as though a thinner atmosphere gave more drag.
✓ 1
Answer B
Answer: B
Exam tip: Find the dependence first, here vt ∝ √(g/ρ). Then work out the combined factor inside the root before taking the square root.
116IBP2S-001-03-10Non-uniform accelerationPaper 1AHard1 mark
At t = 0 a car X passes a stationary car Y at a constant 20 m s−1. At the same instant Y starts from rest with a uniform acceleration of 2.0 m s−2 in the same direction. On a velocity–time graph the two lines cross at t = 10 s. At what time does Y draw level with X?
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The cars are level when their displacements are equal. This is when the areas under their v–t graphs are equal: ½(2.0)t2 = 20t, so t = 20 s. Why the other options are wrong: • 10 s: at this time the cars only have equal velocities, and X is still 100 m ahead. • 14 s: this assumes Y must cover the 200 m that X had travelled by 10 s (t2 = 200). It ignores the fact that X keeps moving. • 17 s: this uses relative motion after t = 10 s but drops the ½. Writing 2.0t′2 = 100 gives t′ ≈ 7.1 s, so the total is about 17 s. Correctly, ½(2.0)t′2 = 100 gives t′ = 10 s, and the total is 20 s.
✓ 1
Answer D
Answer: D
Common error: Treating the crossing point of two v–t lines as the moment the objects meet. Meeting needs equal areas under the graphs, not equal heights.
117IBP2S-001-02-08Kinematics data & measurementPaper 1AHard1 mark
A student measures the speed of a trolley released from rest by timing how long a card on the trolley takes to pass through a light gate. The card length is entered into the software as 5.0 cm, but it is actually 4.8 cm. The student plots v2 against distance travelled s. Compared with the graph obtained using the correct card length, the student's graph has
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Every calculated speed is larger than the true value by the same factor, 5.0/4.8. Every v2 value is therefore multiplied by (5.0/4.8)2 ≈ 1.09. Scaling all y-values by a constant factor steepens the line but leaves the point (0, 0) unchanged. Option B describes a zero (offset) error, not a proportional error. Option C wrongly assumes systematic errors have no effect. Option D reverses the direction of the error.
✓ 1
Answer A
Answer: A
Common error: Not every systematic error shifts the intercept. A proportional (scale) error changes the gradient instead.
118IBP2S-001-02-10Motion graphs & definitionsPaper 1AHard1 mark
An electric car accelerates from rest to a speed v in time T. Its acceleration is greatest at the start and decreases steadily to zero at time T. What is the average velocity of the car during this time?
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The velocity–time graph rises steeply at first and then flattens. It is curved, concave downwards, and lies entirely above the straight line from (0, 0) to (T, v). The area under the curve, which is the displacement, is therefore greater than ½vT. So the average velocity is greater than v/2. Option A applies (u + v)/2, which is valid only for uniform acceleration. Option B confuses decreasing acceleration with low velocity. Option D introduces mass, which is irrelevant in kinematics.
✓ 1
Answer C
Answer: C
Common error: Average velocity = (u + v)/2 holds only for uniform acceleration. Otherwise use displacement ÷ time.
In a film stunt, a small bag is released from rest at the top of a 12 m high wall. At the same instant, a ball is thrown vertically upwards from the foot of the wall at 15 m s−1, directly below the bag. Air resistance is negligible. How long after release do the bag and the ball meet?
Diagram not to scale
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Both objects have the same downward acceleration g, so g cancels in their relative motion. Relative to the bag, the ball moves upward at a constant 15 m s−1 and must close a 12 m gap. The time is t = 12/15 = 0.80 s. Check: the bag falls 4.9 × 0.802 = 3.1 m, and the ball rises 15 × 0.80 − 3.1 = 8.9 m; 3.1 + 8.9 = 12 m. Option A assumes each object covers half the height at 15 m s−1. Option C is u/g, the time for the ball to reach its highest point. Option D is the time for the bag to fall the full 12 m.
✓ 1
Answer B
Answer: B
Exam tip: When two bodies share the same acceleration, use relative velocity. The common acceleration cancels out.
120IBP2S-001-01-10Motion graphs & definitionsPaper 1AHard1 mark
Two autonomous haul trucks travel in the same direction along a straight mine tunnel. The front truck moves at a constant 8.0 m s−1. The rear truck moves at 12 m s−1 and is 20 m behind. The rear truck then begins to decelerate uniformly. What is the minimum deceleration that avoids a collision?
Diagram not to scale
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Work relative to the front truck. The rear truck approaches at 12 − 8.0 = 4.0 m s−1 and has 20 m in which to reduce this relative speed to zero. The minimum deceleration is a = (4.0)2/(2 × 20) = 0.40 m s−2. Option C uses (122 − 8.02) with the 20 m gap, mixing speeds relative to the ground with a relative distance. Option D assumes the rear truck must stop completely within 20 m. Option B uses only the front truck's speed.
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Answer A
Answer: A
Common error: The collision is avoided once the speeds become equal. The rear truck does not need to stop, so use relative velocity together with relative distance.
121A-1B-01Uncertainties & processing dataPaper 1BEasy10 marks
A student measures her reaction time with a ruler-drop test. A partner holds a 30 cm ruler vertically, with its zero mark level with the top of the student's open thumb and finger, and releases it without warning. The student catches the ruler as quickly as possible and the distance d through which it has fallen is read at the top of the thumb. The ruler is graduated in millimetres.
Ten trials are made with the student's dominant hand. The ruler falls freely from rest.
Trial
1
2
3
4
5
6
7
8
9
10
d / cm
19.4
21.0
18.2
20.6
17.9
19.8
22.1
18.8
20.3
19.5
(a)
State the resolution of the ruler.
(1)
(b)
(i)
Calculate the mean value of d.
(1)
(ii)
Estimate the absolute uncertainty in d, using half the range of the readings.
(1)
(c)
Show that the mean reaction time t of the student is about 0.20 s.
(1)
(d)
Determine the reaction time together with its absolute uncertainty.
(3)
(e)
The test is repeated with the non-dominant hand, giving t = 0.213 ± 0.010 s. The student concludes that her reaction time is longer with her non-dominant hand. Discuss whether the data support this conclusion.
(2)
(f)
Suggest one improvement to the procedure that would make the comparison in (e) more reliable.
(1)
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Part (a)
0.1 cm OR 1 mm
✓ 1
Accept ±0.05 cm OR ±0.5 mm. Do not accept 0.1 m or 1 cm.
Part (b)(i)
mean d = 19.76 ≈ 19.8 cm
✓ 1
Accept 19.76 cm.
Part (b)(ii)
½ × (22.1 − 17.9) = 2.1 cm
✓ 1
Accept 2 cm.
Part (c)
t = √(2d/g) = √(2 × 0.1976 / 9.81) = 0.201 s
✓ 1
Must see full substitution OR answer to at least 3 s.f. Use of d in cm (giving 2.0 s) scores 0.
Part (d)
fractional uncertainty in d = 2.1 / 19.8 = 0.106 OR 10.6 %
✓ 1
Allow ECF from (b). Accept 0.11 OR 11 %.
t ∝ d1/2, so fractional uncertainty in t = ½ × 0.106 = 0.053 OR 5.3 %
✓ 1
A candidate who doubles instead of halving (21 %) loses MP2 but may score MP3 by ECF.
t = 0.201 ± 0.011 s OR t = 0.20 ± 0.01 s
✓ 1
MP3 is for matching the precision of value and uncertainty; unit required. Accept Δt = 0.010–0.011 s.
Part (e)
ranges are 0.190–0.212 s «dominant» and 0.203–0.223 s «non-dominant», so the ranges overlap
✓ 1
Accept the equivalent statement that the difference in the means, 0.012 s, is smaller than the combined uncertainty, 0.021 s. Allow ECF from (d).
so the difference is not significant and the data do not support the conclusion
✓ 1
OWTTE. Do not award for 'the non-dominant time is larger, so the conclusion is correct'.
Part (f)
any one specific improvement, eg: many more trials with each hand to reduce the random uncertainty in each mean; alternate the hands between trials to remove a learning or fatigue effect; vary the delay before release so the student cannot anticipate it; keep the same partner and starting position for every trial
✓ 1
Do not accept 'repeat the experiment' or 'be more careful' without a purpose.
Answers: (b)(i) 19.8 cm · (b)(ii) ±2.1 cm · (c) 0.201 s · (d) 0.201 ± 0.011 s (the remaining parts are explanations — see the table above)
A sports-science database publishes the times at which a sprinter passed each 10 m mark in a 100 m race, measured from the starting signal by a laser system. The times are given to the nearest 0.01 s.
The graph shows the average speed in each 10 m interval plotted against the time at the middle of that interval. The error bars show the uncertainty in each average speed.
Distance / m
Split time / s
Time for the 10 m interval / s
Average speed in interval / m s−1
10
2.01
2.01
4.98
20
3.07
1.06
9.43
30
4.04
0.97
40
4.97
0.93
10.8
50
5.90
0.93
10.8
60
6.81
0.91
11.0
70
7.74
0.93
10.8
80
8.69
0.95
10.5
90
9.65
0.96
10.4
100
10.63
0.98
Graph drawn to scale
(a)
Complete the table.
(2)
(b)
Outline why each average speed is plotted at the middle time of its interval.
(1)
(c)
Draw the curve of best fit on the graph.
(1)
(d)
Each split time is uncertain by ±0.01 s. Determine the absolute uncertainty in the average speed for the interval from 50 m to 60 m.
(2)
(e)
The sprinter's coach claims that the sprinter slows down over the last 40 m of the race. Discuss whether the data support this claim.
(2)
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Part (a)
20 m to 30 m: 10 / 0.97 = 10.3 «m s−1»
✓ 1
90 m to 100 m: 10 / 0.98 = 10.2 «m s−1»
✓ 1
Accept answers to 2 or 3 s.f. Award [1] for one correct value.
Part (b)
the average speed over a short interval is approximately the instantaneous speed at the middle of the interval «exactly so for uniform acceleration»
✓ 1
OWTTE.
Part (c)
smooth curve rising steeply, levelling off near 11 m s−1 and falling slightly at the end, passing through the error bars
✓ 1
Do not accept point-to-point joining.
Part (d)
uncertainty in the interval time = ±0.02 s, so fractional uncertainty = 0.02 / 0.91 = 0.022 «2.2 %»
✓ 1
Award [0] for using ±0.01 s: two readings are subtracted.
Δv = 0.022 × 11.0 = 0.24 ≈ 0.2 m s−1
✓ 1
Accept 0.2–0.25 m s−1.
Part (e)
50–60 m: 11.0 ± 0.2 m s−1; 90–100 m: 10.2 ± 0.2 m s−1, so the ranges do not overlap
✓ 1
Allow ECF from (d). Any fair comparison of an interval near 60 m with the last interval.
the decrease is larger than the uncertainties, so the data support the claim
✓ 1
Do not award MP2 without reference to uncertainty.
Answers: (a) 10.3 and 10.2 m s−1 · (d) ±0.2 m s−1(the remaining parts are explanations — see the table above)
A small steel ball rolls down a curved ramp whose lower end is horizontal and level with the edge of a bench. A light gate at the lower end measures the speed v with which the ball leaves the bench. The ball lands on carbon paper on the floor, and the horizontal distance R from the edge of the bench to the landing mark is measured. v is changed by releasing the ball from different points on the ramp.
Each value of R is the mean of five landings and is uncertain by ±0.015 m. Air resistance is negligible. The height H through which the ball falls is the same for every launch.
v / m s−1
R / m
1.18
0.511
1.52
0.643
1.87
0.805
2.24
0.957
2.61
1.127
2.93
1.250
Graph drawn to scale
(a)
Show that R = v √(2H/g).
(1)
(b)
Draw the line of best fit on the graph.
(1)
(c)
Determine the gradient of the line, stating its unit.
(2)
(d)
Determine H using your answer to (c).
(2)
(e)
The uncertainty in the gradient is ±2 %. Determine the absolute uncertainty in H.
(2)
(f)
H measured directly with a metre rule is 0.905 ± 0.005 m. Comment on whether the two values agree.
(1)
(g)
Suggest one change to the method that would reduce the uncertainty in each value of R.
(1)
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Part (a)
vertical motion: H = ½gt² with zero initial vertical velocity, so t = √(2H/g) for every launch AND horizontal motion at constant v gives R = vt = v√(2H/g)
✓ 1
Both the time of flight and R = vt must be seen.
Part (b)
single straight line through all error bars, passing through or very close to the origin
✓ 1
Do not accept a line joined point to point.
Part (c)
gradient = 0.428
✓ 1
Accept 0.418–0.438.
unit: s
✓ 1
Accept m ÷ m s−1. The unit mark is independent.
Part (d)
gradient = √(2H/g), so H = g × gradient²/2
✓ 1
H = 9.81 × 0.428²/2 = 0.898 «m»
✓ 1
Accept 0.86–0.94 m. Allow ECF from (c).
Part (e)
H ∝ gradient², so percentage uncertainty in H = 2 × 2 % = 4 %
✓ 1
ΔH = 0.04 × 0.898 = ±0.04 «m»
✓ 1
Allow ECF from (d). Accept ±0.03–0.04 m.
Part (f)
the ranges 0.86–0.94 m and 0.900–0.910 m overlap, so the values agree
✓ 1
Allow ECF. Must refer to the uncertainties.
Part (g)
any one specific change, eg: record more landings for each release point and take the mean; film the landing on a scale with a high-frame-rate camera
✓ 1
Do not accept 'repeat the experiment' without stating what is averaged.
Answers: (c) 0.428 s · (d) 0.90 m · (e) ±0.04 m (the remaining parts are explanations — see the table above)
A trolley is given a short push up a long, straight ramp and released at t = 0. It moves up the ramp, stops momentarily and then rolls back down. A motion sensor at the bottom of the ramp records the velocity v of the trolley every 0.50 s. Velocity up the ramp (away from the sensor) is positive. Friction is negligible.
The sensor displays velocity to the nearest 0.01 m s−1. The graph shows the data.
t / s
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
v / m s−1
0.85
0.68
0.54
0.40
0.23
0.10
−0.06
−0.22
−0.35
−0.51
−0.67
Graph drawn to scale
(a)
State the resolution of the velocity readings.
(1)
(b)
Draw the line of best fit on the graph.
(1)
(c)
Determine the acceleration of the trolley. State its unit.
(2)
(d)
Show that the trolley is momentarily at rest at about t = 2.8 s.
(1)
(e)
Determine, for the interval from t = 0 to t = 5.0 s, the distance travelled by the trolley and the magnitude of its displacement.
(3)
(f)
The experiment is repeated on a rough ramp, on which friction is not negligible. State and explain how the magnitude of the gradient of the graph after the trolley stops would compare with its magnitude before the trolley stops.
(2)
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Notes
Part (a)
0.01 m s−1
✓ 1
Unit required. Accept ±0.01 m s−1.
Part (b)
single straight line through the points, extending from t = 0 to t = 5.0 s and crossing the t axis
✓ 1
Do not accept two separate lines or a line drawn point to point.
Part (c)
gradient using two well-separated points on the line, eg (−0.66 − 0.84)/5.0
✓ 1
a = −0.30 m s−2 OR 0.30 m s−2 down the ramp
✓ 1
Accept magnitude 0.28–0.32. Unit required for this mark. Sign or direction required.
Part (d)
t = u/|a| = 0.84/0.30 = 2.80 «s» OR line crosses the t axis at 2.80 s
✓ 1
Must see the working or a reading of the intercept to better than 0.1 s. Allow ECF from (c).
Part (e)
area above the axis = ½ × 0.84 × 2.80 = 1.18 m «up the ramp»
✓ 1
Accept 1.12–1.24 m. Allow ECF from their line.
area below the axis = ½ × 0.66 × 2.20 = 0.73 m «down the ramp»
✓ 1
Accept 0.66–0.80 m.
distance = 1.91 m AND displacement = 0.45 m «up the ramp from the release point»
✓ 1
Accept distance 1.8–2.0 m and displacement 0.40–0.50 m. Both needed for this mark.
Part (f)
the magnitude of the gradient would be smaller after the trolley stops
✓ 1
friction acts down the ramp while the trolley moves up, adding to the component of the weight, but acts up the ramp while it moves down, opposing it, so the resultant force is smaller on the way down
✓ 1
OWTTE. MP2 needs the change in the direction of friction.
Answers: (c) −0.30 m s−2 · (d) 2.8 s · (e) distance 1.9 m; displacement 0.45 m (the remaining parts are explanations — see the table above)
125A-1B-32Distance, displacement & averagesPaper 1BEasy10 marks
A bike computer calculates speed from the number of turns the front wheel makes each second and the wheel circumference that the rider enters. A cyclist checks it on a straight, level cycle path on which two marks are 100.0 m apart (measured to ±0.1 m). She rides past the marks at a steady speed while a friend times her with a stopwatch, and she notes the steady speed vd shown on the display, in km h−1. The timing uncertainty, due mainly to the friend's reaction time, is ±0.2 s.
The table shows five runs. The graph shows vd in m s−1 against the measured speed v for runs 1, 2 and 5. The error bars show the uncertainty in v.
Run
t / s
v / m s−1
vd / km h−1
vd / m s−1
1
24.81
4.03
15.4
4.28
2
18.17
5.50
20.8
5.78
3
14.35
26.6
7.39
4
11.76
8.50
32.2
5
10.11
9.89
37.8
10.5
Graph drawn to scale
(a)
Calculate, for run 1, the measured speed v and its absolute uncertainty.
(2)
(b)
Complete the table.
(2)
(c)
Plot the points for runs 3 and 4 on the graph, and draw the line of best fit.
(2)
(d)
Outline how the graph shows that the bike computer has a systematic error that is proportional to the speed rather than a zero error.
(1)
(e)
The circumference entered was 2.210 m. Determine, using the graph, the true circumference of the wheel.
(2)
(f)
Suggest one change to the procedure that would reduce the uncertainty in each measured speed.
(1)
Show mark scheme
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Mark
Notes
Part (a)
v = 100.0/24.81 = 4.03 «m s−1»
✓ 1
fractional uncertainty = 0.2/24.81 + 0.1/100.0 = 0.0091, so Δv = ±0.04 m s−1
✓ 1
Accept ±0.03–0.04 m s−1; ignoring the distance uncertainty gives the same answer to 1 s.f. Award [2] for CNA with unit.
Part (b)
run 3: v = 100.0/14.35 = 6.97 «m s−1»
✓ 1
run 4: vd = 32.2/3.6 = 8.94 «m s−1»
✓ 1
Both to 3 s.f. Award [1] for one correct value.
Part (c)
both points plotted correctly: (6.97, 7.39) and (8.50, 8.94)
✓ 1
Within half a small square. Allow ECF from (b).
single straight line through the error bars and close to all five points, passing close to the origin
✓ 1
Do not accept a line joined point to point.
Part (d)
the line passes through the origin (no intercept) but its gradient is greater than 1, so the display is too high by the same fraction at every speed
✓ 1
OWTTE. Both ideas needed: through origin AND gradient ≠ 1.
Part (e)
gradient = 1.060
✓ 1
Accept 1.03–1.08.
true circumference = 2.210/1.060 = 2.08 m
✓ 1
Accept 2.05–2.15 m. Allow ECF from their gradient. Do not accept multiplying by the gradient.
Part (f)
any one specific change, eg: time over a longer distance (several hundred metres) so the reaction-time uncertainty is a smaller fraction of t; use light gates or a video with a frame counter at each mark instead of a hand-held stopwatch
✓ 1
Do not accept 'repeat the runs' without explaining how this reduces the uncertainty in each speed.
Answers: (a) 4.03 ± 0.04 m s−1 · (b) 6.97 m s−1; 8.94 m s−1 · (e) 2.08 m (the remaining parts are explanations — see the table above)
126A-1B-07Non-uniform accelerationPaper 1BMedium10 marks
A student places a smartphone flat on the floor of a lift and uses its accelerometer app to record the vertical acceleration a while the lift travels upwards from rest to rest. Upwards is positive. The graph shows the smoothed data-logger trace.
Before the lift starts to move, the phone gives a steady reading of +0.06 m s−2.
Graph drawn to scale
(a)
(i)
State the type of error that the reading before the lift starts reveals.
(1)
(ii)
Outline how the student should correct the readings.
(1)
(b)
Outline how the graph shows that the lift moves with constant velocity between t = 3.4 s and t = 12.0 s.
(1)
(c)
Determine the maximum speed of the lift.
(2)
(d)
The value of the offset is known to ±0.02 m s−2. Determine the absolute uncertainty in the maximum speed.
(1)
(e)
The manufacturer states that the lift travels at 1.25 m s−1. Comment on whether the result of (c) agrees with this.
(1)
(f)
The floors of the building are 3.50 m apart. Estimate the distance travelled by the lift and hence deduce the number of floors it rises.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
systematic error «zero / offset error»
✓ 1
Do not accept 'random error'.
Part (a)(ii)
subtract 0.06 m s−2 from every reading
✓ 1
OWTTE, eg re-zero (calibrate) the sensor before the journey and repeat.
Part (b)
the corrected acceleration is zero «reading equals the offset» in that interval, so the velocity does not change
✓ 1
Part (c)
area under the corrected graph from 1.0 s to 3.4 s = ½ × (2.4 + 0.8) × 0.80
✓ 1
Award [1] max for 1.42 m s−1 from the uncorrected trace.
vmax = 1.28 m s−1
✓ 1
Accept 1.24–1.32 m s−1.
Part (d)
Δv = 0.02 × 2.4 = 0.048 ≈ 0.05 m s−1
✓ 1
Accept 0.04–0.05 m s−1.
Part (e)
range 1.23–1.33 m s−1 includes 1.25 m s−1, so the result agrees «within uncertainty»
✓ 1
Allow ECF from (c) and (d).
Part (f)
while speeding up and slowing down the average speed is ½vmax «symmetrical trace» OR uses the area under a trapezium-shaped v–t graph
✓ 1
distance = 1.28 × (8.6 + 1.2 + 1.2) = 14.1 m
✓ 1
Accept 13.6–14.5 m. Allow ECF from (c).
14.1 / 3.50 = 4.0, so the lift rises 4 floors
✓ 1
Must be a whole number.
Answers: (c) 1.28 m s−1 · (d) ±0.05 m s−1 · (f) 14 m, 4 floors (the remaining parts are explanations — see the table above)
127A-1B-09Fluid resistance & terminal speedPaper 1BMedium10 marks
A paper cone is released from rest directly below a motion sensor. The sensor records the distance s fallen by the cone every 0.10 s. The data are shown in the table and plotted on the graph.
t / s
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
s / m
0.000
0.049
0.176
0.363
0.572
0.805
1.035
1.278
1.512
1.755
1.996
2.231
2.474
Graph drawn to scale
(a)
Draw the curve of best fit on the graph.
(1)
(b)
Assuming that the acceleration of the cone is uniform during the first 0.10 s, show that it is about 10 m s−2.
(1)
(c)
Explain, with reference to the forces acting on the cone, why the gradient of the graph becomes constant.
(2)
(d)
Determine the terminal speed of the cone.
(2)
(e)
Calculate the time the cone would take to fall 2.00 m from rest if there were no air resistance, and compare it with the time taken in the experiment.
(2)
(f)
The student suspects that the cone was released a short time after the sensor started recording. State and explain whether this affects the value found in (d).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
smooth curve through «or close to» all points, becoming a straight line after about 0.5 s
✓ 1
Do not accept point-to-point joining.
Part (b)
a = 2s/t² = 2 × 0.049 / 0.10² = 9.8 m s−2
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (c)
air resistance increases as the speed increases while the weight stays constant, so the resultant force «and acceleration» decreases
✓ 1
when air resistance equals the weight the resultant force is zero, so the velocity «gradient of s–t» is constant
✓ 1
Part (d)
gradient of the straight part of the graph, using a large triangle «eg from 0.6 s to 1.2 s»
✓ 1
Award [0] for a single point s/t.
terminal speed = 2.40 m s−1
✓ 1
Accept 2.3–2.5 m s−1.
Part (e)
t = √(2 × 2.00 / 9.81) = 0.64 s
✓ 1
the cone takes about 1.00 s, so air resistance increases the time of fall «by about 0.36 s / a factor of 1.6»
✓ 1
Accept 0.98–1.02 s read from the table or graph.
Part (f)
no effect
✓ 1
Award MP1 only if a valid reason is given.
a delay shifts every reading by the same time, moving the graph sideways without changing the gradient of the straight section
✓ 1
OWTTE.
Answers: (b) 9.8 m s−2 · (d) 2.4 m s−1 · (e) 0.64 s (no air) vs 1.00 s (the remaining parts are explanations — see the table above)
128A-1B-18Equations of motionPaper 1BMedium11 marks
A road-safety agency publishes typical stopping distances for a car on a dry road. The thinking distance is the distance travelled at constant speed during the driver's reaction time; the braking distance dB is the distance travelled after the brakes are applied. The published distances are rounded to the nearest metre.
A student tests the hypothesis that dB = kvn, where v is the speed of the car and k and n are constants.
v / km h−1
v / m s−1
Thinking distance / m
dB / m
lg(v / m s−1)
lg(dB / m)
30
8.33
6
5
0.921
0.699
50
13.9
9
15
1.143
1.176
70
19.4
13
29
90
25.0
17
48
1.398
1.681
110
30.6
21
72
130
36.1
25
100
1.558
2.000
Graph drawn to scale
(a)
Show that the agency assumes a reaction time of about 0.7 s.
(1)
(b)
Complete the table.
(2)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine n.
(2)
(e)
Explain how your value of n is consistent with the car having a constant deceleration while braking.
(2)
(f)
Determine the deceleration assumed by the agency.
(2)
(g)
Suggest why the point for 30 km h−1 lies furthest from the line.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
eg 17 / 25.0 = 0.68 s
✓ 1
Any row. Must see substitution of a thinking distance and a speed in m s−1.
Part (b)
70 km h−1: 1.289 and 1.462
✓ 1
110 km h−1: 1.485 and 1.857
✓ 1
Accept 2 or 3 decimal places. Award [1] for any three correct values.
Part (c)
single straight line with points evenly scattered on either side
✓ 1
Line need not pass through the 30 km h−1 point.
Part (d)
gradient calculated from a large triangle on the line
✓ 1
n = 2.0
✓ 1
Accept 1.9–2.1.
Part (e)
with final speed zero, v² = u² + 2as gives dB = v²/2a
✓ 1
so for constant a, dB ∝ v², that is n = 2, as found
✓ 1
Allow ECF from (d) if the value is close to 2.
Part (f)
uses a = v²/2dB with a point on the line OR intercept = −lg 2a
✓ 1
Award [0] for use of the thinking distance.
a = 6.5 m s−2
✓ 1
Accept 6.2–6.9 m s−2.
Part (g)
rounding to ±0.5 m is a much larger fraction of 5 m «10 %» than of the larger distances, so lg dB is least precise there
✓ 1
OWTTE.
Answers: (a) 0.68 s · (b) 1.289, 1.462; 1.485, 1.857 · (d) n = 2.0 · (f) 6.5 m s−2(the remaining parts are explanations — see the table above)
129A-1B-23Fluid resistance & terminal speedPaper 1BMedium10 marks
Paper cake cases are stacked inside one another and dropped from rest, open side up, below a motion sensor. The sensor records the position of the stack, and the terminal speed v is found from the final, straight part of the distance–time graph. The number N of cases in the stack is varied, so the weight of the stack is proportional to N while its shape stays the same. Each value of v is uncertain by ±0.03 m s−1.
A student tests the hypothesis that the drag force on the stack is proportional to v². The graph shows lg v against lg N, with error bars.
N
v / m s−1
lg N
lg(v / m s−1)
1
1.11
0.000
0.045
2
1.54
0.301
0.188
3
1.92
4
2.18
0.602
0.338
6
2.71
8
3.10
0.903
0.491
Graph drawn to scale
(a)
Explain why, if the hypothesis is correct, the graph of lg v against lg N is a straight line of gradient 0.5.
(2)
(b)
Complete the table.
(2)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine the gradient of the line.
(2)
(e)
Lines of maximum and minimum gradient drawn through the error bars have gradients 0.51 and 0.49. Discuss whether the data support the hypothesis.
(2)
(f)
Suggest how the student could check that each stack had reached its terminal speed before v was measured.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
at the terminal speed the drag equals the weight: kv² = Nmg, where m is the mass of one case
✓ 1
Accept kv² ∝ N.
so v = √(mg/k) × N0.5 and lg v = 0.5 lg N + lg √(mg/k): straight line, gradient 0.5
✓ 1
Must see the logarithmic form or a clear statement that the power becomes the gradient.
Part (b)
lg 3 = 0.477 AND lg 6 = 0.778
✓ 1
Both needed; 3 d.p. Accept 0.477 and 0.778.
lg 1.92 = 0.283 AND lg 2.71 = 0.433
✓ 1
Both needed; 3 d.p.
Part (c)
single straight line passing through all error bars
✓ 1
Do not accept a line forced through the origin.
Part (d)
two well-separated points on the line used «at least half the line apart»
✓ 1
gradient = 0.50
✓ 1
Accept 0.49–0.51.
Part (e)
n = 0.50 ± 0.01 OR the range 0.49–0.51
✓ 1
Accept their gradient ± 0.01.
0.5 lies within this range, so the data support the hypothesis «that drag ∝ v²»
✓ 1
Allow ECF from (d). Do not accept 'the gradient is close to 0.5' without reference to the range.
Part (f)
check that the distance–time graph is a straight line «constant gradient» over the region used, OR repeat with the stack released from a greater height and check that v is unchanged
✓ 1
OWTTE.
Answers: (b) 0.477, 0.283; 0.778, 0.433 · (d) 0.50 · (e) 0.50 ± 0.01 (the remaining parts are explanations — see the table above)
130A-1B-33Equations of motionPaper 1BMedium12 marks
A cart with a battery-driven fan moves along a level track. The fan gives the cart a constant acceleration a. The cart is released from rest at the same point for every run and passes light gate 1, which is fixed, with speed u. Light gate 2 is placed a distance s beyond gate 1 and records the time Δt for which a card of width 50.0 mm on the cart blocks its beam. The speed at gate 2 is v = (card width)/Δt.
The uncertainty in each value of v² is ±0.03 m² s−2. The graph shows v² against s with error bars for all the values of s except 1.00 m.
s / m
Δt / s
v / m s−1
v² / m² s−2
0.20
0.0820
0.610
0.372
0.40
0.0687
0.728
0.530
0.60
0.0587
0.852
0.726
0.80
0.0531
0.942
0.887
1.00
0.0479
1.20
0.0446
1.121
1.257
Graph drawn to scale
(a)
Explain why a graph of v² against s is expected to be a straight line, and state what its gradient and its intercept on the v² axis represent.
(2)
(b)
Complete the table for s = 1.00 m.
(1)
(c)
Plot the point for s = 1.00 m, with its error bar, and draw the line of best fit on the graph.
(2)
(d)
Determine a.
(2)
(e)
By drawing lines of maximum and minimum gradient, determine the absolute uncertainty in a.
(2)
(f)
Show that u is about 0.4 m s−1.
(1)
(g)
The student later finds that the card is 48.5 mm wide, not 50.0 mm. State and explain the effect of this on the value of a found in (d).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
v² = u² + 2as, which has the form y = c + mx with u and a the same for every run
✓ 1
gradient = 2a AND intercept = u²
✓ 1
Both needed.
Part (b)
v = 0.0500/0.0479 = 1.044 «m s−1» AND v² = 1.090 «m² s−2»
✓ 1
Both needed. Accept 1.04 and 1.09.
Part (c)
point plotted at (1.00, 1.090) with an error bar of ±0.03 m² s−2
✓ 1
Within half a small square. Allow ECF from (b).
single straight line through all the error bars, not forced through the origin
✓ 1
Part (d)
gradient = 0.895 «m s−2»
✓ 1
Accept 0.86–0.94.
a = gradient/2 = 0.448 ≈ 0.45 m s−2
✓ 1
Accept 0.43–0.47 m s−2. Allow ECF from their gradient.
Part (e)
lines of maximum and minimum gradient drawn through all the error bars
✓ 1
Δa = ¼ × (difference of the two gradients) ≈ ±0.03 m s−2
✓ 1
Accept 0.01–0.04 m s−2. Allow ECF. The factor ¼ = ½ (half the range) × ½ (a = gradient/2).
Part (f)
intercept = 0.18 «m² s−2», so u = √0.18 = 0.43 «m s−1»
✓ 1
Must see the intercept read from the graph or calculated. Accept intercept 0.15–0.21 m² s−2.
Part (g)
every value of v is too large by the factor 50.0/48.5, so every v² is too large by (50.0/48.5)² = 1.063
✓ 1
so the gradient, and therefore a, is too large by about 6 %
✓ 1
Award MP2 for 'a is overestimated' with a correct reason. Accept a corrected value of about 0.42 m s−2.
Answers: (b) 1.044 m s−1; 1.090 m² s−2 · (d) 0.45 m s−2 · (e) ±0.03 m s−2 · (f) 0.43 m s−1(the remaining parts are explanations — see the table above)
A spring launcher fires a light plastic ball from floor level across a level floor. The launch speed u is the same for every shot, and the launch angle θ above the horizontal is set with a protractor to ±1°. Each range R is the mean of five shots and is uncertain by ±0.02 m.
If air resistance is negligible, R = (u²/g) sin 2θ. The graph shows R against sin 2θ with error bars.
θ / °
sin 2θ
R / m
15
0.500
0.891
20
0.643
1.120
30
0.866
1.533
40
0.985
1.728
50
0.985
1.746
60
0.866
1.515
70
0.643
1.138
75
0.500
0.873
Graph drawn to scale
(a)
Outline how the data show that two launch angles that add up to 90° give the same range.
(1)
(b)
Draw the line of best fit on the graph.
(1)
(c)
Determine u using the graph.
(3)
(d)
The lines of maximum and minimum gradient have gradients 1.81 m and 1.72 m. Determine the absolute uncertainty in u.
(2)
(e)
A light gate at the launcher gives u = 4.31 ± 0.04 m s−1. Discuss, with reference to your answers to (c) and (d), whether this agrees with the graph value, and suggest a reason for any difference.
(2)
(f)
Explain why the ±1° uncertainty in θ has the smallest effect on R when θ is close to 45°.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
pairs such as 15° and 75°, 20° and 70°, 30° and 60°, 40° and 50° have the same sin 2θ and their ranges are equal within ±0.02 m «the uncertainty»
✓ 1
OWTTE. Must quote at least one pair with its ranges.
Part (b)
single straight line through all the error bars, passing close to the origin
✓ 1
Part (c)
gradient = u²/g
✓ 1
gradient = 1.76 m
✓ 1
Accept 1.71–1.81 m.
u = √(9.81 × 1.76) = 4.16 m s−1
✓ 1
Accept 4.10–4.22 m s−1. Allow ECF from their gradient.
Part (d)
fractional uncertainty in gradient = ½(1.81 − 1.72)/1.76 = 0.026
✓ 1
fractional uncertainty in u is half of this, so Δu = ±0.05 m s−1
✓ 1
Accept ±0.04–0.06 m s−1. MP2 needs the halving for the square root.
Part (e)
graph range 4.11–4.21 m s−1 and gate range 4.27–4.35 m s−1 do not overlap, so they do not agree
✓ 1
Allow ECF from (c) and (d).
air resistance on the light ball reduces each range, so the graph gives a value of u that is too small
✓ 1
OWTTE. Do not accept a vague 'experimental error' or 'human error'.
Part (f)
sin 2θ has its maximum at θ = 45°, so it changes very little when θ changes by 1° near 45°
✓ 1
OWTTE. Accept the gradient of sin 2θ against θ is zero at 45°.
Answers: (c) 4.16 m s−1 · (d) ±0.05 m s−1(the remaining parts are explanations — see the table above)
132A-1B-35Distance, displacement & averagesPaper 1BMedium10 marks
Tectonic plates move slowly across the surface of the Earth. A geodesy database publishes the position of a permanent GPS station fixed to bedrock on one plate. The table gives the north component of the position of the station on 1 January of each year, relative to a fixed reference point chosen by the database. Seasonal movements of the ground make each value uncertain by ±15 mm.
The graph shows the north position against time with error bars.
Year
North position / mm
2014
3
2015
51
2016
122
2017
167
2018
219
2019
290
2020
344
2021
392
2022
462
2023
510
2024
574
Graph drawn to scale
(a)
Draw the line of best fit on the graph.
(1)
(b)
Determine the north component of the average velocity of the station, in mm yr−1.
(2)
(c)
Lines of maximum and minimum gradient through the error bars have gradients 59.7 mm yr−1 and 54.8 mm yr−1. State the north component of the average velocity together with its absolute uncertainty.
(1)
(d)
The east component of the average velocity, found in the same way, is 32 ± 2 mm yr−1. Determine the magnitude and the direction of the average velocity of the station.
(2)
(e)
A plate-motion model predicts a speed of 67 mm yr−1 for this station. Discuss whether the data agree with the model.
(2)
(f)
Show that the magnitude of the average velocity is about 2 × 10−9 m s−1.
(1)
(g)
Outline why the distance travelled by the station during the ten years is greater than the magnitude of its displacement.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
single straight line through all or almost all of the error bars
✓ 1
Part (b)
gradient calculated using two well-separated points on the line
✓ 1
57 mm yr−1
✓ 1
Accept 55–59 mm yr−1. Award [2] for CNA.
Part (c)
57 ± 2 mm yr−1
✓ 1
Uncertainty = half the difference of the gradients = 2.45. Accept ±2 OR ±3 mm yr−1. Allow ECF from (b). Value and uncertainty to matching precision.
Part (d)
magnitude = √(57² + 32²) = 65 mm yr−1
✓ 1
Accept 63–68 mm yr−1. Allow ECF.
direction tan−1(32/57) = 29° east of north
✓ 1
Accept 28°–31° E of N, OR N 29° E, OR 61° north of east. Reference direction required.
Part (e)
largest and smallest speeds consistent with the uncertainties: √(59² + 34²) = 68 and √(55² + 30²) = 63 mm yr−1
✓ 1
Accept any valid estimate of the uncertainty in the magnitude, eg ±3 mm yr−1.
67 mm yr−1 lies within this range, so the data agree with the model
✓ 1
Allow ECF. Do not accept a bare 'agrees' without reference to the uncertainty.
Part (f)
65 × 10−3 m ÷ (365 × 24 × 3600 s) = 2.1 × 10−9 m s−1
✓ 1
Must see the conversion of years to seconds.
Part (g)
the station moves back and forth with the seasons «and changes direction», so the path length is longer than the straight line from its start position to its end position
✓ 1
OWTTE.
Answers: (b) 57 mm yr−1 · (c) 57 ± 2 mm yr−1 · (d) 65 mm yr−1, 29° east of north · (f) 2.1 × 10−9 m s−1(the remaining parts are explanations — see the table above)
133A-1B-38Equations of motionPaper 1BMedium10 marks
Sports scientists estimate the height of a vertical jump from the flight time T, the time between take-off and landing recorded by a contact mat on which the athlete stands. The mat records T to the nearest 0.001 s. It is assumed that the centre of mass of the athlete is at the same height at take-off and at landing, and air resistance is negligible.
One athlete makes five jumps.
Jump
1
2
3
4
5
T / s
0.512
0.498
0.521
0.505
0.509
(a)
Show that the height h through which the centre of mass rises is given by h = gT²/8.
(1)
(b)
Calculate the mean flight time and its absolute uncertainty, using half the range of the readings.
(2)
(c)
Determine h, together with its absolute uncertainty.
(3)
(d)
A video analysis of the same jumps gives h = 0.290 ± 0.010 m. Discuss whether the two methods agree.
(2)
(e)
Athletes often land with their knees and ankles more bent than at take-off. Explain how this accounts for your answer to (d).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
the centre of mass rises for T/2 and then falls from rest through h in T/2, so h = ½g(T/2)² = gT²/8
✓ 1
Accept the equivalent using v = gT/2 at take-off and h = v²/2g.
Part (b)
mean T = 0.509 s
✓ 1
ΔT = ½ × (0.521 − 0.498) = 0.0115 ≈ ±0.012 s
✓ 1
Accept ±0.011–0.012 s.
Part (c)
h = 9.81 × 0.509²/8 = 0.318 m
✓ 1
fractional uncertainty in h = 2 × 0.0115/0.509 = 0.045 «4.5 %»
✓ 1
MP2 is for doubling the fractional uncertainty in T. Allow ECF from (b).
h = 0.318 ± 0.014 m OR 0.32 ± 0.01 m
✓ 1
MP3 is for matching the precision of value and uncertainty. Accept Δh = 0.014–0.015 m.
Part (d)
mat range 0.303–0.332 m; video range 0.280–0.300 m; the ranges do not overlap
✓ 1
Allow ECF from (c).
so the methods do not agree: the mat method gives a larger height
✓ 1
OWTTE.
Part (e)
the centre of mass is lower at landing than at take-off, so the athlete is in the air for longer than the time needed to rise and fall through h
✓ 1
T is too large, so gT²/8 overestimates h — a systematic error consistent with the mat value being larger
✓ 1
OWTTE. MP2 needs the direction of the effect.
Answers: (b) 0.509 ± 0.012 s · (c) 0.318 ± 0.014 m (the remaining parts are explanations — see the table above)
134A-1B-02Distance, displacement & averagesPaper 1BHard11 marks
A trolley is released from rest at the same point on a straight inclined track for every run. A light gate is fixed at a point P further down the track. Cards of different length L are mounted on the trolley in turn, always with their leading edge in the same place, so the leading edge of every card reaches P with the same instantaneous speed u. The timer records the time Δt for which each card blocks the beam and the average speed vavg = L/Δt is calculated.
The uncertainty in each value of vavg is ±0.010 m s−1. The graph shows vavg against Δt with error bars.
L / m
Δt / s
vavg / m s−1
0.025
0.0384
0.651
0.050
0.0747
0.669
0.075
0.1075
0.100
0.1400
0.714
0.150
0.1971
0.200
0.2517
0.795
Graph drawn to scale
(a)
The trolley has a constant acceleration a. Show that vavg = u + ½aΔt.
(1)
(b)
Complete the table by calculating the two missing values of vavg.
(2)
(c)
Draw the line of best fit on the graph.
(1)
(d)
(i)
Determine u, the instantaneous speed of the trolley at P.
(1)
(ii)
By drawing lines of maximum and minimum gradient, determine the absolute uncertainty in u.
(2)
(e)
Determine the acceleration of the trolley.
(2)
(f)
(i)
The light-gate software reports L/Δt as “the speed at P”. Explain why this value is always greater than u in this experiment.
(1)
(ii)
A student suggests using a card only 5 mm long so that L/Δt is closer to u. Suggest a disadvantage of this.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
s = ut + ½at² with s = L and t = Δt, so L/Δt = u + ½aΔt
✓ 1
Accept the argument that for uniform acceleration the average velocity equals the velocity at the mid-time, u + a(Δt/2).
Part (b)
0.075/0.1075 = 0.698 «m s−1»
✓ 1
0.150/0.1971 = 0.761 «m s−1»
✓ 1
Both values to 3 s.f. for the second mark. Award [1] for one correct value.
Part (c)
single straight line passing through all error bars
✓ 1
Do not accept a line joined point to point or a line forced through the origin.
Part (d)(i)
u = intercept on the vavg axis = 0.621 «m s−1»
✓ 1
Accept 0.612–0.630 m s−1.
Part (d)(ii)
lines of maximum and minimum gradient drawn through all error bars, extrapolated to Δt = 0
✓ 1
Δu = ½ × difference of their intercepts ≈ 0.011 m s−1, so u = 0.62 ± 0.01 m s−1
✓ 1
Accept 0.005–0.015 m s−1. Allow ECF from their lines.
Part (e)
gradient = ½a; gradient read from line ≈ 0.69 m s−2
✓ 1
a = 2 × gradient = 1.38 m s−2
✓ 1
Accept 1.30–1.46 m s−2. Award [1] for the gradient alone.
Part (f)(i)
the trolley speeds up while the card passes P, so the average speed over Δt exceeds the speed of the leading edge at P
✓ 1
OWTTE. Accept a numerical illustration, eg the 0.100 m card gives 0.714 m s−1, about 15 % more than u.
Part (f)(ii)
the percentage uncertainty in L «and in Δt» becomes very large, eg 0.5 mm in 5 mm is 10 %
✓ 1
OWTTE.
Answers: (b) 0.698 and 0.761 m s−1 · (d)(i) 0.62 m s−1 · (d)(ii) ±0.01 m s−1 · (e) 1.4 m s−2(the remaining parts are explanations — see the table above)
A student films a small ball thrown upwards at an angle and analyses the video with tracking software. A metre rule placed in the picture is used to set the scale. The software gives the horizontal displacement x and the vertical displacement y of the ball from its release point at time t after release. Air resistance is negligible.
The uncertainty in each value of y is ±0.010 m. The graph shows y/t plotted against t; the error bars show the uncertainty in y/t.
t / s
x / m
y / m
(y/t) / m s−1
0.10
0.305
0.390
3.90
0.20
0.599
0.674
3.37
0.30
0.904
0.887
0.40
1.199
0.989
2.47
0.50
1.507
1.015
2.03
0.60
1.806
0.930
1.55
0.70
2.104
0.772
0.80
2.409
0.504
0.63
Graph drawn to scale
(a)
(i)
Outline how the data show that the horizontal component of velocity is constant.
(1)
(ii)
Show that the horizontal component of velocity is about 3.0 m s−1.
(1)
(b)
Complete the table.
(2)
(c)
The vertical launch velocity is uy. Explain why a graph of y/t against t is a straight line, and state what its gradient represents.
(2)
(d)
Draw the line of best fit on the graph.
(1)
(e)
Determine the acceleration of free fall g from the graph.
(2)
(f)
The lines of maximum and minimum gradient give an uncertainty in g of ±0.2 m s−2. Comment on the result of (e).
(1)
(g)
The metre rule was held closer to the camera than the plane in which the ball moved. Explain how this accounts for the result.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
x increases by equal amounts «≈ 0.30 m» in equal time intervals
✓ 1
OWTTE.
Part (a)(ii)
ux = 2.409 / 0.80 = 3.01 m s−1
✓ 1
Must see a full substitution OR answer to 3 s.f. Any row or a gradient of x against t is acceptable.
Part (b)
0.887/0.30 = 2.96
✓ 1
0.772/0.70 = 1.10
✓ 1
Award [1] for one correct value. Accept 3 s.f.
Part (c)
y = uyt − ½gt² so y/t = uy − ½gt, which is linear in t
✓ 1
gradient = −g/2 «intercept = uy»
✓ 1
Accept 'gradient = −½g'.
Part (d)
single straight line through all error bars «drawn with a ruler»
✓ 1
Part (e)
gradient read from their line ≈ −4.63 m s−2
✓ 1
Sign of the gradient not required for MP1.
g = −2 × gradient = 9.3 m s−2
✓ 1
Accept 9.0–9.45 m s−2.
Part (f)
range 9.1–9.5 m s−2 does not include 9.81 m s−2, so there is a systematic error «not explained by random error»
✓ 1
Allow ECF from (e).
Part (g)
the rule appears larger than it would in the plane of the ball, so the scale gives too few metres per pixel and every measured distance of the ball is too small by the same factor
✓ 1
g «and ux, uy» are proportional to the distance scale, so g is too small «by about 6 %»
✓ 1
OWTTE.
Answers: (a)(ii) 3.01 m s−1 · (b) 2.96 and 1.10 m s−1 · (e) 9.3 m s−2(the remaining parts are explanations — see the table above)
A steel ball is held by an electromagnet above a trapdoor. Switching off the current releases the ball and starts an electronic timer; the ball opens the trapdoor and stops the timer. The vertical distance h from the bottom of the ball to the trapdoor is measured with a metre rule and the time t is recorded three times for each height.
The graph shows the mean time against √h. The error bars show the uncertainty in each mean time, taken as half the range of the readings.
h / m
√(h / m)
t1 / s
t2 / s
t3 / s
mean t / s
0.250
0.500
0.241
0.236
0.239
0.239
0.400
0.632
0.295
0.300
0.302
0.299
0.550
0.742
0.349
0.343
0.350
0.700
0.837
0.390
0.394
0.388
0.391
0.850
0.922
0.426
0.431
0.424
0.427
1.000
1.000
0.468
0.463
0.465
0.465
Graph drawn to scale
(a)
For h = 0.550 m, calculate the mean time and its absolute uncertainty.
(2)
(b)
The ball starts to fall a short time t0 after the timer starts. Explain why a graph of t against √h is expected to be a straight line, and state what its gradient and its intercept on the t axis represent.
(2)
(c)
Draw the line of best fit on the graph.
(1)
(d)
Determine, using the graph, the acceleration of free fall g.
(2)
(e)
The uncertainty in the gradient is ±2.0 %. Determine the absolute uncertainty in your value of g.
(2)
(f)
Comment on your result in the light of the accepted value g = 9.81 m s−2.
(1)
(g)
Identify the systematic error suggested by the graph, and explain why it does not affect the value of g found in (d).
Accept ±0.0035 s OR ±0.004 s. Do not accept the resolution 0.001 s as the uncertainty.
Part (b)
h = ½g(t − t0)², so t = √(2/g) √h + t0, which has the form y = mx + c
✓ 1
Accept the argument starting from s = ut + ½at² with u = 0.
gradient = √(2/g) AND intercept = t0, the delay before the ball starts to fall
✓ 1
Both needed. OWTTE.
Part (c)
single straight line through all error bars, extending to √h = 0
✓ 1
Do not accept a line forced through the origin or joined point to point.
Part (d)
gradient from line ≈ 0.450 «s m−1/2»
✓ 1
Accept 0.440–0.462.
g = 2/gradient² = 2/0.450² = 9.88 «m s−2»
✓ 1
Accept 9.4–10.3 m s−2. Allow ECF from their gradient. Award [1] for g = 1/gradient² (half the correct value).
Part (e)
g ∝ (gradient)−2, so percentage uncertainty in g = 2 × 2.0 % = 4.0 %
✓ 1
Δg = 0.040 × 9.88 = ±0.4 «m s−2» so g = 9.9 ± 0.4 m s−2
✓ 1
Allow ECF from (d). A candidate who uses 2.0 % (not doubled) scores [1] max.
Part (f)
the range 9.5–10.3 m s−2 includes 9.81 m s−2, so the result is consistent with the accepted value
✓ 1
Allow ECF from (d) and (e). Must refer to the uncertainty range.
Part (g)
the line meets the t axis at about +0.014 s «instead of the origin», so every time includes a constant extra time: the timer starts before the ball begins to fall «delay in the magnetic field collapsing»
✓ 1
Accept intercept 0.003–0.024 s. OWTTE.
the same delay is added to every value of t, so the line is shifted parallel to itself and the gradient, from which g is found, is unchanged
✓ 1
OWTTE.
Answers: (a) 0.347 ± 0.004 s · (d) g ≈ 9.9 m s−2 · (e) ±0.4 m s−2 · (g) intercept ≈ 0.014 s (the remaining parts are explanations — see the table above)
A nozzle of internal diameter 4.00 ± 0.05 mm points vertically upwards. Water leaves the nozzle at speed v and rises to a maximum height h above the nozzle, measured against a vertical scale. The volume flow rate Q is found by collecting the water in a measuring cylinder for a measured time; each value of Q is uncertain by ±2 %. The speed v = Q/A, where A is the cross-sectional area of the nozzle. Each value of h is uncertain by ±0.02 m.
A student tests the hypothesis that h = kvn, where k and n are constants. The graph shows lg(h/m) against lg(v/m s−1) with error bars for every row except v = 3.02 m s−1.
Q / cm³ s−1
v / m s−1
h / m
lg(v/m s−1)
lg(h/m)
25.2
0.21
0.303
−0.678
31.2
2.48
0.31
0.394
−0.509
37.9
3.02
0.47
43.9
3.49
0.62
0.543
−0.208
50.5
4.02
0.83
0.604
−0.081
62.6
4.98
1.16
0.697
0.064
75.6
6.02
1.53
0.780
0.185
Graph drawn to scale
(a)
Show that the speed for the first row is about 2.0 m s−1.
(1)
(b)
Calculate the percentage uncertainty in v.
(2)
(c)
Show that, if air resistance is negligible, lg h = 2 lg v − lg(2g).
(1)
(d)
Complete the table for v = 3.02 m s−1.
(1)
(e)
Plot the point for v = 3.02 m s−1 and draw the line of best fit for the points with v < 4.5 m s−1.
(2)
(f)
Determine n, and state whether your value supports the hypothesis that n = 2.
(2)
(g)
Determine, using the intercept of your line, a value for g.
(2)
(h)
Suggest why the points for the two highest speeds lie below the line.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A = π × (2.00 × 10−3)² = 1.257 × 10−5 m², so v = 25.2 × 10−6/1.257 × 10−5 = 2.005 «m s−1»
✓ 1
Must see full substitution OR answer to at least 3 s.f. Using the diameter instead of the radius scores 0.
Part (b)
percentage uncertainty in A = 2 × (0.05/4.00 × 100) = 2.5 %
✓ 1
percentage uncertainty in v = 2.0 + 2.5 = 4.5 %
✓ 1
Accept 4.5 % OR 5 %. Allow ECF for MP2.
Part (c)
v² = 2gh «from v² = u² + 2as with zero speed at the top», so h = v²/2g; taking logs: lg h = 2 lg v − lg(2g)
✓ 1
Must see the kinematic equation and the log step.
Part (d)
lg v = 0.480 AND lg h = −0.328
✓ 1
Both needed. Accept 2 or 3 decimal places.
Part (e)
point plotted at (0.480, −0.328)
✓ 1
Within half a small square. Allow ECF from (d).
single straight line through the error bars of the first five points, extended to lg v = 0
✓ 1
Do not accept a line forced through the last two points.
Part (f)
n = gradient = 1.99
✓ 1
Accept 1.9–2.1.
value is consistent with n = 2, so the hypothesis is supported «for v < 4.5 m s−1»
✓ 1
Allow ECF from their gradient.
Part (g)
intercept = −lg(2g) = −1.285
✓ 1
Accept −1.24 to −1.34. Allow a value calculated from a point on the line with n = 2.
g = 101.285/2 = 9.6 m s−2
✓ 1
Accept 8.7–11 m s−2. Allow ECF from their intercept.
Part (h)
air resistance on the jet «and its break-up into drops» is greater at higher speeds, so the water rises less than v²/2g
✓ 1
OWTTE.
Answers: (a) 2.01 m s−1 · (b) 4.5 % · (d) 0.480, −0.328 · (f) 1.99 · (g) 9.6 m s−2(the remaining parts are explanations — see the table above)
138A-1B-37Non-uniform accelerationPaper 1BHard12 marks
The live video of a rocket launch shows telemetry from the launch vehicle. A student records the speed v displayed (in km h−1, to the nearest 1 km h−1) at times t after lift-off read from the on-screen clock, which shows whole seconds only. The student tests whether the data fit a relationship v = ktn for this part of the flight, where k and n are constants.
The graph shows lg(v/km h−1) against lg(t/s) for all the times except 20 s and 50 s. The horizontal error bars show an uncertainty of ±0.5 s in t.
t / s
v / km h−1
lg(t/s)
lg(v/km h−1)
10
113
1.000
2.053
20
303
30
552
1.477
2.742
40
827
1.602
2.918
50
1157
60
1500
1.778
3.176
70
1868
1.845
3.271
Graph drawn to scale
(a)
Calculate the average acceleration of the vehicle during the first 60 s, in m s−2.
(2)
(b)
Complete the table.
(2)
(c)
Plot the points for t = 20 s and t = 50 s on the graph, and draw the line of best fit.
(2)
(d)
Determine n.
(2)
(e)
Explain what your value of n shows about the acceleration of the vehicle during the first 70 s.
(2)
(f)
Suggest one reason for the change in acceleration that you described in (e).
(1)
(g)
Outline why the horizontal error bar is largest for the first point.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
v = 1500/3.6 = 417 «m s−1»
✓ 1
average acceleration = 417/60 = 6.9 m s−2
✓ 1
Award [2] for CNA. Accept 6.9–7.0 m s−2.
Part (b)
t = 20 s: 1.301 AND 2.481
✓ 1
t = 50 s: 1.699 AND 3.063
✓ 1
Accept 2 or 3 decimal places. Award [1] for one correct row.
Part (c)
both points plotted correctly: (1.301, 2.481) and (1.699, 3.063)
✓ 1
Within half a small square. Allow ECF from (b).
single straight line passing through or close to all the points
✓ 1
Part (d)
gradient calculated using two well-separated points on the line
✓ 1
n = 1.44
✓ 1
Accept 1.38–1.52. The unit of v does not affect the gradient.
Part (e)
n = 1 would mean uniform acceleration «v ∝ t»; n > 1 means v increases faster than in proportion to t
✓ 1
so the acceleration «the gradient of the v–t graph» increases with time
✓ 1
OWTTE. Allow ECF from (d) for a value of n < 1.
Part (f)
the mass of the vehicle decreases as fuel is burned while the thrust stays about the same, so the resultant force per unit mass increases
✓ 1
Accept thrust increasing as the air pressure falls, OR the drag/weight component along the path decreasing. OWTTE.
Part (g)
the time uncertainty ±0.5 s is the same for every reading, so it is the largest fraction «5 %» of t when t = 10 s, and Δlg t ≈ 0.43 Δt/t is largest
✓ 1
OWTTE.
Answers: (a) 6.9 m s−2 · (b) 1.301, 2.481; 1.699, 3.063 · (d) 1.44 (the remaining parts are explanations — see the table above)
An electric scooter and its rider, of total mass 95 kg, travel along a straight, level cycle path. Figure 1 shows how the velocity v of the scooter varies with time t.
Figure 2 is a set of axes for part (c).
Figure 1 — graph drawn to scaleFigure 2 — axes for (c)
(a)
(i)
State what is represented by the gradient of a velocity–time graph.
(1)
(ii)
Calculate the acceleration of the scooter during the first 4.0 s.
(1)
(b)
(i)
Show that the scooter travels about 70 m in the 15 s shown.
(1)
(ii)
Calculate the average speed of the scooter for the 15 s.
(1)
(c)
Sketch, on Figure 2, a graph to show how the acceleration a of the scooter varies with time from t = 0 to t = 15 s.
(3)
(d)
(i)
Calculate the magnitude of the resultant force on the scooter and rider between t = 12 s and t = 15 s.
(1)
(ii)
The rider brakes between t = 12 s and t = 15 s. Explain why the force exerted by the brakes is smaller than your answer to (d)(i).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
acceleration
✓ 1
Do not accept “speed” or “change in velocity”.
Part (a)(ii)
a = 6.0 / 4.0 = 1.5 m s−2
✓ 1
Part (b)(i)
area under the graph = ½ × 4.0 × 6.0 + 8.0 × 6.0 + ½ × 3.0 × 6.0 = 12 + 48 + 9 «= 69 m»
✓ 1
Must see full substitution OR answer to 3 s.f. Accept the trapezium ½(8.0 + 15) × 6.0.
Part (b)(ii)
69 / 15 = 4.6 m s−1
✓ 1
Allow ECF from (b)(i). Accept 4.7 m s−1 from 70 m.
Part (c)
horizontal line at a = +1.5 m s−2 from 0 to 4.0 s
✓ 1
Allow ECF from (a)(ii).
a = 0 «line along the time axis» from 4.0 s to 12 s
✓ 1
horizontal line at a = −2.0 m s−2 from 12 s to 15 s, below the time axis
✓ 1
Vertical joining lines may be shown or omitted. Do not award MP3 for a positive value.
Part (d)(i)
F = ma = 95 × 2.0 = 190 N
✓ 1
Accept 1.9 × 102 N.
Part (d)(ii)
air resistance/rolling friction also act on the scooter opposite to its velocity «in the same direction as the braking force»
✓ 1
the resultant force is the sum of the braking force and these resistive forces, so the brakes provide less than 190 N
✓ 1
OWTTE. MP2 only scores if MP1 scores.
Answers: (a)(ii) 1.5 m s−2 · (b)(i) 69 m · (b)(ii) 4.6 m s−1 · (d)(i) 190 N (the remaining parts are explanations — see the table above)
A garden sprinkler at ground level sends out a jet of water at a speed of 13 m s−1 at an angle of 32° above the horizontal. The ground is level. Each small mass of water in the jet may be treated as a projectile, and air resistance is negligible.
Diagram not to scale
(a)
(i)
Show that the vertical component of the initial velocity of the water is about 7 m s−1.
(1)
(ii)
Calculate the horizontal component of the initial velocity of the water.
(1)
(b)
Calculate the maximum height reached by the water.
(2)
(c)
A flower bed, in line with the jet, extends from 14.0 m to 17.0 m from the sprinkler. Determine whether the water lands in the flower bed.
(3)
(d)
State and explain the speed of the water as it reaches the ground.
(2)
(e)
Calculate the height of the water above the ground at a horizontal distance of 6.0 m from the sprinkler.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
uy = 13 sin 32° «= 6.89 m s−1»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (a)(ii)
ux = 13 cos 32° = 11.0 m s−1
✓ 1
Accept 11 m s−1.
Part (b)
v² = u² − 2gh with v = 0 at the top: h = 6.89² / (2 × 9.81)
✓ 1
Accept h = uyt − ½gt² with t = uy/g.
h = 2.42 m
✓ 1
Accept 2.4–2.5 m «using 7 m s−1». Award [2] for CNA.
Part (c)
time of flight = 2 × 6.89 / 9.81 = 1.40 s
✓ 1
Allow use of symmetry: twice the time to the highest point.
horizontal distance = 11.02 × 1.404 = 15.5 m
✓ 1
Accept 15.4–15.7 m. Allow ECF from (a).
15.5 m is between 14.0 m and 17.0 m, so the water lands in the flower bed
✓ 1
The decision must be consistent with the candidate’s distance. Do not award MP3 for a bare “yes”.
Part (d)
13 m s−1 «same as the launch speed»
✓ 1
horizontal component unchanged AND vertical component has the same magnitude as at launch «by symmetry» OR water returns to the same height so the gravitational potential energy and hence the kinetic energy are the same as at launch
✓ 1
OWTTE.
Part (e)
time to reach 6.0 m: t = 6.0 / 11.0 = 0.545 s
✓ 1
Allow ECF from (a)(ii).
height = 6.89 × 0.545 − ½ × 9.81 × 0.545² = 2.30 m
✓ 1
Accept 2.3–2.4 m «using 7 m s−1». Award [2] for CNA. Award [1 max] for 3.76 m, which omits the ½gt² term.
Answers: (a)(i) 6.89 m s−1 · (a)(ii) 11.0 m s−1 · (b) 2.42 m · (c) 15.5 m; yes · (d) 13 m s−1 · (e) 2.30 m(the remaining parts are explanations — see the table above)
141A-2-11Distance, displacement & averagesPaper 2Easy11 marks
A survey drone flies at a constant height above a field. Starting from point B it flies 400 m due east at a constant speed of 10 m s−1 (leg 1), then 300 m due north at a constant speed of 6.0 m s−1 (leg 2). It then hovers for 20 s before flying in a straight line directly back to B at a constant speed of 12.5 m s−1.
Figure 1 shows legs 1 and 2 drawn to scale.
Figure 1 — drawn to scale on a square grid
(a)
(i)
Draw, on Figure 1, an arrow to represent the displacement of the drone from B at the end of leg 2.
(1)
(ii)
Calculate the magnitude and direction of this displacement.
(2)
(b)
(i)
Show that the whole flight, including the hover, lasts 150 s.
(1)
(ii)
Calculate the average speed of the drone for the whole flight.
(2)
(iii)
State the average velocity of the drone for the whole flight.
(1)
(c)
(i)
Calculate the magnitude of the average velocity of the drone for legs 1 and 2 together.
(1)
(ii)
Explain why your answer to (c)(i) is less than the average speed for legs 1 and 2.
(2)
(iii)
State the instantaneous speed of the drone at t = 60 s after it leaves B.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
single straight arrow from B to the end of leg 2 «(400 m, 300 m)», arrowhead at the end of leg 2
✓ 1
Part (a)(ii)
√(400² + 300²) = 500 m
✓ 1
tan−1(300/400) = 37° north of east
✓ 1
Accept a bearing of 053° or 53° east of north. A direction is needed for MP2.
Must see the return time calculated from the 500 m displacement.
Part (b)(ii)
total distance = 400 + 300 + 500 = 1200 m
✓ 1
Allow ECF from (a)(ii).
average speed = 1200 / 150 = 8.0 m s−1
✓ 1
Award [2] for CNA.
Part (b)(iii)
zero «the displacement for the whole flight is zero»
✓ 1
Part (c)(i)
500 / 90 = 5.6 m s−1
✓ 1
Allow ECF from (a)(ii).
Part (c)(ii)
average speed uses the distance travelled «700 m» but average velocity uses the displacement «500 m» over the same time
✓ 1
the displacement is the straight line from B, which is shorter than the path flown «because the drone changed direction»
✓ 1
OWTTE.
Part (c)(iii)
6.0 m s−1
✓ 1
The drone is on leg 2 from 40 s to 90 s.
Answers: (a)(ii) 500 m, 37° north of east · (b)(ii) 8.0 m s−1 · (c)(i) 5.6 m s−1 · (c)(iii) 6.0 m s−1(the remaining parts are explanations — see the table above)
An autonomous robot moves along a straight aisle in a warehouse. Figure 1 shows how its displacement s from its charging dock varies with time t. Displacement away from the dock is positive.
Figure 1 — graph drawn to scaleFigure 2 — axes for (d)
(a)
State the displacement of the robot from the dock at t = 28 s.
(1)
(b)
Calculate the total distance travelled by the robot in the 28 s.
(1)
(c)
Calculate, for the whole 28 s, the average speed of the robot and the magnitude of its average velocity.
(2)
(d)
Sketch, on Figure 2, the velocity–time graph of the robot for the 28 s.
(3)
(e)
Suggest why the displacement–time graph of the real robot cannot have a sharp corner at t = 8 s.
(1)
(f)
The robot and its load have a total mass of 45 kg. In practice the robot slows from 1.5 m s−1 to rest in 0.50 s. Calculate the magnitude of the average resultant force on the robot while it slows down.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
9.0 m away from the dock «+9.0 m»
✓ 1
Direction or sign required.
Part (b)
12 + 9 + 6 = 27 m
✓ 1
Part (c)
average speed = 27/28 = 0.96 m s−1
✓ 1
average velocity = 9/28 = 0.32 m s−1
✓ 1
Part (d)
horizontal lines at +1.5 m s−1 from 0 to 8 s and from 24 s to 28 s
✓ 1
v = 0 from 8 s to 12 s and from 18 s to 24 s
✓ 1
horizontal line at −1.5 m s−1 from 12 s to 18 s
✓ 1
Do not award if drawn above the t-axis.
Part (e)
a sharp corner means an instantaneous change of velocity, which would need an infinite acceleration «and force»
✓ 1
OWTTE.
Part (f)
a = 1.5/0.50 = 3.0 m s−2
✓ 1
Accept Δp = 67.5 kg m s−1.
F = 45 × 3.0 = 135 N
✓ 1
Award [2] for CNA.
Answers: (a) +9.0 m · (b) 27 m · (c) 0.96 m s−1; 0.32 m s−1 · (f) 135 N (the remaining parts are explanations — see the table above)
143A-2-43Distance, displacement & averagesPaper 2Easy13 marks
An average-speed enforcement system on a motorway uses two cameras. The length of road between the cameras is 4.80 km; the straight-line distance between them is 4.62 km. Each camera reads the number plate of a car and records the time, to the nearest second, on synchronised clocks. The speed limit is 100 km h−1 (27.8 m s−1).
A car passes the first camera at 10:41:12 and the second camera at 10:44:05.
Figure 1 — axes for (e)(ii)
(a)
Determine whether the average speed of the car between the cameras was above the speed limit.
(2)
(b)
Calculate the magnitude of the average velocity of the car between the cameras.
(1)
(c)
Outline why the system must use the length of the road rather than the straight-line distance.
(1)
(d)
Outline why the driver may have broken the speed limit even though the answer to (a) is correct.
(1)
(e)
The car travelled the first half of the road at a constant 32.0 m s−1 and the second half at a lower constant speed. Ignore the time taken to change speed.
(i)
Calculate the speed of the car in the second half of the road.
(2)
(ii)
Sketch, on Figure 1, the speed–time graph of the car between the cameras. Add values to your graph where appropriate.
(2)
(f)
Discuss, with reference to the precision of the recorded times, whether the system could fine a driver who passed the cameras 172 s apart.
(3)
(g)
Suggest one advantage, for road safety, of an average-speed system compared with a single camera that measures the speed of a car at one point.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
t = 2 min 53 s = 173 s; v = 4800/173 = 27.7 m s−1
✓ 1
= 99.9 km h−1, which is «just» below the limit, so not above
✓ 1
MP2 needs a comparison in consistent units.
Part (b)
4620/173 = 26.7 m s−1
✓ 1
Part (c)
speed is distance travelled divided by time, and the car travels along the road, not along the straight line «which would underestimate its speed»
✓ 1
OWTTE.
Part (d)
the average speed does not show the instantaneous speed: the car could have been above the limit for part of the time and below it for the rest
✓ 1
Part (e)(i)
time for first half = 2400/32.0 = 75 s, so the second half takes 173 − 75 = 98 s
✓ 1
v = 2400/98 = 24.5 m s−1
✓ 1
Award [2] for CNA.
Part (e)(ii)
horizontal line at 32.0 m s−1 from 0 to 75 s
✓ 1
Allow ECF from (e)(i).
horizontal line at 24.5 m s−1 from 75 s to 173 s
✓ 1
Both lines on the correct side of the speed-limit line.
Part (f)
«nominal» average speed = 4800/172 = 27.9 m s−1 = 100.5 km h−1, just above the limit
✓ 1
each recorded time is uncertain by ±0.5 s, so the interval is 172 ± 1 s and the average speed lies between 99.9 and 101.1 km h−1
✓ 1
Accept a percentage uncertainty of about 0.6 %.
the range includes speeds below the limit, so the system cannot show that the driver exceeded the limit «a fine needs a tolerance above the limit»
✓ 1
MP3 is for a conclusion consistent with the range.
Part (g)
drivers cannot avoid detection by slowing down only near the camera, so the speed is kept down along the whole length of road
✓ 1
OWTTE.
Answers: (a) 27.7 m s−1 = 99.9 km h−1: not above · (b) 26.7 m s−1 · (e)(i) 24.5 m s−1(the remaining parts are explanations — see the table above)
A television camera hangs from cables above a football pitch. Motors move it along a straight horizontal line between the two goals. Figure 1 shows how the velocity v of the camera varies with time t during a 40 s sequence. Velocity towards the north goal is positive. At t = 0 the camera is at its starting position.
Figure 2 is a set of axes for (e).
Figure 1 — graph drawn to scaleFigure 2 — axes for (e)
(a)
(i)
State what is represented by the area between a velocity–time graph and the time axis.
(1)
(ii)
Calculate the acceleration of the camera between t = 12 s and t = 16 s.
(1)
(b)
Determine the displacement of the camera from its starting position at t = 40 s.
(2)
(c)
(i)
Calculate the average speed of the camera for the 40 s.
(2)
(ii)
Calculate the magnitude of the average velocity of the camera for the 40 s.
(1)
(d)
State the greatest distance of the camera from its starting position. Outline how Figure 1 shows when the camera is at this greatest distance.
(2)
(e)
Sketch, on Figure 2, a graph to show how the displacement s of the camera from its starting position varies with t from t = 0 to t = 40 s.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
displacement «change in position»
✓ 1
Do not accept “distance”: areas below the axis count as negative.
Part (a)(ii)
a = (0 − 6.0)/4.0 = −1.5 m s−2
✓ 1
Sign required, OR “1.5 m s−2 towards the south goal”.
Part (b)
areas: +72 m «12 + 48 + 12» from 0 to 16 s AND −48 m «6 + 36 + 6» from 20 s to 35 s
✓ 1
Award this mark for either area seen with a correct sign.
displacement = 72 − 48 = 24 m «towards the north goal»
✓ 1
Direction or sign required. Award [2] for CNA.
Part (c)(i)
distance = 72 + 48 = 120 m
✓ 1
Allow ECF from (b).
average speed = 120/40 = 3.0 m s−1
✓ 1
Award [2] for CNA. Award [1 max] for 0.60 m s−1.
Part (c)(ii)
24/40 = 0.60 m s−1
✓ 1
Allow ECF from (b).
Part (d)
72 m
✓ 1
Allow ECF from (b).
the camera moves away while v is positive and back while v is negative, so it is furthest away from 16 s to 20 s, when v = 0 «between the positive and negative sections»
✓ 1
OWTTE. Accept any time from 16 s to 20 s.
Part (e)
from 0 to 12 s: curve starting with zero gradient at the origin, then a straight line from «4 s, 12 m» to «12 s, 60 m»
✓ 1
curve with decreasing gradient to 72 m at 16 s, then horizontal at 72 m from 16 s to 20 s
✓ 1
falls from 72 m to 24 m between 20 s and 35 s «straight from 23 s to 32 s» AND horizontal at 24 m from 35 s to 40 s
✓ 1
Allow ECF from (b). Shape and values both needed.
Answers: (a)(ii) −1.5 m s−2 · (b) 24 m north · (c)(i) 3.0 m s−1 · (c)(ii) 0.60 m s−1 · (d) 72 m (the remaining parts are explanations — see the table above)
145A-2-57Velocity & accelerationPaper 2Easy13 marks
A manufacturer tests tennis balls by releasing each ball from rest so that its bottom is 2.54 m above a concrete floor. A ball passes the test if it then rebounds to a height between 1.35 m and 1.47 m. Air resistance is negligible unless stated otherwise. Take upwards as positive.
Figure 1 is a set of axes for (e).
Figure 1 — axes for (e)
(a)
Show that the ball reaches the floor with a speed of about 7.1 m s−1.
(1)
(b)
A ball leaves the floor with a speed of 5.20 m s−1. Determine whether this ball passes the test.
(2)
(c)
The ball is in contact with the floor for 4.5 ms. Calculate the magnitude and direction of the average acceleration of the ball during the contact.
(2)
(d)
Outline why the greatest acceleration of the ball during the contact is larger than your answer to (c).
(1)
(e)
Sketch, on Figure 1, a graph to show how the velocity v of the ball varies with time t from its release until it reaches the top of its rebound. Add values to your graph where appropriate.
(3)
(f)
Calculate, for the motion from release to the top of the rebound, the distance travelled by the ball and the magnitude of its displacement.
(2)
(g)
The same balls are tested in a city at high altitude, where the density of the air is much lower. The balls rebound slightly higher. Suggest why.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
v = √(2 × 9.81 × 2.54) «= 7.06 m s−1»
✓ 1
Must see full substitution OR answer to 3 s.f. Accept energy conservation.
Part (b)
rebound height = 5.202/(2 × 9.81) = 1.38 m
✓ 1
Accept v = 0 at the top with v2 = u2 + 2as.
1.35 m < 1.38 m < 1.47 m, so the ball passes
✓ 1
The decision must follow from the height. Do not award for a bare “yes”.
Part (c)
Δv = +5.20 − (−7.06) = 12.3 m s−1
✓ 1
Allow ECF from (a).
a = 12.3/4.5 × 10−3 = 2.7 × 103 m s−2 upwards
✓ 1
Direction required. Award [1 max] for 4.1 × 102 m s−2, which ignores the reversal of the velocity.
Part (d)
the force from the floor «and so the acceleration» changes during the contact, from zero to a maximum at greatest compression; the average is less than this maximum
✓ 1
OWTTE.
Part (e)
straight line from the origin with negative gradient to about −7.1 m s−1 at about 0.72 s
✓ 1
«almost» vertical rise to +5.2 m s−1 at about 0.72 s
✓ 1
Allow ECF from (a) and (b).
straight line with the same gradient as the first, reaching v = 0 at about 1.25 s
✓ 1
Accept 1.2–1.3 s. Lines must be parallel.
Part (f)
distance = 2.54 + 1.38 = 3.92 m
✓ 1
Allow ECF from (b).
displacement = 2.54 − 1.38 = 1.16 m «downwards»
✓ 1
Part (g)
the air resistance is smaller during the fall, so the ball reaches the floor faster «and leaves it faster»
✓ 1
OWTTE.
the air resistance is smaller during the rise, so the ball loses less speed and rises higher
✓ 1
ALTERNATIVE for one mark: the pressure difference between the gas inside the ball and the air outside is larger, so the ball is «stiffer and» bouncier.
Answers: (b) 1.38 m, passes · (c) 2.7 × 103 m s−2 upwards · (f) 3.92 m; 1.16 m (the remaining parts are explanations — see the table above)
146A-2-64Distance, displacement & averagesPaper 2Easy12 marks
Thousands of robotic floats drift in the oceans and measure the temperature and salinity of the water. Each float repeats a 10.0-day cycle. It sinks from the surface to a depth of 1000 m and drifts with the ocean currents at that depth for about 9 days. It then sinks further, to a depth of 2000 m, at a constant speed of 0.050 m s−1, and finally rises vertically to the surface at a constant speed of 0.100 m s−1, taking measurements on the way up. At the surface it sends its data to a satellite and the cycle starts again.
Figure 1 is a set of axes for (d).
Figure 1 — axes for (d). Positions below the surface are negative.
(a)
Calculate the total vertical distance travelled by the float in one cycle.
(1)
(b)
State the vertical component of the displacement of the float over one complete cycle.
(1)
(c)
Show that the float takes about 5.6 hours to rise from a depth of 2000 m to the surface.
(1)
(d)
Draw, on Figure 1, a graph to show how the vertical position of the float varies with time from the moment it leaves the depth of 1000 m until it reaches the surface.
(3)
(e)
In one cycle the float comes to the surface 24.0 km from the point where it last left the surface. Calculate the magnitude of the average velocity of the float for this cycle.
(2)
(f)
Explain why the average speed of the float for the cycle is greater than the magnitude of its average velocity.
(2)
(g)
An oceanographer divides 24.0 km by 9 days to find the velocity of the current at a depth of 1000 m. Suggest two reasons why this gives only an estimate of that velocity.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
1000 + 1000 + 2000 = 4000 m
✓ 1
Part (b)
zero «the float returns to the surface»
✓ 1
Part (c)
t = 2000/0.100 = 2.0 × 104 s = 2.0 × 104/3600 «= 5.56 h»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (d)
two straight lines «constant speed in each stage»: one sloping down from (0, −1000 m) to the lowest point, then one sloping up to 0
✓ 1
Do not accept curves.
lowest point at −2000 m at about 5.6 h
✓ 1
Accept 5.5–5.7 h.
reaches 0 «the surface» at about 11.1 h, so the rising line is twice as steep as the sinking line
✓ 1
Accept 11.0–11.2 h. Allow ECF from (c).
Part (e)
t = 10.0 × 24 × 3600 = 8.64 × 105 s
✓ 1
v = 24.0 × 103/8.64 × 105 = 2.8 × 10−2 m s−1
✓ 1
Accept 2.4 km day−1. Award [2] for CNA.
Part (f)
average speed = total distance travelled ÷ time; average velocity = displacement ÷ time
✓ 1
the distance includes the 4000 m of vertical travel «and any horizontal changes of direction», whereas the displacement is only the straight line of 24.0 km between the two surface points
✓ 1
OWTTE.
Part (g)
the float also drifts horizontally while sinking, rising and at the surface, where the currents are different
✓ 1
the current may change its speed or direction during the 9 days, so 24.0 km/9 days is only an average «the displacement is less than the distance drifted»
✓ 1
Accept any two distinct, valid reasons.
Answers: (a) 4000 m · (b) 0 · (e) 2.8 × 10−2 m s−1(the remaining parts are explanations — see the table above)
The “speed” of a golf green is measured by letting a ball roll down a short grooved ramp. The ball leaves the ramp and rolls onto the level green at 1.83 m s−1. On one green the ball rolls 3.35 m in a straight line before it stops. Assume that the deceleration of the ball on the green is uniform.
Figure 1 is a set of axes for (d).
Figure 1 — axes for (d)
(a)
Show that the deceleration of the ball is about 0.50 m s−2.
(1)
(b)
Calculate the time for which the ball rolls on the green.
(1)
(c)
Calculate the speed of the ball 2.0 s after it leaves the ramp.
(1)
(d)
Sketch, on Figure 1, a graph to show how the distance s rolled by the ball varies with time t until it stops.
(2)
(e)
A golfer putts a ball on this green towards a hole 5.0 m away on level ground. The ball drops into the hole only if it reaches the hole with a speed of 1.3 m s−1 or less. Determine the range of initial speeds for which the ball drops into the hole.
(3)
(f)
On a slower green the ball from the ramp rolls only 2.30 m. Calculate the least initial speed for which a putt reaches a hole 5.0 m away on this green.
(2)
Show mark scheme
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Mark
Notes
Part (a)
a = 1.832/(2 × 3.35) «= 0.500 m s−2»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)
t = 1.83/0.50 = 3.66 s
✓ 1
Accept 3.7 s. Accept t = 2s/u.
Part (c)
v = 1.83 − 0.50 × 2.0 = 0.83 m s−1
✓ 1
Allow ECF from (a).
Part (d)
curve from the origin whose gradient decreases
✓ 1
Do not accept a straight line.
gradient zero «horizontal» at about 3.7 s and s = 3.35 m, and horizontal afterwards if drawn
✓ 1
Allow ECF from (b).
Part (e)
least speed: the ball just reaches the hole, u = √(2 × 0.50 × 5.0) = 2.24 m s−1
✓ 1
greatest speed: u2 = 1.32 + 2 × 0.50 × 5.0
✓ 1
range 2.24 m s−1 to 2.59 m s−1
✓ 1
Accept 2.2–2.6 m s−1. Award [3] for CNA.
Part (f)
a = 1.832/(2 × 2.30) = 0.728 m s−2
✓ 1
u = √(2 × 0.728 × 5.0) = 2.70 m s−1
✓ 1
Accept 2.7 m s−1. Award [2] for CNA.
Answers: (b) 3.66 s · (c) 0.83 m s−1 · (e) 2.24–2.59 m s−1 · (f) 2.70 m s−1(the remaining parts are explanations — see the table above)
A sports scientist uses a radar gun to record the velocity v of a sprinter at one-second intervals after the start of a race.
t / s: 0, 1.0, 2.0, 3.0, 4.0, 5.0 v / m s−1: 0, 4.6, 7.4, 9.0, 9.8, 10.2
Data from the table
(a)
State how the data show that the acceleration of the sprinter is not uniform.
(1)
(b)
Calculate the average acceleration of the sprinter during the first 2.0 s.
(1)
(c)
Treat each one-second interval as a trapezium (straight lines joining the data points). Estimate the distance travelled by the sprinter in the first 5.0 s.
(2)
(d)
The true velocity–time graph is a smooth curve through these points. Explain whether the true distance is greater or smaller than the trapezium estimate from the previous part.
(2)
Show mark scheme
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Notes
Part (a)
The increase in velocity in each second is not the same; it gets smaller (4.6, 2.8, 1.6, 0.8, 0.4 m s−1)
The gradient (acceleration) decreases with time, so the curve lies above the straight lines joining the points
✓ 1
The trapezia therefore leave out some area, so the true distance is slightly greater than the estimate
✓ 1
Exam tip: For distance from tabulated velocities, add the trapezium areas strip by strip. Then decide whether the straight lines lie above or below the real curve.
149IBP2S-001-05-12Motion graphs & definitionsPaper 2Easy5 marks
Doppler ultrasound is used to measure the speed of blood in a patient's aorta during one heartbeat. A simplified model has two stages: • the speed rises uniformly from 0 to 1.2 m s−1 in 0.10 s; • the speed then falls uniformly back to 0 in a further 0.20 s.
Simplified model
(a)
Calculate the acceleration of the blood during the first 0.10 s.
(1)
(b)
Determine the distance the blood moves along the aorta during this 0.30 s.
(2)
(c)
In reality the v–t graph is a smooth curve with a rounded peak. State the acceleration of the blood at the instant of peak speed, and explain your answer.
(2)
Show mark scheme
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Mark
Notes
Part (a)
a = 1.2 / 0.10 = 12 m s−2
✓ 1
Part (b)
Uses the area under the v–t graph: ½ × 0.30 × 1.2. Equivalent methods also earn this mark: the stage distances ½ × 1.2 × 0.10 = 0.06 m and ½ × 1.2 × 0.20 = 0.12 m, or average speed 0.60 m s−1 × 0.30 s
✓ 1
= 0.18 m
✓ 1
Part (c)
The acceleration is zero
✓ 1
The gradient of the v–t graph, which equals the acceleration, is zero at a maximum
✓ 1
Common error: Do not apply a single suvat equation across both stages, because the acceleration changes. Either find the area of the whole triangle or treat each stage separately and add the two distances.
150IBP2S-001-00-11Non-uniform accelerationPaper 2Easy6 marks
An electric scooter starts from rest on a straight, level path. Its velocity is recorded every second:
t / s: 0, 1.0, 2.0, 3.0, 4.0, 5.0, 6.0 v / m s−1: 0, 1.8, 3.6, 5.4, 6.8, 7.6, 8.0
Data from the table
(a)
Calculate the acceleration of the scooter during the first 3.0 s.
(2)
(b)
Outline how the data show that the acceleration decreases after t = 3.0 s.
(2)
(c)
Estimate the distance travelled in the first 6.0 s.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Use a = Δv / Δt = (5.4 − 0) / 3.0 . a = 1.8 m s−2 .
✓ 2
Part (b)
Up to 3.0 s the velocity rises by 1.8 m s−1 each second; after that it rises by 1.4, then 0.8, then 0.4 m s−1 . The increase in velocity per second gets smaller, so the gradient of the v–t graph, which is the acceleration, decreases .
✓ 2
Part (c)
Distance = area under the v–t graph, for example by adding trapezia for each 1.0 s interval . Distance ≈ 29 m (accept 28–30 m) .
✓ 2
Exam tip: When velocity data are in a table, find the distance by adding trapezium areas between successive readings.
151IBP2S-001-04-11Fluid resistance & terminal speedPaper 2Easy5 marks
A badminton shuttlecock of mass 5.0 g is dropped from a high balcony. It reaches a terminal speed of 6.8 m s−1. Model the drag force on the shuttlecock as F = kv2, where v is its speed. Buoyancy is negligible. Take g = 9.8 m s−2.
(a)
State the resultant force on the shuttlecock when it is moving at terminal speed.
(1)
(b)
Calculate k.
(2)
(c)
Calculate the speed of the shuttlecock when its acceleration is 4.9 m s−2.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The resultant force is zero, because the drag balances the weight
✓ 1
Part (b)
At terminal speed kvt2 = mg, so k = 0.0050 × 9.8 / 6.82
✓ 1
k = 1.1 × 10−3 kg m−1 (accept 1.06 × 10−3)
✓ 1
Part (c)
a = g/2, so the drag equals mg/2, giving kv2 = ½kvt2; or numerically kv2 = m(g − a) = 0.0245 N using the value of k found earlier, allowing error carried forward
✓ 1
v = 6.8/√2 = 4.8 m s−1
✓ 1
Exam tip: Convert grams to kilograms before substituting. A mass left as 5.0 instead of 0.0050 makes k 1000 times too large.
152IBP2S-001-03-11Motion graphs & definitionsPaper 2Easy4 marks
A hospital lift starts from rest and accelerates uniformly to 1.6 m s−1 in 2.0 s. It then moves at constant velocity for 8.0 s. Finally it decelerates uniformly to rest in 2.0 s.
(a)
Calculate the acceleration of the lift during the first 2.0 s.
(1)
(b)
Determine the total distance travelled by the lift.
(2)
(c)
Calculate the average speed of the lift for the whole journey.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
a = 1.6/2.0 = 0.80 m s−2
✓ 1
Part (b)
Uses the area under the v–t graph (or suvat for each phase): two triangles each of ½ × 2.0 × 1.6 = 1.6 m, plus a rectangle of 8.0 × 1.6 = 12.8 m
✓ 1
Total distance = 16.0 m (accept 16 m)
✓ 1
Part (c)
Average speed = 16.0/12.0 = 1.33 m s−1 (accept 1.3 m s−1; allow error carried forward from the total distance found earlier)
✓ 1
Exam tip: Average speed uses the total time, here 12 s, including the acceleration and deceleration phases.
154IBP2S-001-02-13Distance, displacement & averagesPaper 2Easy6 marks
In a sports fitness test, an athlete runs back and forth between two lines 20.0 m apart. She starts at line A, runs to line B, and keeps turning at each line. After completing 7 lengths she stops, 42 s after starting.
(a)
Outline the difference between distance and displacement.
(2)
(b)
State the distance travelled and the displacement of the athlete when she stops.
(2)
(c)
Calculate her average speed and the magnitude of her average velocity.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Distance is the total length of path travelled and is a scalar
✓ 1
Displacement is the change in position (straight line from start to finish) and is a vector, with magnitude and direction
✓ 1
Part (b)
Distance = 7 × 20.0 = 140 m
✓ 1
Displacement = 20.0 m from A towards B (she finishes at line B)
✓ 1
Part (c)
Average speed = 140/42 = 3.3 m s−1
✓ 1
Average velocity = 20.0/42 = 0.48 m s−1; ECF from (b)
✓ 1
Exam tip: Count the lengths carefully. After an odd number of lengths the athlete is at the far line, so her displacement is not zero.
155IBP2S-001-01-11Motion graphs & definitionsPaper 2Easy6 marks
A sprinter in a 100 m race accelerates uniformly from rest, reaching 10.8 m s−1 in 3.6 s. She then runs the rest of the race at this constant speed.
(a)
Calculate the acceleration of the sprinter.
(2)
(b)
Calculate the distance she covers while accelerating.
(2)
(c)
Determine her total time for the 100 m race.
(2)
Show mark scheme
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Mark
Notes
Part (a)
a = (v − u)/t = 10.8/3.6 ; a = 3.0 m s−2
✓ 2
Part (b)
s = ½at2 = ½ × 3.0 × 3.62, or s = ½(0 + 10.8) × 3.6 ; s = 19.4 m . ECF from (a)
✓ 2
Part (c)
time at constant speed = (100 − 19.4)/10.8 = 7.46 s ; total time = 3.6 + 7.46 = 11.1 s . ECF from (b)
✓ 2
Exam tip: Split motion into stages and choose the right equation for each. The suvat equations apply only to the accelerating stage; the constant-speed stage needs only distance = speed × time.
An aircraft starts from rest at one end of a straight, level runway of length 1800 m. It accelerates uniformly at 2.0 m s−2 and leaves the ground when its speed reaches 70 m s−1.
If the pilot abandons the take-off, the brakes give the aircraft a constant deceleration of 3.5 m s−2. The pilot’s reaction time may be neglected.
Figure 1 — axes for (b)(ii)
(a)
(i)
Calculate the time taken for the aircraft to reach its take-off speed.
(1)
(ii)
Show that the distance needed for take-off is about 1.2 km.
(1)
(b)
On one flight the pilot abandons the take-off when the speed is 50 m s−1.
(i)
Calculate the total distance travelled by the aircraft from the start until it stops.
(2)
(ii)
Sketch, on Figure 1, the velocity–time graph for this flight from the start until the aircraft stops. Add values to your graph where appropriate.
(2)
(c)
Determine the greatest speed at which the take-off can be abandoned so that the aircraft stops before the end of the runway.
(3)
(d)
On a wet runway the braking deceleration is smaller, but the acceleration is unchanged. State and explain the effect on your answer to (c).
(2)
Show mark scheme
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Notes
Part (a)(i)
t = 70/2.0 = 35 s
✓ 1
Part (a)(ii)
s = 70²/(2 × 2.0) = 1225 «m»
✓ 1
ALTERNATIVE: s = ½ × 2.0 × 35². Must see full substitution OR answer to 3 s.f.
Part (b)(i)
accelerating: 50²/(2 × 2.0) = 625 m AND braking: 50²/(2 × 3.5) = 357 m
✓ 1
total = 982 m
✓ 1
Accept 980 m. Award [2] for CNA. Award [1 max] for 625 m or 357 m alone.
Part (b)(ii)
straight line from the origin to 50 m s−1 at 25 s
✓ 1
straight, steeper line from that point down to zero at 39 s
✓ 1
Accept 39–40 s. Allow ECF from (b)(i).
Part (c)
total distance = v²/(2 × 2.0) + v²/(2 × 3.5) = 1800 m
✓ 1
v² = 4580 «m² s−2»
✓ 1
v = 68 m s−1
✓ 1
Award [3] for CNA. Award [0] for 112 m s−1 «whole runway used for braking».
Part (d)
at any speed the braking distance v²/2a is greater, so less of the runway is left for accelerating
✓ 1
so the greatest speed at which the take-off can be abandoned is lower
✓ 1
MP2 only scores if MP1 scores.
Answers: (a)(i) 35 s · (a)(ii) 1225 m · (b)(i) 982 m · (c) 68 m s−1(the remaining parts are explanations — see the table above)
A delivery drone flies horizontally at a constant velocity of 12 m s−1 at a height of 30 m above level ground. It releases a package, which falls to the ground. Air resistance on the package is negligible unless stated otherwise.
Figure 1 — drawn to scale
(a)
(i)
State the horizontal and the vertical components of the velocity of the package at the instant of release.
(1)
(ii)
Show that the package takes about 2.5 s to reach the ground.
(1)
(iii)
Calculate the horizontal distance travelled by the package before it lands.
(1)
(b)
Determine the velocity of the package just before it lands.
(3)
(c)
Sketch, on Figure 1, the path of the package as seen by an observer at rest on the ground. Mark with a cross the position of the drone at the instant the package lands.
(2)
(d)
The package can survive an impact at a speed of up to 25 m s−1. Determine the greatest height from which the drone, flying at 12 m s−1, can release it safely.
(3)
(e)
In practice air resistance on the package is not negligible. State and explain where the package lands relative to the point directly below the drone at that instant.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
horizontal: 12 m s−1 AND vertical: zero
✓ 1
Both needed.
Part (a)(ii)
t = √(2 × 30 / 9.81) «= 2.47 s»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (a)(iii)
12 × 2.47 = 29.7 m
✓ 1
Accept 30 m from 2.5 s.
Part (b)
vy = 9.81 × 2.47 = 24.3 m s−1
✓ 1
Accept √(2 × 9.81 × 30).
v = √(12² + 24.3²) = 27.1 m s−1
✓ 1
Allow ECF from (a)(ii).
tan−1(24.3/12) = 64° below the horizontal
✓ 1
Direction must be stated as below the horizontal «or 26° to the vertical». Accept 63°–64°.
Part (c)
curve from the release point, horizontal at the start and becoming steeper, reaching the ground at ≈ 30 m
✓ 1
Allow ECF from (a)(iii). Do not accept a straight line.
cross on the drone’s path vertically above the landing point
✓ 1
The drone and the package have the same horizontal velocity throughout.
Award [1 max] for 31.9 m, which ignores the horizontal component of the velocity.
Part (e)
behind the drone «short of the point below it»
✓ 1
air resistance reduces the horizontal velocity of the package while the drone keeps moving at 12 m s−1
✓ 1
MP2 only scores if MP1 scores.
Answers: (a)(ii) 2.47 s · (a)(iii) 29.7 m · (b) 27.1 m s−1 at 64° below the horizontal · (d) 24.5 m (the remaining parts are explanations — see the table above)
159A-2-14Fluid resistance & terminal speedPaper 2Medium14 marks
A badminton shuttlecock of mass 5.0 g is hit from point O with a speed of 14 m s−1 at 40° above the horizontal. The dashed curve in Figure 1 shows, to scale, the path the shuttlecock would follow if there were no air resistance.
In reality the air resistance on a shuttlecock is large and increases with its speed. A shuttlecock dropped from a great height reaches a terminal speed of 6.8 m s−1.
Figure 1 — graph drawn to scale
(a)
(i)
Calculate the weight of the shuttlecock.
(1)
(ii)
State the magnitude of the air resistance on the shuttlecock when it falls vertically at its terminal speed, and explain your answer.
(2)
(b)
Sketch, on Figure 1, the path of the shuttlecock when air resistance acts. It leaves O with the same velocity as before.
(3)
(c)
At the highest point of its real path the shuttlecock is moving horizontally. State and explain the direction of its acceleration at this point.
(2)
(d)
Explain, in terms of the forces acting, why the shuttlecock takes longer to fall from its highest point than to rise to it.
(3)
(e)
A player argues that, because the range of a projectile is proportional to the square of its launch speed, hitting the shuttlecock twice as fast will make it travel four times as far. Evaluate this argument.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
W = 5.0 × 10−3 × 9.81 = 4.9 × 10−2 N
✓ 1
Part (a)(ii)
4.9 × 10−2 N «equal to the weight»
✓ 1
Allow ECF from (a)(i).
the velocity is constant so the resultant force is zero «Newton’s first law»
✓ 1
MP2 needs the reason, not just “terminal speed”.
Part (b)
lower maximum height than the dashed curve
✓ 1
shorter range «lands closer to O»
✓ 1
asymmetric path: the highest point is reached before half of the horizontal distance AND the descent is steeper than the ascent «almost vertical near the end»
✓ 1
Both features needed for MP3. Do not award MP3 for a symmetric curve.
Part (c)
weight acts vertically downwards AND air resistance acts horizontally, opposite to the velocity
✓ 1
so the acceleration is directed downwards and backwards «below the horizontal, opposite to the motion», with a magnitude greater than g
✓ 1
Do not accept “vertically downwards”. MP2 only scores if MP1 scores.
Part (d)
on the way up the air resistance has a downward component that adds to the weight, so the vertical deceleration is greater than g
✓ 1
on the way down the air resistance has an upward component that opposes the weight, so the downward acceleration is less than g
✓ 1
the same vertical distance is covered with a smaller average vertical speed on the way down, so the fall takes longer
✓ 1
OWTTE.
Part (e)
the argument holds without air resistance: time of flight and horizontal velocity are each proportional to the launch speed
✓ 1
air resistance increases with speed, so a faster shuttlecock experiences a much larger resistive force and loses speed more rapidly
✓ 1
so the range increases by much less than four times «the argument does not apply to a shuttlecock»
✓ 1
MP3 is for a conclusion supported by the effect of air resistance.
Answers: (a)(i) 4.9 × 10−2 N · (a)(ii) 4.9 × 10−2 N (the remaining parts are explanations — see the table above)
160A-2-29Distance, displacement & averagesPaper 2Medium13 marks
A diver jumps from a springboard 3.0 m above the surface of the water in a pool. She leaves the board moving vertically upwards at 4.2 m s−1, rises, falls into the water and slows down until she momentarily stops below the surface. Treat the diver as a point; air resistance is negligible.
The graph shows how her height above the water surface varies with time t from the moment she leaves the board. Upwards is positive.
Graph drawn to scale
(a)
Using the graph, determine the distance travelled by the diver, and her displacement, between leaving the board and reaching her lowest point.
(3)
(b)
(i)
Draw a tangent to the graph at t = 1.00 s and use it to determine the velocity of the diver at this instant.
(2)
(ii)
Calculate the speed of the diver as she reaches the water.
(2)
(c)
Calculate the average velocity of the diver between leaving the board and reaching the water.
(2)
(d)
Describe, with reference to the graph, how the acceleration of the diver changes after she enters the water.
(2)
(e)
Estimate the average acceleration of the diver while she is slowing down in the water.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
greatest height 3.9 m, so she rises 0.9 m above the board
✓ 1
Accept 3.8–4.0 m.
distance = 0.9 + 3.9 + 2.4 = 7.2 m
✓ 1
Accept 7.0–7.4 m.
displacement = 5.4 m downwards «−5.4 m»
✓ 1
Direction or sign required. Accept 5.3–5.5 m.
Part (b)(i)
tangent drawn touching the curve at t = 1.00 s
✓ 1
gradient = −5.6 m s−1 «5.6 m s−1 downwards»
✓ 1
Accept 5.2–6.0 m s−1 downwards. The equations of motion give v = 4.2 − 9.81 × 1.00 = −5.61 m s−1.
Part (b)(ii)
v² = 4.2² + 2 × 9.81 × 3.0
✓ 1
Accept energy conservation. Accept v = u − gt with t read from the graph.
v = 8.7 m s−1
✓ 1
Accept 8.75 m s−1.
Part (c)
displacement = −3.0 m in 1.32 s
✓ 1
Accept 1.3 s from the graph.
average velocity = −2.3 m s−1 «2.3 m s−1 downwards»
✓ 1
Direction or sign required. Accept (u + v)/2 = (4.2 − 8.7)/2, valid because the acceleration is uniform in the air.
Part (d)
the acceleration is upwards «opposite to the velocity»: the gradient of the graph decreases in magnitude to zero
✓ 1
the magnitude of the acceleration decreases «as her speed falls», so the acceleration is non-uniform «it is greatest just after she enters the water»
✓ 1
Part (e)
time to stop ≈ 2.1 − 1.3 = 0.8 s
✓ 1
Accept 0.7–0.9 s.
a ≈ 8.7/0.8 = 11 m s−2 «upwards»
✓ 1
Accept 9.5–13 m s−2. Allow ECF from (b)(ii).
Answers: (a) 7.2 m; 5.4 m downwards · (b)(i) −5.6 m s−1 · (b)(ii) 8.7 m s−1 · (c) −2.3 m s−1 · (e) 11 m s−2(the remaining parts are explanations — see the table above)
A mountain-bike rider leaves the lip of a ramp at a speed of 9.0 m s−1 in a direction 12° below the horizontal. The lip is 2.2 m above level ground. Beyond the lip there is a ditch whose far edge is 5.0 m horizontally from the lip (Figure 1). Treat the rider and bicycle as a point. Air resistance is negligible unless stated otherwise.
Figure 1 — Diagram NOT accurately drawnFigure 2 — axes for (f)
(a)
Calculate the horizontal and vertical components of the velocity of the rider at the lip.
(2)
(b)
Show that the rider takes about 0.51 s to fall the 2.2 m to the level of the ground.
(1)
(c)
Determine whether the rider clears the ditch.
(2)
(d)
Explain why, at the same speed, the rider would travel further horizontally before reaching the ground if the lip were horizontal.
(2)
(e)
Determine the minimum speed, in the same direction of 12° below the horizontal, with which the rider must leave the lip to clear the ditch.
(3)
(f)
Sketch, on Figure 2, how the vertical component of velocity vy of the rider varies with time t from leaving the lip at 9.0 m s−1 until reaching the level of the ground. Take upwards as positive.
(2)
(g)
Outline one effect of air resistance on the horizontal distance travelled by the rider.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
horizontal: 9.0 cos 12° = 8.80 m s−1
✓ 1
vertical: 9.0 sin 12° = 1.87 m s−1 downwards
✓ 1
Direction or a negative sign «upwards positive» required.
Part (b)
2.2 = 1.87t + ½ × 9.81 × t² leading to t = 0.506 «s»
✓ 1
Must see full substitution OR answer to 3 s.f. Accept a check by substitution: 1.87 × 0.51 + 4.905 × 0.51² = 2.2 m. Award [0] if the vertical component is ignored.
Part (c)
x = 8.80 × 0.506 = 4.45 m
✓ 1
Allow ECF from (a) and (b).
4.45 m < 5.0 m so the rider does not clear the ditch
✓ 1
MP2 needs a comparison with 5.0 m. Do not award MP2 for a value from a horizontal launch «6.0 m».
Part (d)
the horizontal component would be larger «9.0 m s−1 instead of 8.80 m s−1»
✓ 1
there would be no initial downward component of velocity, so the time to fall the same height would be longer «0.67 s instead of 0.51 s»
✓ 1
The time-of-flight idea is essential for MP2.
Part (e)
x = v cos 12° × T = 5.0 and 2.2 = v sin 12° × T + ½gT², so 2.2 = 5.0 tan 12° + 4.905T²
✓ 1
Accept any valid method, including trial and improvement.
T = 0.482 s
✓ 1
v = 5.0/(cos 12° × 0.482) = 10.6 m s−1
✓ 1
Accept 10.5–10.7 m s−1. Award [3] for CNA.
Part (f)
straight line with a negative gradient «−9.81 m s−2» starting at vy ≈ −1.9 m s−1 at t = 0
✓ 1
Do not award if the line starts at zero.
line ends at t ≈ 0.51 s with vy ≈ −6.8 m s−1
✓ 1
Allow ECF from (b). Accept −6.5 to −7.1 m s−1.
Part (g)
air resistance reduces the horizontal component of velocity during the flight, so the horizontal distance is smaller
✓ 1
OWTTE. Do not accept “it slows the rider” without reference to the horizontal distance.
Answers: (a) 8.80 m s−1; 1.87 m s−1 downwards · (b) 0.506 s · (c) 4.45 m: does not clear · (e) 10.6 m s−1(the remaining parts are explanations — see the table above)
Two lorries, L and F, travel in a “platoon” along a straight, level motorway. F follows L at the same speed of 22.0 m s−1, with a gap of 15.0 m between the front of F and the back of L. The brakes of the two lorries are linked by a wireless signal.
At t = 0, L brakes with a constant deceleration of 5.5 m s−2. The signal makes F begin to brake 0.30 s later. F is more heavily loaded, and its greatest deceleration is 4.4 m s−2. Both lorries brake until they stop.
Figure 1 is a set of axes for (c).
Figure 1 — axes for (c)
(a)
Calculate the distance travelled by L from t = 0 until it stops.
(1)
(b)
Show that F travels about 62 m from t = 0 until it stops.
(2)
(c)
Sketch, on Figure 1, the speed–time graphs of L and of F from t = 0 until both lorries have stopped. Label each graph.
(3)
(d)
State what is represented by the area between the two graphs.
(1)
(e)
Determine whether F collides with L.
(2)
(f)
Calculate the longest delay of the signal for which F would not collide with L. Assume that the decelerations are unchanged.
(2)
(g)
Explain why, in this situation, the smallest gap between the lorries occurs when F stops.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
22.02/(2 × 5.5) = 44.0 m
✓ 1
Part (b)
during the delay: 22.0 × 0.30 = 6.6 m
✓ 1
braking: 22.02/(2 × 4.4) = 55.0 m, so total = 61.6 m «≈ 62 m»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (c)
L: straight line from 22 m s−1 at t = 0 to zero at 4.0 s
✓ 1
F: horizontal at 22 m s−1 from 0 to 0.30 s, then a straight line to zero at 5.3 s
✓ 1
Allow ECF from (b).
F's braking line is less steep than L's AND F's graph is never below L's; both labelled
✓ 1
Part (d)
the decrease in the gap between the lorries «the extra distance travelled by F compared with L»
✓ 1
OWTTE.
Part (e)
F needs 61.6 − 44.0 = 17.6 m more than L to stop
✓ 1
Allow ECF from (a) and (b).
17.6 m > 15.0 m, so F collides with L
✓ 1
The decision must be consistent with the candidate's values.
Part (f)
22.0 × t + 55.0 − 44.0 = 15.0
✓ 1
Allow ECF from (a) and (b).
t = 0.18 s
✓ 1
Award [2] for CNA.
Part (g)
F starts braking later and decelerates less, so at every instant F is moving at least as fast as L
✓ 1
Accept reference to the graphs: F's graph is never below L's.
so the gap decreases throughout and is smallest when F «the second lorry» stops «comparing stopping distances is enough»
✓ 1
OWTTE. MP2 only scores if MP1 scores.
Answers: (a) 44.0 m · (e) collides (17.6 m > 15.0 m) · (f) 0.18 s (the remaining parts are explanations — see the table above)
163A-2-60Fluid resistance & terminal speedPaper 2Medium13 marks
A weather-research aircraft flying horizontally at 150 m s−1 at an altitude of 9000 m releases a dropsonde, a small instrument package that falls on a parachute and measures the atmosphere on its way down. Figure 1 shows how the vertical component of the velocity of the dropsonde varies with time t for the first 30 s after release. The parachute opens at the instant of release.
The table gives the steady downward speed of the dropsonde measured as it passes lower altitudes.
Figure 2 shows, to scale, the path the dropsonde would follow if there were no air resistance.
Altitude / m
6000
3000
0
Steady downward speed / m s−1
20.2
17.2
14.8
Figure 1 — graph drawn to scaleFigure 2 — graph drawn to scale; distance below the release point increases downwards
(a)
State the vertical component of the acceleration of the dropsonde at the instant of release. Explain your answer.
(2)
(b)
State the terminal speed of the dropsonde at an altitude of about 9000 m.
(1)
(c)
Explain, with reference to the forces on the dropsonde, why the vertical component of its velocity becomes constant.
(2)
(d)
Estimate the vertical distance fallen by the dropsonde in the first 20 s.
(2)
(e)
Explain why the speed of the dropsonde decreases as it falls to lower altitudes, where the density of the air is greater.
(2)
(f)
Estimate the time taken by the dropsonde to fall from 9000 m to sea level.
(2)
(g)
Sketch, on Figure 2, the path of the dropsonde with air resistance acting.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
9.81 m s−2 «downwards»
✓ 1
Accept g.
the velocity is horizontal at release, so the air resistance is horizontal and has no vertical component; the only vertical force is the weight
✓ 1
OWTTE. MP2 only scores if MP1 scores.
Part (b)
24 m s−1
✓ 1
Accept 23.5–24.5 m s−1.
Part (c)
air resistance increases as the speed increases until its upward «vertical» component equals the weight
✓ 1
the resultant vertical force is then zero, so there is no vertical acceleration
✓ 1
OWTTE.
Part (d)
area under the graph from 0 to 20 s, eg by counting squares OR a triangle plus rectangles
✓ 1
≈ 420 m
✓ 1
Accept 390–450 m.
Part (e)
at the same speed the air resistance is greater in denser air, so it becomes greater than the weight
✓ 1
the resultant force «and acceleration» is upwards, so the dropsonde slows down until air resistance again equals the weight at a lower terminal speed
✓ 1
OWTTE.
Part (f)
uses distance ÷ average speed for each section, with the value from (b) at 9000 m, eg 3000/22.1 + 3000/18.7 + 3000/16.0 OR 9000/19.4
✓ 1
Allow ECF from (b).
≈ 480 s «8 minutes»
✓ 1
Accept 450–500 s. The time taken to reach the terminal speed may be ignored.
Part (g)
starts horizontally at the release point and curves downwards more steeply than the dashed line, lying below it
✓ 1
becomes «almost» vertical, at a horizontal distance much less than the dashed path «below 200 m»
✓ 1
Do not award MP2 for a path that reaches the bottom of the grid beyond 300 m.
Answers: (a) 9.81 m s−2 · (b) 24 m s−1 · (d) ≈ 420 m · (f) ≈ 480 s (the remaining parts are explanations — see the table above)
In an amusement-park drop tower, a gondola carrying riders is released from rest at the top of a vertical track. It falls freely through 42.0 m. Magnetic brakes then give it a uniform deceleration so that it comes to rest after a further 18.0 m. Air resistance and friction are negligible during the free fall. Take upwards as positive.
Figure 1 is a set of axes for (d).
Figure 1 — axes for (d)
(a)
Show that the speed of the gondola at the end of the free fall is about 29 m s−1.
(1)
(b)
Calculate the time for which the gondola falls freely.
(1)
(c)
Calculate the magnitude of the deceleration of the gondola during braking, as a multiple of g.
(2)
(d)
Sketch, on Figure 1, a graph to show how the acceleration a of the gondola varies with time t from release until it comes to rest. Add values to your graph where appropriate.
(3)
(e)
Calculate the average velocity of the gondola from release until it comes to rest.
(2)
(f)
Calculate the force exerted by the seat on a rider of mass 62 kg during braking.
(2)
(g)
In a real magnetic brake the braking force is proportional to the speed of the gondola. Suggest how the acceleration of the gondola during braking would then differ from your graph in (d).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
v = √(2 × 9.81 × 42.0) «= 28.7 m s−1»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)
t = 28.7/9.81 = 2.93 s
✓ 1
Accept √(2 × 42.0/9.81). Accept 2.9 s.
Part (c)
a = 28.72/(2 × 18.0) = 22.9 m s−2
✓ 1
Allow ECF from (a).
= 2.33g
✓ 1
Accept 2.3g. Accept the ratio 42.0/18.0 with a valid argument. Award [2] for CNA.
Part (d)
horizontal line at a = −9.81 m s−2 from 0 to 2.9 s
✓ 1
Allow ECF from (b).
horizontal line at a = +22.9 m s−2, above the time axis, starting at 2.9 s
✓ 1
Allow ECF from (c).
braking line ends at about 4.2 s «2.93 + 1.25 s» AND a = 0 afterwards if shown
✓ 1
Accept 4.1–4.3 s. Vertical joining lines may be shown or omitted.
Part (e)
total time = 2.93 + 28.7/22.9 = 4.18 s
✓ 1
Allow ECF from (b) and (c).
average velocity = 60.0/4.18 = 14.4 m s−1 downwards
✓ 1
Direction or negative sign required. Award [2] for CNA.
Part (f)
N − mg = ma OR N = 62 × (9.81 + 22.9)
✓ 1
Allow ECF from (c).
N = 2.0 × 103 N «upwards»
✓ 1
Award [1 max] for 1.4 × 103 N, which omits the weight.
Part (g)
the deceleration is greatest at the start of braking and then decreases as the speed decreases «non-uniform»
✓ 1
the deceleration tends to zero / the gondola approaches a small constant speed instead of stopping, so another brake is needed to stop it
✓ 1
OWTTE. Accept “for the same braking distance the greatest deceleration must be larger than 2.3g”.
Answers: (b) 2.93 s · (c) 22.9 m s−2 = 2.33g · (e) 14.4 m s−1 downwards · (f) 2.0 × 103 N (the remaining parts are explanations — see the table above)
A coastguard team uses a line-throwing launcher to fire a light rope to a vessel in difficulty. The launcher is on the top of a vertical cliff, 25.0 m above the sea. The projectile, which carries the end of the rope, leaves the launcher at 32.0 m s−1 at 30° above the horizontal, towards the vessel. Treat the projectile as a point. Air resistance and the pull of the rope are negligible unless stated otherwise.
Diagram not to scale
(a)
(i)
Show that the vertical component of the initial velocity of the projectile is 16.0 m s−1.
(1)
(ii)
Calculate the horizontal component of the initial velocity.
(1)
(b)
Calculate the maximum height of the projectile above the sea.
(2)
(c)
Show that the projectile reaches sea level about 4.4 s after it is fired.
(2)
(d)
The vessel is 105 m from the foot of the cliff. For the rope to fall across the vessel, the projectile must travel at least 105 m horizontally before it reaches sea level. Determine whether it does so.
(2)
(e)
Calculate the speed of the projectile as it reaches sea level.
(2)
(f)
The manufacturer states that the drag of the rope and the air reduce the horizontal distance travelled by about 20 %. Comment on whether the rope will reach the vessel, and suggest one change the team could make.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
uy = 32.0 sin 30° «= 16.0 m s−1»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (a)(ii)
ux = 32.0 cos 30° = 27.7 m s−1
✓ 1
Part (b)
height above the cliff top = 16.02/(2 × 9.81) = 13.0 m
✓ 1
height above the sea = 25.0 + 13.0 = 38.0 m
✓ 1
Award [1 max] for 13.0 m alone. Award [2] for CNA.
Part (c)
−25.0 = 16.0t − ½ × 9.81 × t2 OR 4.905t2 − 16.0t − 25.0 = 0
✓ 1
Signs must be consistent.
t = 4.42 «s» «positive root»
✓ 1
Must see full substitution OR answer to 3 s.f. Accept a check by substitution of 4.4 s giving a displacement of about −25 m.
Part (d)
x = 27.7 × 4.42 = 122 m
✓ 1
Allow ECF from (a)(ii) and (c).
122 m > 105 m, so it travels far enough «the rope falls across the vessel»
✓ 1
The decision must be consistent with the candidate's distance.
A fireworks shell is fired vertically upwards from a launch tube of length 0.60 m. The shell starts from rest at the bottom of the tube and accelerates uniformly, leaving the top of the tube at 50 m s−1. A time fuse is lit as the shell leaves the tube; after 4.2 s it makes the shell burst. Air resistance is negligible unless stated otherwise. Take upwards as positive.
Speed of sound in air = 340 m s−1. Figure 1 is a set of axes for (f).
Figure 1 — axes for (f)
(a)
(i)
Calculate the acceleration of the shell in the tube.
(2)
(ii)
Calculate the time the shell spends in the tube.
(1)
(b)
Show that, without air resistance, the shell would rise about 130 m above the top of the tube.
(1)
(c)
Calculate, without air resistance, the height and the velocity of the shell when it bursts.
(2)
(d)
An observer standing beside the launch tube sees the flash of the burst directly overhead and hears the bang 0.30 s later.
(i)
Calculate the height at which the shell bursts.
(1)
(ii)
Outline why the time taken by the light to reach the observer can be ignored.
(1)
(e)
Explain why the height in (d)(i) is less than the height in (c).
(2)
(f)
Sketch, on Figure 1, the velocity–time graph of the shell from leaving the tube until it bursts: (1) without air resistance, labelled N, and (2) with air resistance, labelled R.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
v2 = u2 + 2as: 502 = 2 × a × 0.60
✓ 1
a = 2.1 × 103 m s−2
✓ 1
Accept 2080 m s−2. Award [2] for CNA.
Part (a)(ii)
t = 2 × 0.60/50 = 0.024 s
✓ 1
Accept t = v/a. Allow ECF from (a)(i).
Part (b)
h = 502/(2 × 9.81) «= 127 m»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (c)
s = 50 × 4.2 − ½ × 9.81 × 4.22 = 123 m
✓ 1
Accept 120–124 m.
v = 50 − 9.81 × 4.2 = 8.8 m s−1 upwards
✓ 1
Direction or positive sign required.
Part (d)(i)
340 × 0.30 = 102 m
✓ 1
Accept 100 m.
Part (d)(ii)
light travels about a million times faster than sound, so it takes only about 3 × 10−7 s
✓ 1
Accept a comparison of speeds without the time.
Part (e)
air resistance acts downwards on the rising shell, adding to its weight
✓ 1
so the deceleration is greater than g and the shell has a smaller velocity and height at each instant
✓ 1
OWTTE. Accept “the shell reaches a lower height and stops rising sooner”.
Part (f)
N: straight line from 50 m s−1 at t = 0 to about 9 m s−1 at 4.2 s
✓ 1
Allow ECF from (c).
R: starts at 50 m s−1 and lies below N at all later times
✓ 1
R: curve whose gradient is steepest at the start and becomes less steep «tending towards the gradient of N» as the speed decreases, ending near zero at 4.2 s
✓ 1
Accept R ending anywhere between about −5 and +5 m s−1.
Answers: (a)(i) 2.1 × 103 m s−2 · (a)(ii) 0.024 s · (c) 123 m; 8.8 m s−1 up · (d)(i) 102 m (the remaining parts are explanations — see the table above)
167A-2-67Velocity & accelerationPaper 2Medium12 marks
Wildlife researchers fit a cheetah with a tracking collar that combines satellite positioning with motion sensors. The table shows the distance s of the cheetah from its starting point at time t as it starts a chase from rest along a straight line.
t / s
0.00
0.25
0.50
0.75
1.00
1.25
1.50
s / m
0.00
0.28
1.13
2.53
4.50
7.03
10.13
Data from the table
(a)
Deduce, using the data, whether the acceleration of the cheetah is uniform during the first 1.50 s.
(2)
(b)
Determine the acceleration of the cheetah.
(2)
(c)
Calculate the speed of the cheetah at t = 1.50 s.
(1)
(d)
Calculate the average speed of the cheetah for the first 1.50 s, and explain why it is half of your answer to (c).
(2)
(e)
(i)
The top speed of the cheetah is 26 m s−1. Calculate the distance the cheetah would need to reach this speed if its acceleration stayed the same.
(2)
(ii)
In practice the cheetah needs a much greater distance to reach its top speed. Suggest why.
(1)
(f)
Calculate the average speed of the cheetah between t = 1.25 s and t = 1.50 s, and explain why it is less than your answer to (c).
(2)
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Notes
Part (a)
a valid test applied to at least three rows, eg s/t2 = 4.48, 4.52, 4.50, 4.50 … m s−2 OR the distances in successive 0.25 s intervals increase by a constant 0.56 m OR doubling t from 0.50 s to 1.00 s multiplies s by 4
✓ 1
s/t2 is constant, so s ∝ t2, as s = ½at2 predicts for uniform acceleration from rest: the acceleration is uniform
✓ 1
OWTTE. MP2 only scores if MP1 scores.
Part (b)
a = 2s/t2 = 2 × 10.13/1.502 OR a = 0.56/0.252
✓ 1
Any row, or the constant from (a).
a = 9.0 m s−2
✓ 1
Accept 8.9–9.1 m s−2. Award [2] for CNA.
Part (c)
v = 9.0 × 1.50 = 13.5 m s−1
✓ 1
Allow ECF from (b).
Part (d)
10.13/1.50 = 6.75 m s−1
✓ 1
Accept 6.8 m s−1.
for uniform acceleration the average speed is (u + v)/2, and u = 0
✓ 1
OWTTE.
Part (e)(i)
s = v2/2a = 262/(2 × 9.0)
✓ 1
Allow ECF from (b).
s = 38 m
✓ 1
Accept 37.6 m. Award [2] for CNA.
Part (e)(ii)
the acceleration decreases as the speed increases, eg because air resistance increases with speed or the cheetah cannot push backwards on the ground as hard at high speed
✓ 1
OWTTE.
Part (f)
(10.13 − 7.03)/0.25 = 12.4 m s−1
✓ 1
the cheetah speeds up during the interval, so its average speed over the interval is less than its speed at the end «it equals the speed at the middle of the interval, 1.375 s»
✓ 1
OWTTE.
Answers: (b) 9.0 m s−2 · (c) 13.5 m s−1 · (d) 6.75 m s−1 · (e)(i) 38 m · (f) 12.4 m s−1(the remaining parts are explanations — see the table above)
168A-2-68Non-uniform accelerationPaper 2Medium13 marks
In a school competition, model cars of mass 50 g are driven along a straight, level 20.0 m track by a jet of gas from a small carbon dioxide cartridge. The thrust of the gas decreases steadily and the cartridge is empty after 0.50 s. Figure 1 shows how the speed v of one car varies with time t from the start.
Figure 2 is a set of axes for (d).
Figure 1 — graph drawn to scaleFigure 2 — axes for (d)
(a)
The cartridge is empty at t = 0.50 s. Explain why the speed of the car is greatest before this time.
(2)
(b)
Determine the time taken by the car to travel the 20.0 m track.
(3)
(c)
Calculate the average speed of the car over the 20.0 m track.
(1)
(d)
Sketch, on Figure 2, a graph to show how the acceleration of the car varies with time from t = 0 to t = 1.4 s.
(3)
(e)
Explain why, after the cartridge is empty, the car slows down with a deceleration that decreases.
(2)
(f)
A second car has the same body shape and the same cartridge, but its mass is 75 g. Assume that the resistive force on each car depends only on its speed.
(i)
Deduce how the initial acceleration of the second car compares with that of the first car.
(1)
(ii)
State and explain how the deceleration of the second car after its cartridge is empty compares with that of the first car at the same speed.
(1)
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Notes
Part (a)
the speed is greatest when the acceleration «gradient» is zero, ie when the resultant force is zero: the decreasing thrust has fallen to equal the resistive forces
✓ 1
OWTTE.
this happens at about 0.47 s; after it the thrust is less than the resistive forces, so the car slows down although gas is still leaving the cartridge
✓ 1
Accept 0.45–0.49 s. MP2 only scores if MP1 scores.
Part (b)
distance = area under the graph, eg by counting squares OR by splitting it into trapezia
✓ 1
area up to 1.00 s ≈ 17.5 m, so the last ≈ 2.5 m is covered at about 19 m s−1
✓ 1
Accept 16.5–18.5 m. Accept any valid method that finds where the area reaches 20.0 m.
t ≈ 1.13 s
✓ 1
Accept 1.10–1.18 s. Award [1 max] for 20.0/21.3 = 0.94 s, which assumes the greatest speed throughout.
Part (c)
20.0/1.13 = 17.7 m s−1
✓ 1
Allow ECF from (b). Accept 16.9–18.2 m s−1.
Part (d)
positive acceleration starting at about 90 m s−2 «50–100 m s−2» and decreasing to zero at about 0.47 s
✓ 1
Accept zero anywhere from 0.45 s to 0.50 s. Allow ECF from (a).
negative acceleration from about 0.5 s to 1.4 s
✓ 1
the negative acceleration has a small magnitude «about 3–4 m s−2» that decreases slowly
✓ 1
Accept a line that looks horizontal below the axis for this mark only if its value is between −2 and −10 m s−2.
Part (e)
after 0.50 s the only horizontal forces are air resistance and friction, which act backwards, so the car decelerates
✓ 1
air resistance decreases as the speed decreases, so the resultant force «and the deceleration» decreases
✓ 1
OWTTE.
Part (f)(i)
smaller, about 2/3 of that of the first car «≈ 60 m s−2»: the same thrust acts on a larger mass «a = F/m» and there is no resistive force at t = 0
✓ 1
Accept 55–65 m s−2 if a value is given.
Part (f)(ii)
smaller: at the same speed the same resistive force acts on a larger mass
✓ 1
OWTTE. Do not accept “the car is heavier so it slows down more”.
Answers: (b) ≈ 1.13 s · (c) ≈ 17.7 m s−1(the remaining parts are explanations — see the table above)
169IBP-A1-N1-07Motion graphs & definitionsPaper 2Medium5 marks
A cheetah starts from rest and accelerates uniformly in a straight line, reaching 28 m s−1 after 4.0 s. It then runs at a constant 28 m s−1 for a further 6.0 s.
(a)
Calculate the acceleration of the cheetah during the first 4.0 s.
(1)
(b)
Determine the distance the cheetah covers in the 10.0 s.
(2)
(c)
Calculate the average speed of the cheetah over the 10.0 s and explain why it is less than 28 m s−1.
(2)
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Notes
Part (a)
a = 28/4.0 = 7.0 m s−2
✓ 1
Part (b)
Area under v–t graph: ½ × 4.0 × 28 = 56 m and 28 × 6.0 = 168 m
✓ 1
Total = 220 m (224 m)
✓ 1
Part (c)
Average speed = 224/10.0 = 22 m s−1 (22.4 m s−1); allow ECF from (b)
✓ 1
During the first 4.0 s the cheetah's speed is below 28 m s−1, so it covers less distance in that time than it would at 28 m s−1
✓ 1
Exam tip: Sketching a quick v–t graph for a two-stage journey turns a distance question into finding two simple areas: a triangle and a rectangle.
170IBP-A1-N1-10Motion graphs & definitionsPaper 2Medium6 marks
A speedboat leaves a harbour and moves in a straight line. Its displacement–time graph is a smooth curve that starts at the origin with zero gradient and passes through (10 s, 20 m) and (20 s, 80 m). A tangent drawn to the curve at t = 20 s passes through the points (10 s, 0 m) and (30 s, 160 m). The acceleration of the boat is uniform.
Not to scale
(a)
Calculate the average velocity of the boat between t = 0 and t = 20 s.
(1)
(b)
Determine the instantaneous velocity of the boat at t = 20 s.
(2)
(c)
Explain, with reference to the graph, why your answer to (b) is greater than your answer to (a).
(1)
(d)
Deduce the acceleration of the boat.
(2)
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Part (a)
80/20 = 4.0 m s−1
✓ 1
Part (b)
Gradient of tangent = (160 − 0)/(30 − 10)
✓ 1
= 8.0 m s−1
✓ 1
Part (c)
The gradient of the curve increases with time, so the boat is speeding up; the average over 0–20 s includes the earlier, slower motion
✓ 1
Part (d)
Zero gradient at the origin means the initial velocity is zero
✓ 1
a = (8.0 − 0)/20 = 0.40 m s−2; allow ECF from (b)
—
ors = ½at2 with u = 0: 80 = ½a(20)2, so a = 0.40 m s−2
✓ 1
Exam tip: For an instantaneous value from a curved s–t graph, use the gradient of the tangent at that instant, taking two points on the tangent that are far apart.
171IBP-A1-N2-06Equations of motionPaper 2Medium4 marks
A space probe, far from any planet or star, moves in a straight line. Its ion engine gives it a uniform acceleration that increases its speed from 3.00 km s−1 to 3.45 km s−1 in 5.0 days. (1 day = 86 400 s)
(a)
Calculate the acceleration of the probe.
(2)
(b)
Calculate the distance travelled by the probe during the 5.0 days.
(2)
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Notes
Part (a)
Δv = 450 m s−1 and t = 5.0 × 86 400 = 4.32 × 105 s
✓ 1
a = 450 / 4.32 × 105 = 1.0 × 10−3 m s−2 (1.04 × 10−3 m s−2)
✓ 1
Part (b)
s = ½(u + v)t = ½(3000 + 3450) × 4.32 × 105
✓ 1
s = 1.4 × 109 m (1.39 × 109 m)
✓ 1
Exam tip: Convert every quantity to SI units (m, s) before substituting. Here that means km s⁻¹ to m s⁻¹ and days to seconds.
172IBP-A1-N2-07Equations of motionPaper 2Medium5 marks
A glider sits on a long, straight air track that is tilted slightly. The glider is pushed so that it moves up the slope at 0.80 m s−1. After the push it has a constant acceleration of 0.25 m s−2 directed down the slope, whichever way it is moving. Friction is negligible. Take up the slope as positive.
(a)
Calculate the time from the end of the push until the glider is momentarily at rest.
(1)
(b)
Calculate the distance the glider moves up the slope before it stops.
(2)
(c)
Determine the velocity of the glider 4.0 s after the end of the push.
(2)
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Notes
Part (a)
t = 0.80 / 0.25 = 3.2 s
✓ 1
Part (b)
0 = 0.802 + 2(−0.25)s
✓ 1
s = 1.3 m (1.28 m)
✓ 1
Part (c)
v = 0.80 + (−0.25)(4.0) = −0.20 m s−1
✓ 1
i.e. 0.20 m s−1 down the slope
✓ 1
Common error: Taking the acceleration as positive because the glider is going up. The acceleration is down the slope throughout, so with up as positive it is −0.25 m s⁻² both before and after the turning point.
173IBP-A1-N2-09Equations of motionPaper 2Medium5 marks
A delivery robot in a warehouse starts from rest and accelerates uniformly at 0.60 m s−2 along a straight aisle for a distance of 3.0 m.
(a)
Calculate the speed of the robot at the end of the 3.0 m.
(2)
(b)
Calculate the time taken to cover the 3.0 m.
(1)
(c)
The robot then decelerates uniformly to rest in a further 2.5 s. Calculate the distance it travels while decelerating.
(2)
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Notes
Part (a)
v2 = 0 + 2 × 0.60 × 3.0 = 3.6
✓ 1
v = 1.9 m s−1 (1.90 m s−1)
✓ 1
Part (b)
t = v/a = 1.897 / 0.60 = 3.2 s (3.16 s)
✓ 1
Part (c)
s = ½(u + v)t = ½ × 1.897 × 2.5
✓ 1
s = 2.4 m (2.37 m)
✓ 1
Exam tip: Carry the unrounded value (1.897 m s⁻¹) into later parts and round only the final answer. Rounding early can shift the last significant figure.
An astronaut on the Moon throws a small hammer vertically upwards with a speed of 4.8 m s−1. The acceleration of free fall on the Moon is 1.6 m s−2 and there is no atmosphere.
(a)
Calculate the time taken for the hammer to reach its highest point.
(2)
(b)
Calculate the maximum height reached above the point of release.
(2)
(c)
The astronaut catches the hammer at the height from which it was thrown. State the total time the hammer is in flight.
(1)
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Notes
Part (a)
use of v = u + at with v = 0 and a = −1.6 m s−2
✓ 1
t = 4.8/1.6 = 3.0 s
✓ 1
Part (b)
s = u2/(2g) ors = ½(u + v)t
✓ 1
s = 4.82/3.2 = 7.2 m
✓ 1
Part (c)
motion is symmetric, so total time = 2 × 3.0 = 6.0 s
✓ 1
Exam tip: Free-fall equations are not tied to Earth. Replace 9.8 with whatever acceleration of free fall the question gives, and keep the sign convention the same.
A bolt falls from rest from the top of a scaffold, 40 m above the ground on a building site. Air resistance is negligible and g = 9.8 m s−2.
(a)
Calculate the time taken for the bolt to reach the ground.
(2)
(b)
Calculate the speed of the bolt just before it hits the ground.
(2)
(c)
Determine the distance the bolt falls during the final second of its fall.
(2)
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Notes
Part (a)
40 = ½ × 9.8 × t2
✓ 1
t = √(80/9.8) = 2.9 s (2.86 s)
✓ 1
Part (b)
v2 = 0 + 2 × 9.8 × 40 orv = 9.8 × 2.86
✓ 1
v = 28 m s−1 (28.0 m s−1)
✓ 1
Part (c)
distance fallen in first (2.86 − 1.00) s = ½ × 9.8 × 1.862 = 16.9 m
✓ 1
distance in final second = 40 − 16.9 = 23 m (23.1 m)
✓ 1
Alternative: s = vt − ½gt2 = 28.0 × 1.0 − 4.9 = 23.1 m
—
Exam tip: Carry unrounded values (such as 2.857 s) into later parts. Rounding the time to 2.9 s before part (c) would give 22 m instead of 23 m for the final second.
176IBP-A1-N5-05Kinematics data & measurementPaper 2Medium5 marks
A wind-up toy car pulls a paper tape through a ticker-timer that makes 50 dots per second. The tape is cut into consecutive sections, each containing five dot-to-dot spaces. The section lengths are: section: 1, 2, 3, 4 length / cm: 1.2, 2.0, 2.8, 3.6
Dots at the ends of each section; not to scale
(a)
State the time interval represented by each section.
(1)
(b)
Calculate the average speed of the car during section 1 and during section 4.
(2)
(c)
Determine the acceleration of the car.
(2)
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Notes
Part (a)
5 × (1/50) s = 0.10 s
✓ 1
Part (b)
section 1: 0.012 / 0.10 = 0.12 m s−1
✓ 1
section 4: 0.036 / 0.10 = 0.36 m s−1
✓ 1
Part (c)
average speeds apply at the mid-times of the sections, which are 3 × 0.10 = 0.30 s apart
✓ 1
a = (0.36 − 0.12) / 0.30 = 0.80 m s−2
✓ 1
Exam tip: An average speed over an interval is best treated as the instantaneous speed at the middle of that interval. Use the time between the mid-points when you find an acceleration.
177IBP-A1-N5-06Kinematics data & measurementPaper 2Medium5 marks
In a factory, a box of length 0.30 m is carried along a straight, speeding-up conveyor belt. It passes through two light gates. The box blocks gate 1 for 0.25 s and gate 2 for 0.15 s. The time between the middles of the two blocked intervals is 1.6 s. Assume the acceleration of the box is uniform.
Diagram not to scale
(a)
Calculate the speed of the box at each gate.
(2)
(b)
Determine the acceleration of the box.
(2)
(c)
Suggest one change to the apparatus that would make each calculated speed closer to an instantaneous speed.
(1)
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Notes
Part (a)
gate 1: 0.30 / 0.25 = 1.2 m s−1
✓ 1
gate 2: 0.30 / 0.15 = 2.0 m s−1
✓ 1
Part (b)
a = (2.0 − 1.2) / 1.6
✓ 1
= 0.50 m s−2
✓ 1
Part (c)
use a shorter object / attach a narrow card to the box to block the beam, so the speed changes less while the beam is blocked
✓ 1
Exam tip: A light gate gives an average speed over the time the beam is blocked. The shorter the blocking object, the closer this is to an instantaneous speed.
178IBP-A1-N5-09Kinematics data & measurementPaper 2Medium6 marks
A GPS collar on a cheetah records its speed as it starts a chase from rest along a straight line: t / s: 0, 1.0, 2.0, 3.0, 4.0, 5.0 v / m s−1: 0, 7.0, 13.0, 18.0, 21.0, 22.0
Data from the table
(a)
Calculate the average acceleration of the cheetah during the first 2.0 s.
(1)
(b)
Describe how the acceleration of the cheetah changes during the 5.0 s. Refer to the data.
(2)
(c)
Estimate the distance travelled by the cheetah in the 5.0 s.
(2)
(d)
Hence determine the average speed of the cheetah over the 5.0 s.
(1)
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Notes
Part (a)
a = 13.0 / 2.0 = 6.5 m s−2
✓ 1
Part (b)
the acceleration decreases (but stays positive)
✓ 1
the speed increase in each successive second gets smaller: 7, 6, 5, 3, 1 m s−1
✓ 1
Part (c)
distance = area under the v–t graph, e.g. by trapezia: 1.0 × (0 + 7.0)/2 + …
✓ 1
= 3.5 + 10 + 15.5 + 19.5 + 21.5 = 70 m
✓ 1
Part (d)
70 / 5.0 = 14 m s−1
✓ 1
Common error: The average speed is total distance ÷ total time. It is not the mean of the first and last speeds unless the acceleration is uniform.
179IBP2S-001-05-14Fluid resistance & terminal speedPaper 2Medium8 marks
A badminton shuttlecock experiences a large air resistance force, which increases with its speed. When dropped, it reaches a terminal speed of 6.8 m s−1 after falling only a short distance.
(a)
Outline why the acceleration of the dropped shuttlecock decreases as it falls.
(2)
(b)
Calculate the time taken for the shuttlecock to fall a further 3.4 m once it is moving at terminal speed.
(1)
(c)
A player hits the shuttlecock vertically upwards at 20 m s−1. State and explain how the magnitude of its acceleration just after being hit compares with g.
(2)
(d)
Taking upwards as positive, describe the velocity–time graph for the shuttlecock's motion from being hit until it falls well below its launch point.
(3)
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Notes
Part (a)
As its speed increases, the air resistance (upwards) increases
✓ 1
The resultant downward force (weight − drag) decreases, so the acceleration decreases
✓ 1
Part (b)
t = 3.4 / 6.8 = 0.50 s
✓ 1
Part (c)
It is greater than g
✓ 1
Both the weight and the air resistance act downwards, opposing the upward motion
✓ 1
Part (d)
Starts at +20 m s−1 with a steep negative gradient, whose magnitude decreases as the speed falls
✓ 1
Crosses v = 0, where the gradient equals −g (−9.8 m s−2)
✓ 1
Becomes negative and levels off, approaching −6.8 m s−1 asymptotically (gradient tending to zero)
✓ 1
Common error: The v–t line does not become horizontal at the top of the flight. Only the velocity is zero there; the gradient is −g.
180IBP2S-001-00-12Projectile motionPaper 2Medium8 marks
During an eruption, a volcano ejects a lava bomb from its vent at 60 m s−1 at 55° above the horizontal. The bomb lands on a slope 40 m below the level of the vent. Air resistance is negligible.
Diagram not to scale
(a)
Calculate the maximum height of the bomb above the vent.
(2)
(b)
Determine the time of flight of the bomb.
(3)
(c)
Calculate the horizontal distance from the vent to the landing point.
(1)
(d)
Calculate the speed of the bomb just before it lands.
(2)
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Part (a)
uy = 60 sin55° = 49.1 m s−1 . h = uy2 / 2g = 49.12 / 19.6 = 123 m .
✓ 2
Part (b)
Taking upwards as positive: −40 = 49.1t − 4.9t2 . Solve the quadratic 4.9t2 − 49.1t − 40 = 0, taking the positive root . t = 10.8 s .
✓ 3
Part (c)
x = (60 cos55°)(10.8) = 34.4 × 10.8 ≈ 370 m; ECF from (b) .
✓ 1
Part (d)
Horizontal component is unchanged at 34.4 m s−1; vy2 = 49.12 + 2 × 9.8 × 40, giving vy = 56.6 m s−1 . v = √(34.42 + 56.62) = 66 m s−1 (allow √(602 + 2 × 9.8 × 40)) .
✓ 2
Common error: When the landing point is below the launch point, the vertical displacement is negative. Using +40 m gives a wrong quadratic.
181IBP2S-001-00-13Kinematics data & measurementPaper 2Medium8 marks
A rover on Mars tests a spring-loaded sampler. The sampler fires a small capsule vertically upwards at 24 m s−1. The capsule returns to its launch point 13.0 s later. The Martian atmosphere has a negligible effect on the capsule.
(a)
Show that the gravitational acceleration on Mars is about 3.7 m s−2.
(2)
(b)
Calculate the maximum height reached by the capsule.
(2)
(c)
The sampler is tilted to fire at 40° above the horizontal, at the same speed, across level ground. Determine the horizontal range of the capsule.
(3)
(d)
Outline why neglecting air resistance is reasonable on Mars but would not be for the same capsule on Earth.
(1)
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Notes
Part (a)
The displacement is zero after 13.0 s: 0 = 24 × 13.0 − ½g(13.0)2, or the time to the top is 6.5 s with v = 0 . g = 48 / 13.0 = 3.69 m s−2 ≈ 3.7 m s−2 .
✓ 2
Part (b)
h = u2 / 2g = 242 / (2 × 3.69) . h = 78 m .
✓ 2
Part (c)
Time of flight = 2 × 24 sin40° / 3.69 = 8.36 s . Horizontal component = 24 cos40° = 18.4 m s−1 . Range = 18.4 × 8.36 ≈ 154 m (accept 150–156 m); ECF from (a) .
✓ 3
Part (d)
The Martian atmosphere has a very low density, so the drag force is much smaller than on Earth .
✓ 1
Exam tip: In a 'show that' question, give the answer to one more significant figure than the value quoted, here 3.69 m s⁻².
182IBP2S-001-00-14Motion graphs & definitionsPaper 2Medium8 marks
In a drop-tower ride, a car is released from rest and falls freely for 2.8 s. Magnetic brakes then slow it with uniform deceleration, bringing it to rest after a further 22 m of vertical travel. Air resistance is negligible.
(a)
Calculate the speed of the car at the end of the free fall.
(1)
(b)
Calculate the distance fallen during the free fall.
(1)
(c)
Determine the magnitude of the deceleration during braking, as a multiple of g.
(2)
(d)
Calculate the total time from release to rest.
(2)
(e)
Taking downwards as positive, describe the shape of the velocity–time graph for the whole ride. Refer to the gradients of the graph.
(2)
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Notes
Part (a)
v = gt = 9.8 × 2.8 = 27 m s−1 (27.4 m s−1) .
✓ 1
Part (b)
s = ½gt2 = ½ × 9.8 × 2.82 = 38 m .
✓ 1
Part (c)
a = v2 / 2s = 27.42 / (2 × 22) = 17 m s−2; allow ECF from (a) . This is about 1.7g (accept 1.7–1.8g) .
✓ 2
Part (d)
Braking time = 2s / v = 44 / 27.4 = 1.6 s, or v / a; allow ECF . Total time = 2.8 + 1.6 = 4.4 s .
✓ 2
Part (e)
From the origin, a straight line with constant positive gradient of 9.8 m s−2, reaching 27 m s−1 at 2.8 s . Then a straight line with a steeper, negative gradient of about −17 m s−2, reaching zero velocity at 4.4 s .
✓ 2
Exam tip: Using s = (u + v)t / 2 gives the braking time directly, without first finding the deceleration.
183IBP2S-001-04-12Fluid resistance & terminal speedPaper 2Medium8 marks
A quality-control laboratory tests engine oil by timing a small steel sphere as it sinks through a tall cylinder of the oil. The sphere is released from rest at the oil surface. A video analysis gives the distance x fallen at time t:
t / s: 0.00, 0.10, 0.20, 0.30, 0.40, 0.50, 0.60 x / cm: 0.0, 1.2, 3.6, 6.4, 9.3, 12.2, 15.1
Data from the table
(a)
Outline how the data show that the sphere reaches terminal speed.
(2)
(b)
Determine the terminal speed of the sphere.
(2)
(c)
The oil is warmed, which reduces its viscosity. Explain the effect on the terminal speed of the sphere.
(2)
(d)
The original oil is used again. Calculate the total time for the sphere to fall 0.50 m from release.
(2)
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Notes
Part (a)
From t = 0.30 s onwards the sphere falls the same distance (2.9 cm) in each 0.10 s interval
✓ 1
Equal displacements in equal times mean the velocity is constant, so the acceleration is zero
✓ 1
Part (b)
v = (15.1 − 6.4) cm / (0.60 − 0.30) s, or any valid interval after 0.30 s
✓ 1
v = 0.29 m s−1
✓ 1
Part (c)
Lower viscosity means less drag at any given speed
✓ 1
The drag must reach a higher speed before it balances the weight minus the upthrust, so the terminal speed increases
✓ 1
Part (d)
Remaining distance after 0.60 s is 0.50 − 0.151 = 0.349 m, covered at 0.29 m s−1 in 1.2 s; ECF from (b)
✓ 1
Total time = 0.60 + 1.2 = 1.8 s
✓ 1
Exam tip: Find the terminal speed only from the region of constant gradient. Including the early, accelerating readings gives a value that is too low.
184IBP2S-001-04-13Fluid resistance & terminal speedPaper 2Medium9 marks
A trap machine launches a clay target from ground level at 24 m s−1, at 35° above the horizontal, over level ground. At first, treat the target as a projectile with negligible air resistance.
(a)
Calculate the maximum height reached by the target.
(2)
(b)
Calculate the horizontal range of the target.
(2)
(c)
In reality, air resistance acts on the target. Explain why the horizontal component of its velocity now decreases, whereas it was constant in the model.
(2)
(d)
Describe how the actual path of the target differs from the path in the model.
(3)
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Notes
Part (a)
Vertical component uy = 24 sin 35° = 13.8 m s−1
✓ 1
h = uy2/(2g) = 13.82/(2 × 9.8) = 9.7 m
✓ 1
Part (b)
Time of flight t = 2uy/g = 2 × 13.8/9.8 = 2.81 s, using the vertical component found for the maximum height; OR substitution into R = u2 sin 2θ/g = 242 sin 70°/9.8
✓ 1
Range = 24 cos 35° × 2.81 = 19.7 × 2.81 = 55 m (55.2 m)
✓ 1
Part (c)
In the model there is no horizontal force, so the horizontal acceleration is zero and the horizontal velocity is constant
✓ 1
Drag acts opposite to the velocity, so it has a horizontal component that opposes the motion and causes a horizontal deceleration
✓ 1
Part (d)
The maximum height is lower
✓ 1
The range is shorter
✓ 1
The path is asymmetric: the descent is steeper than the ascent, so the highest point lies more than halfway along the horizontal range
✓ 1
Common error: A wrong range usually comes from using sin 35° for the horizontal component, or from forgetting to double the time taken to reach the top.
185IBP2S-001-04-15Non-uniform accelerationPaper 2Medium10 marks
A drag-racing car of total mass 1200 kg crosses the finish line at 120 m s−1. It then deploys a braking parachute. The drag force from the parachute is F = kv2, where k = 0.80 kg m−1. Ignore all other resistive forces.
(a)
Calculate the deceleration of the car immediately after the parachute opens.
(2)
(b)
Explain why the deceleration of the car decreases as it slows down.
(2)
(c)
Describe the shape of the velocity–time graph for the car from the moment the parachute opens.
(2)
(d)
A student estimates the distance needed to slow from 120 m s−1 to 60 m s−1 using the equations for uniform acceleration, with the deceleration from (a). Calculate this estimate.
(2)
(e)
State and explain whether the actual distance is greater or less than the estimate in (d).
(2)
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Notes
Part (a)
F = 0.80 × 1202 = 1.15 × 104 N
✓ 1
a = F/m = 11520/1200 = 9.6 m s−2
✓ 1
Part (b)
The drag force depends on v2, so it falls as the speed falls
✓ 1
The mass is constant, so a smaller resultant force gives a smaller deceleration, since a = F/m
✓ 1
Part (c)
The curve starts at 120 m s−1 and the velocity decreases continuously
✓ 1
The gradient is steepest at the start and its magnitude decreases, so the curve flattens out and approaches the time axis gradually, with no sudden stop
✓ 1
Part (d)
s = (u2 − v2)/(2a) = (1202 − 602)/(2 × 9.6), allow ECF from (a)
✓ 1
s = 563 m, accept 560 m
✓ 1
Part (e)
The actual distance is greater
✓ 1
After the first instant the real deceleration is always less than 9.6 m s−2, so a larger distance is needed to lose the same speed. For example, a smaller a in u2 − v2 = 2as means a larger s
✓ 1
Exam tip: The equations of uniform acceleration apply only when the acceleration is constant. When the drag depends on speed, they give only an estimate, and you should say whether that estimate is too high or too low.
187IBP2S-001-03-13Velocity & accelerationPaper 2Medium7 marks
A tennis ball is released from rest 1.8 m above a hard floor. It rebounds with a speed equal to 70% of its impact speed. Air resistance is negligible and the contact time with the floor is very short.
(a)
Calculate the speed of the ball just before it hits the floor.
(2)
(b)
Taking downwards as positive, describe the velocity–time graph of the ball's motion from release until it reaches the top of its first bounce. Include key values of velocity and time.
(3)
(c)
Determine the maximum height reached by the ball after the bounce.
(2)
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Notes
Part (a)
v2 = 2gh = 2 × 9.8 × 1.8
✓ 1
v = 5.9 m s−1
✓ 1
Part (b)
A straight line from the origin with positive gradient 9.8 m s−2, reaching 5.9 m s−1 at about 0.61 s
✓ 1
An almost vertical drop to −4.2 m s−1 at impact
✓ 1
A straight line with the same positive gradient (parallel to the first), rising from −4.2 m s−1 to zero at about 1.0 s
✓ 1
Part (c)
Rebound speed = 0.70 × 5.94 = 4.16 m s−1, then h = v2/2g; or h = 0.702 × 1.8; or the area under the second line of the graph (allow ECF from part 1)
✓ 1
h = 0.88 m (accept 0.87–0.88 m)
✓ 1
Exam tip: Both straight sections of the graph have the same gradient, because the acceleration is g both before and after the bounce. Because height is proportional to speed squared, a 70% speed gives 0.70² = 49% of the drop height.
189IBP2S-001-02-12Kinematics data & measurementPaper 2Medium5 marks
A track cyclist is timed as she passes posts placed every 20.0 m along a straight section of a velodrome. The uncertainty in each time reading is ±0.1 s. The uncertainty in the post positions is negligible.
Position / m
0.0
20.0
40.0
60.0
80.0
Time / s
0.0
4.0
7.2
10.0
12.6
(a)
Calculate the average speed of the cyclist over the whole 80.0 m.
(1)
(b)
Show, using the data, that the cyclist is accelerating.
(2)
(c)
Determine the absolute uncertainty in the average speed for the section from 60.0 m to 80.0 m.
(2)
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Notes
Part (a)
80.0 ÷ 12.6 = 6.3 m s−1
✓ 1
Part (b)
Time for each 20.0 m section: 4.0, 3.2, 2.8 and 2.6 s, or section average speeds 5.0, 6.3, 7.1 and 7.7 m s−1 (at least two correct values)
✓ 1
Equal distances are covered in shorter times, so the average speed rises (overall from 5.0 to 7.7 m s−1) and she is accelerating
✓ 1
Part (c)
Time interval = 12.6 − 10.0 = 2.6 s, with uncertainty ±0.2 s because the uncertainties of the two readings add
✓ 1
Fractional uncertainty = 0.2/2.6 ≈ 0.077. v = 20.0/2.6 = 7.7 m s−1, so Δv ≈ 0.077 × 7.7 ≈ 0.6 m s−1. Also accept a max/min method: 20/2.4 = 8.3 and 20/2.8 = 7.1, so the half-range is ≈ 0.6 m s−1
✓ 1
Common error: A time interval is the difference of two readings. Its absolute uncertainty is the sum of both reading uncertainties, not just one of them.
190IBP2S-001-02-14Kinematics data & measurementPaper 2Medium7 marks
Engineers use an evacuated drop tower to create short periods of free fall for experiments. An experiment capsule is released from rest. It falls a height h = (120.0 ± 0.5) m in a time t = (4.95 ± 0.02) s. Air resistance is negligible.
(a)
Show that these data give a value for the acceleration of free fall of about 9.8 m s−2.
(2)
(b)
Determine the absolute uncertainty in the value of g.
(3)
(c)
Calculate the average speed of the capsule during the fall, and explain why its speed just before impact is twice this value.
191IBP2S-001-02-15Projectile motionPaper 2Medium7 marks
A tennis ball machine fires balls horizontally from a height of 2.45 m above level ground. A coach measures the horizontal range R for five balls fired at the same machine setting.
Trial
1
2
3
4
5
R / m
14.2
14.6
14.0
14.4
14.3
Assume air resistance is negligible unless stated otherwise.Diagram not to scale
(a)
Determine the mean range and its absolute uncertainty.
(2)
(b)
Calculate the time of flight of a ball.
(1)
(c)
Determine the launch speed of the balls, with its absolute uncertainty. Assume the uncertainty in the height is negligible.
(2)
(d)
In reality, air resistance acts on the balls. Explain why the true launch speed is likely to be greater than the value found in (c).
(2)
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Notes
Part (a)
Mean = 71.5/5 = 14.3 m
✓ 1
Uncertainty = half the range of the data = (14.6 − 14.0)/2 = ±0.3 m
✓ 1
Part (b)
t = √(2 × 2.45/9.8) = 0.71 s
✓ 1
Part (c)
u = 14.3/0.707 = 20.2 m s−1
✓ 1
Percentage uncertainty = 0.3/14.3 ≈ 2 %, so u = (20.2 ± 0.4) m s−1; ECF from (a) and (b)
✓ 1
Part (d)
Air resistance decreases the horizontal velocity throughout the flight, so the average horizontal speed is less than the launch speed
✓ 1
The vertical speeds are small, so the time of flight changes very little; R/t therefore gives the (lower) average horizontal speed, which underestimates the launch speed
✓ 1
Exam tip: For a horizontal launch, the time of flight depends only on the height. Calculate it first from the vertical motion.
192IBP2S-001-01-14Equations of motionPaper 2Medium7 marks
On a drop-tower fairground ride, the passenger car is released from rest and falls freely for 2.5 s. Magnetic brakes then bring it uniformly to rest over a further 18 m. Air resistance is negligible.
(a)
Calculate the speed of the car when the brakes are applied.
(1)
(b)
Calculate the distance fallen during free fall.
(2)
(c)
Calculate the magnitude of the deceleration during braking.
(2)
(d)
Determine the total time from release until the car comes to rest.
(2)
Show mark scheme
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Notes
Part (a)
v = gt = 9.8 × 2.5 = 24.5 m s−1
✓ 1
Part (b)
s = ½gt2 = ½ × 9.8 × 2.52 ; s = 30.6 m
✓ 2
Part (c)
a = v2/(2s) = 24.52/(2 × 18) ; a = 17 m s−2 (16.7 m s−2) . ECF from (a)
✓ 2
Part (d)
braking time = 2s/(u + v) = 2 × 18/24.5 = 1.47 s ; total time = 2.5 + 1.47 = 4.0 s (3.97 s) . ECF from (a)
✓ 2
Exam tip: The final speed of the first stage is the initial speed of the second stage. Carry it forward unrounded to avoid rounding errors.
193IBP2S-001-01-15Kinematics data & measurementPaper 2Medium6 marks
In a cathode-ray device, an electron enters a uniform electric field at 2.0×105 m s−1. It is accelerated in a straight line and leaves the field 0.040 m further on at 6.0×106 m s−1.
(a)
State one assumption needed to apply the equations of uniformly accelerated motion to the electron.
(1)
(b)
Calculate the acceleration of the electron.
(2)
(c)
Outline why the effect of gravity on the electron can be ignored.
(1)
(d)
Calculate the time the electron spends in the field.
(2)
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Notes
Part (a)
The acceleration (the force on the electron) is constant throughout the field . Also accept: the speed is small enough for non-relativistic mechanics to apply.
✓ 1
Part (b)
a = (v2 − u2)/(2s) = ((6.0×106)2 − (2.0×105)2)/(2 × 0.040) ; a = 4.5×1014 m s−2
✓ 2
Part (c)
g (about 10 m s−2) is smaller than the electrical acceleration by a factor of about 5×1013, so its effect is negligible . Accept any valid comparison showing g is many orders of magnitude smaller. ECF from (b)
✓ 1
Part (d)
t = 2s/(u + v) = 2 × 0.040/(6.2×106) ; t = 1.3×10−8 s . Allow t = (v − u)/a using the value from (b): ECF from (b)
✓ 2
Common error: In the calculator, enter (6.0×10⁶)² with brackets. Squaring only the 10⁶ gives an answer that is wrong by a large factor.
A service lift (elevator) in an office building moves upwards at a constant 2.0 m s−1. A loose bolt detaches from the lift ceiling, 2.5 m above the lift floor. Air resistance is negligible.
(a)
Explain why the time taken for the bolt to reach the floor is the same as it would be if the lift were stationary.
(2)
(b)
Calculate the time taken for the bolt to reach the floor.
(2)
(c)
Determine the velocity of the bolt relative to the building just before it hits the floor.
(1)
(d)
Determine the displacement of the bolt relative to the building between leaving the ceiling and hitting the floor.
(2)
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Notes
Part (a)
At the moment it detaches, the bolt has the same velocity as the lift, so its velocity relative to the floor is zero ; the floor moves at constant velocity, so relative to the floor the bolt accelerates at g, exactly as in a stationary lift
✓ 2
Part (b)
2.5 = ½ × 9.8 × t2 ; t = 0.71 s (0.714 s)
✓ 2
Part (c)
v = 2.0 − 9.8 × 0.714 = −5.0 m s−1, i.e. 5.0 m s−1 downwards . ECF from (b)
✓ 1
Part (d)
s = ut + ½at2 = 2.0 × 0.714 − ½ × 9.8 × 0.7142 ; s = −1.1 m, i.e. 1.1 m downwards (1.07 m) . ECF from (b)
✓ 2
Exam tip: State which reference frame you are using. The time is easiest to find relative to the lift, while the displacement and velocity relative to the building need u = +2.0 m s⁻¹.
In a relay race the incoming runner X runs at a constant speed of 9.0 m s−1. The outgoing runner Y waits at the start of a 20 m exchange zone. Y starts from rest at the instant that X is a distance d behind Y, and accelerates uniformly at 2.5 m s−2 until reaching 9.0 m s−1. Y then runs at this constant speed.
The baton is passed at the instant X draws level with Y. Both runners move along the same straight line.
Figure 1 — axes for (d)
(a)
Calculate the time taken for Y to reach a speed of 9.0 m s−1.
(1)
(b)
Show that, during this time, X runs about 16 m further than Y.
(2)
(c)
The coach sets d = 12 m. Determine the distance from the start of the exchange zone at which the baton is passed.
(3)
(d)
Sketch, on Figure 1, velocity–time graphs for X and for Y from t = 0 to t = 5.0 s when d = 12 m. Ignore any change in motion caused by the exchange. Label each graph.
(3)
(e)
(i)
Explain why, if X has not drawn level with Y by t = 3.6 s, X will never draw level with Y.
(2)
(ii)
The coach considers increasing d to 18 m. Determine whether the baton can then be passed.
(2)
(f)
Suggest one advantage of choosing d = 15 m rather than d = 12 m.
(1)
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Notes
Part (a)
t = 9.0 / 2.5 = 3.6 s
✓ 1
Part (b)
X: 9.0 × 3.6 = 32.4 m AND Y: ½ × 2.5 × 3.6² = 16.2 m
✓ 1
Must see both displacements.
difference = 32.4 − 16.2 «= 16.2 m»
✓ 1
Accept the area between the two velocity–time graphs, ½ × 3.6 × 9.0. Must see full substitution OR answer to 3 s.f.
Part (c)
draws level when 9.0t = 12 + ½ × 2.5t²
✓ 1
Accept 1.25t² − 9.0t + 12 = 0.
t = 1.77 s «smaller root»
✓ 1
The larger root is later than 3.6 s, after Y stops accelerating, so it is rejected; X has already drawn level.
distance = ½ × 2.5 × 1.77² = 3.9 m
✓ 1
Accept 3.8–4.0 m. Award [3] for CNA.
Part (d)
X: horizontal line at 9.0 m s−1 from 0 to 5.0 s
✓ 1
Y: straight line from the origin reaching 9.0 m s−1 at 3.6 s
✓ 1
Y: horizontal at 9.0 m s−1 from 3.6 s, coinciding with X’s line; both graphs labelled
✓ 1
Allow ECF from (a) for the time.
Part (e)(i)
before this time X is faster than Y, so the gap closes «by the area between the graphs»
✓ 1
after 3.6 s the speeds are equal so the gap stays constant «and is never closed»
✓ 1
OWTTE.
Part (e)(ii)
the greatest distance X can gain is 16.2 m
✓ 1
Allow ECF from (b).
16.2 m < 18 m so X never draws level AND the baton cannot be passed
✓ 1
ALTERNATIVE: 1.25t² − 9.0t + 18 = 0 has no real solution «81 − 90 < 0» so X never draws level.
Part (f)
the exchange happens later when Y is running faster «6.6 m s−1 instead of 4.4 m s−1», so less time is lost passing the baton
✓ 1
Accept “still within the zone «at 8.6 m» but closer to X’s speed”. OWTTE.
Answers: (a) 3.6 s · (b) 16.2 m · (c) 3.9 m (the remaining parts are explanations — see the table above)
196A-2-15Non-uniform accelerationPaper 2Hard12 marks
A maglev test vehicle of mass 2.4 × 104 kg starts from rest at t = 0 and moves along a straight, horizontal track. Figure 1 shows how its acceleration a varies with time t for the first 30 s.
Figure 1 — graph drawn to scaleFigure 2 — axes for (b)
(a)
(i)
Calculate the speed of the vehicle at t = 8.0 s.
(1)
(ii)
Deduce the maximum speed of the vehicle.
(2)
(b)
Sketch, on Figure 2, the variation with time of the velocity of the vehicle from t = 0 to t = 30 s.
(3)
(c)
Explain why the speed of the vehicle continues to increase until t = 20 s even though its acceleration decreases from t = 8.0 s onwards.
(2)
(d)
Calculate the resultant force on the vehicle at t = 14 s.
(2)
(e)
Outline why the equation s = ut + ½at² cannot be used to find the distance travelled between t = 8.0 s and t = 20 s, and suggest how this distance could be found.
(2)
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Notes
Part (a)(i)
area = 1.5 × 8.0 = 12 m s−1
✓ 1
Part (a)(ii)
area of the triangle from 8.0 s to 20 s = ½ × 12 × 1.5 = 9.0 m s−1
✓ 1
maximum speed = 12 + 9.0 = 21 m s−1
✓ 1
Allow ECF from (a)(i). Award [2] for CNA.
Part (b)
straight line from the origin to 12 m s−1 at 8.0 s
✓ 1
Allow ECF from (a).
curve from 8.0 s to 20 s with decreasing gradient, reaching 21 m s−1 at 20 s
✓ 1
horizontal line at 21 m s−1 from 20 s to 30 s, joined smoothly «zero gradient at 20 s»
✓ 1
Do not award MP3 if the curve has a maximum before 20 s or falls after 20 s.
Part (c)
the acceleration is positive «in the direction of motion» until 20 s, so the speed keeps increasing
✓ 1
a decreasing acceleration means the speed increases at a smaller rate; after 20 s a = 0 so the speed is constant
✓ 1
OWTTE.
Part (d)
a = 0.75 m s−2 «read from the graph»
✓ 1
F = 2.4 × 104 × 0.75 = 1.8 × 104 N
✓ 1
Allow ECF for a.
Part (e)
the equation applies only to uniform acceleration, and a changes during this interval
✓ 1
the distance is the area under the velocity–time graph «e.g. by counting squares»
✓ 1
Answers: (a)(i) 12 m s−1 · (a)(ii) 21 m s−1 · (d) 1.8 × 104 N (the remaining parts are explanations — see the table above)
A rescue helicopter rises vertically at a constant speed of 4.0 m s−1. When the helicopter is 45 m above the surface of the sea, a crew member releases a smoke marker from rest relative to the helicopter. The marker falls into the sea. Air resistance is negligible unless stated otherwise. Take upwards as positive.
Figure 1 — axes for (d)
(a)
(i)
State the magnitude and direction of the velocity of the marker, relative to the sea, immediately after it is released.
(1)
(ii)
Calculate the maximum height of the marker above the sea.
(2)
(b)
Show that the marker reaches the sea about 3.5 s after it is released.
(1)
(c)
Determine the height of the helicopter above the sea at the instant the marker reaches the sea.
(2)
(d)
Sketch, on Figure 1, the velocity–time graph of the marker from release until it reaches the sea. Add values to your graph where appropriate.
(3)
(e)
(i)
Show that, at a time t after release, the vertical distance between the helicopter and the marker is ½gt², whatever the upward speed of the helicopter.
(2)
(ii)
Hence describe the motion of the marker as seen by the crew member in the helicopter.
(1)
(f)
In practice air resistance acts on the marker. State and explain whether the time taken for the marker to reach the sea is more or less than your answer to (b).
(2)
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Notes
Part (a)(i)
4.0 m s−1 upwards
✓ 1
Direction required. Do not accept zero «the marker shares the velocity of the helicopter».
Part (a)(ii)
height gained after release = u²/2g = 4.0²/(2 × 9.81) = 0.82 m
✓ 1
maximum height = 45 + 0.82 = 45.8 m
✓ 1
Award [1] max for 0.82 m alone. Award [2] for CNA.
Part (b)
−45 = 4.0t − ½ × 9.81 × t² leading to t = 3.46 «s»
✓ 1
Must see full substitution OR answer to 3 s.f. Accept a check by substitution of t = 3.5 s giving s ≈ −46 m. Award [0] for s = +45 m or u = 0 used.
Part (c)
the helicopter rises 4.0 × 3.46 = 13.9 m while the marker is in the air
✓ 1
Allow ECF from (b).
height = 45 + 13.9 = 58.9 m
✓ 1
Accept 58.8–59.0 m. Award [2] for CNA.
Part (d)
straight line with negative gradient starting at +4.0 m s−1 at t = 0
✓ 1
line crosses the t-axis at t ≈ 0.41 s
✓ 1
Accept 0.4 s.
line ends at t ≈ 3.5 s with v ≈ −30 m s−1
✓ 1
Allow ECF from (b). Accept −29 to −31 m s−1. Do not award if the line is drawn beyond the splash-down time.
Part (e)(i)
height of helicopter above the release point = ut; height of marker above the release point = ut − ½gt²
✓ 1
Both expressions needed.
difference = ut − (ut − ½gt²) = ½gt², so u cancels
✓ 1
Part (e)(ii)
the marker accelerates downwards from rest at g, as if it had been dropped from a hovering helicopter
✓ 1
OWTTE. Do not accept “it falls” without reference to starting from rest or uniform acceleration.
Part (f)
air resistance opposes the motion, so during the fall the downward acceleration is less than g
✓ 1
OWTTE.
the fall makes up most of the time, so the marker takes longer «than 3.46 s» to reach the sea
✓ 1
Accept the argument that the marker is slower at every point of the descent. MP2 only scores if MP1 scores.
Answers: (a)(i) 4.0 m s−1 upwards · (a)(ii) 45.8 m · (b) 3.46 s · (c) 58.9 m · (d) v = 0 at 0.41 s; v = −30 m s−1 at 3.46 s (the remaining parts are explanations — see the table above)
An archerfish catches insects by knocking them off overhanging leaves with a jet of water. The fish's mouth is at the water surface, at point O. The water leaves O at a speed of 3.8 m s−1 at an angle of 62° above the horizontal. Each small mass of water in the jet may be treated as a projectile, and air resistance is negligible.
An insect sits on a leaf 0.30 m above the water surface. Refractive index of water = 1.33.
Diagram not to scale
(a)
(i)
Show that the vertical component of the initial velocity of the water is about 3.4 m s−1.
(1)
(ii)
Calculate the horizontal component of the initial velocity of the water.
(1)
(b)
(i)
Show that the water first reaches a height of 0.30 m above the surface about 0.11 s after it leaves O.
(2)
(ii)
Determine the later time at which the water is again 0.30 m above the surface, and explain why there are two such times.
(2)
(iii)
The fish hits the insect with water that is still rising. Calculate the horizontal distance between O and the insect.
(1)
(c)
Calculate the speed of the water as it hits the insect.
(2)
(d)
Suggest one advantage, for the fish, of hitting the insect with water that is still rising rather than with water that is falling.
(1)
(e)
The eye of the fish is just below O. The fish sees the insect by light that travels from the insect to O and then enters the water.
(i)
Calculate the angle between the vertical and the path of this light in the water.
(2)
(ii)
Explain why the fish does not see the insect in its true direction.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
uy = 3.8 sin 62° «= 3.36 m s−1»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (a)(ii)
ux = 3.8 cos 62° = 1.78 m s−1
✓ 1
Accept 1.8 m s−1.
Part (b)(i)
0.30 = 3.36t − ½ × 9.81 × t2 OR 4.905t2 − 3.36t + 0.30 = 0
✓ 1
Signs must be correct.
t = 0.106 «s» «smaller root»
✓ 1
Must see full substitution OR answer to 3 s.f. Accept a check by substitution: t = 0.106 s gives y = 0.300 m.
Part (b)(ii)
t = 0.578 s «larger root, OR time of flight 2 × 3.36/9.81 = 0.684 s minus 0.106 s by symmetry»
✓ 1
Accept 0.57–0.58 s.
the water passes this height once on the way up and once on the way down
✓ 1
OWTTE.
Part (b)(iii)
x = 1.78 × 0.106 = 0.19 m
✓ 1
Accept 0.188–0.190 m. Allow ECF from (a)(ii) and (b)(i).
the water arrives much sooner «0.11 s rather than 0.58 s», so the insect has less time to move away OR the water pushes the insect upwards, off the leaf
✓ 1
Do not accept “the water is moving faster”: at the same height the speed is the same on the way up and on the way down.
Part (e)(i)
angle of incidence in air: tan−1(0.19/0.30) = 32°
✓ 1
Allow ECF from (b)(iii).
sin r = sin 32°/1.33, so r = 24°
✓ 1
Accept 23–24°. Award [2] for CNA.
Part (e)(ii)
the light refracts towards the normal as it enters the water, so the insect appears in a direction closer to the vertical «higher» than it really is
✓ 1
OWTTE.
Answers: (a)(ii) 1.78 m s−1 · (b)(ii) 0.578 s · (b)(iii) 0.19 m · (c) 2.92 m s−1 · (e)(i) 24° (the remaining parts are explanations — see the table above)
When a spacecraft lands on the Moon, the exhaust from its rocket engine blasts rock fragments and dust grains away from the landing site. The Moon has no atmosphere.
Mass of the Moon = 7.35 × 1022 kg; radius of the Moon = 1.74 × 106 m. Treat the surface near the landing site as flat and the gravitational field there as uniform.
Diagram not to scale
(a)
Show that the gravitational field strength at the surface of the Moon is about 1.6 N kg−1.
(1)
(b)
A rock fragment leaves the surface at 45 m s−1 at 20° above the horizontal.
(i)
Calculate the time for which the fragment is in flight.
(2)
(ii)
Calculate the horizontal distance from its launch point at which the fragment lands.
(1)
(c)
Fragments leave the surface with speeds of up to 45 m s−1 in all directions above the horizontal. Deduce the shortest distance from the landing site at which equipment is safe from fragments that travel directly to it.
(3)
(d)
State and explain the factor by which the time of flight and the range in (b) would be smaller for the same launch on an Earth without air.
(2)
(e)
On the real Earth, a fine dust grain blasted out at 45 m s−1 lands within a few metres. Explain why.
(2)
(f)
Explain why, on the Moon, a dust grain and a rock fragment that leave the same point with the same velocity land at the same place.
(1)
Show mark scheme
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Mark
Notes
Part (a)
g = GM/r2 = 6.67 × 10−11 × 7.35 × 1022/(1.74 × 106)2 «= 1.62 N kg−1»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)(i)
uy = 45 sin 20° = 15.4 m s−1
✓ 1
t = 2 × 15.4/1.62 = 19.0 s
✓ 1
Allow use of 1.6 N kg−1 «19.2 s». Award [2] for CNA.
Part (b)(ii)
45 cos 20° × 19.0 = 803 m
✓ 1
Accept 790–815 m. Allow ECF from (b)(i).
Part (c)
range = u cos θ × 2u sin θ/g = 2u2 sin θ cos θ/g «= u2 sin 2θ/g»
✓ 1
this is greatest when sin 2θ = 1, θ = 45°
✓ 1
Accept a comparison of ranges at several angles that identifies 45°.
Rmax = 452/1.62 = 1.25 × 103 m
✓ 1
Accept 1.2–1.3 km. Award [1 max] for 803 m.
Part (d)
for a fixed launch velocity the time of flight 2uy/g is proportional to 1/g, so it is smaller by 9.81/1.62 = 6.1
✓ 1
Accept 6.
the horizontal velocity is the same, so the range is also smaller by a factor of 6.1
✓ 1
OWTTE.
Part (e)
the air resistance on a very small grain is large compared with its weight «it has a very small terminal speed»
✓ 1
so its velocity, including the horizontal component, is rapidly reduced and it then falls slowly, almost vertically
✓ 1
OWTTE.
Part (f)
the only force is the weight, so both have the same acceleration g «independent of mass» and follow the same path
✓ 1
Do not accept “no air resistance” alone.
Answers: (b)(i) 19.0 s · (b)(ii) 803 m · (c) 1.25 × 103 m · (d) 6.1 (the remaining parts are explanations — see the table above)
200A-2-65Non-uniform accelerationPaper 2Hard13 marks
In a safety test, a cycle helmet is fitted on a metal model of a head (a headform). The headform is released from rest so that the helmet falls 1.50 m onto a flat steel anvil. An accelerometer inside the headform records its acceleration a from the moment the helmet touches the anvil (t = 0). Figure 1 shows a simplified version of the record. Upwards is positive. Air resistance is negligible.
Figure 1 — graph drawn to scale
(a)
Show that the helmet reaches the anvil at a speed of about 5.4 m s−1.
(1)
(b)
(i)
State what is represented by the area under an acceleration–time graph.
(1)
(ii)
Determine the change in velocity of the headform during the impact.
(2)
(iii)
Hence determine the speed with which the headform rebounds from the anvil.
(2)
(c)
Calculate the average acceleration of the headform during the impact.
(1)
(d)
To pass the test, the greatest acceleration of the headform must not be more than 250g. Determine whether this helmet passes.
(2)
(e)
Explain how the crushable foam liner of the helmet reduces the greatest acceleration of the head.
(2)
(f)
The test is repeated on the same helmet, whose liner was crushed in the first impact. The change in velocity during the impact is the same. Sketch, on Figure 1, a possible graph for this second impact.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
v = √(2 × 9.81 × 1.50) «= 5.42 m s−1»
✓ 1
Must see full substitution OR answer to 3 s.f.
Part (b)(i)
the change in velocity
✓ 1
Do not accept “velocity” or “speed”.
Part (b)(ii)
area of the triangle = ½ × 7.0 × 10−3 × 1900
✓ 1
Accept counting of squares.
Δv = 6.65 m s−1 «upwards»
✓ 1
Accept 6.4–6.9 m s−1.
Part (b)(iii)
Δv = v − u with u = −5.42 m s−1, so v = −5.42 + 6.65
✓ 1
Allow ECF from (a) and (b)(ii).
v = 1.2 m s−1 «upwards»
✓ 1
Award [1 max] for 6.65 m s−1 or 12.1 m s−1.
Part (c)
6.65/7.0 × 10−3 = 950 m s−2 «upwards»
✓ 1
Allow ECF from (b)(ii). Accept 9.5 × 102 m s−2.
Part (d)
greatest acceleration = 1900 m s−2 = 1900/9.81 = 194g
✓ 1
Accept 190–195g. Do not accept use of the average acceleration from (c).
194g < 250g, so the helmet passes
✓ 1
The decision must be consistent with the candidate's value.
Part (e)
as the foam crushes, the head is brought to rest «and reversed» over a longer distance and a longer time
✓ 1
the same change in velocity in a longer time means a smaller «average and greatest» acceleration
✓ 1
OWTTE. MP2 only scores if MP1 scores.
Part (f)
a pulse of shorter duration than 7.0 ms with a higher peak than 1900 m s−2
✓ 1
the area under the new graph is about the same as under the original graph «eg 3.5 ms and 3800 m s−2»
✓ 1
Judge by eye; the area must be clearly not smaller by more than about a quarter.
Answers: (b)(ii) 6.65 m s−1 · (b)(iii) 1.2 m s−1 · (c) 950 m s−2 · (d) 194g, passes (the remaining parts are explanations — see the table above)
A car travels at a constant 15.0 m s−1 towards traffic lights on a straight, level road. When the car is a distance D from the stop line, the light changes to amber. The light stays amber for 3.0 s and then turns red. The driver can either stop before the stop line or keep moving at constant speed and cross the stop line before the light turns red.
The reaction time of the driver is 0.80 s, and when the driver brakes, the car decelerates uniformly at 4.0 m s−2. Treat the car as a point. Figure 1 is a set of axes for (c).
Figure 1 — axes for (c)
(a)
Calculate the distance travelled by the car during the reaction time.
(1)
(b)
Show that the driver can stop before the stop line only if D is at least about 40 m.
(1)
(c)
Sketch, on Figure 1, the speed–time graph of the car from the moment the light changes until the car stops, for the driver who stops. Add values to your graph where appropriate.
(2)
(d)
Calculate the greatest value of D for which the car crosses the stop line before the light turns red, without braking.
(1)
(e)
Hence state the range of values of D for which the driver can safely choose either to stop or to keep going.
(1)
(f)
Another car approaches the same lights at 20.0 m s−1. Its driver has the same reaction time and the car the same deceleration. Determine the range of values of D for which this driver can neither stop before the line nor cross it before the light turns red.
(3)
(g)
Calculate the shortest duration of the amber light for which this range would not exist at 20.0 m s−1.
(1)
(h)
Explain why the range of values of D in (f) becomes wider as the approach speed increases.
straight line from (0.80 s, 15.0 m s−1) to zero at 4.55 s
✓ 1
Accept 4.5–4.6 s. Allow ECF from (a).
Part (d)
15.0 × 3.0 = 45.0 m
✓ 1
Part (e)
40.1 m ≤ D ≤ 45.0 m
✓ 1
Allow ECF from (b) and (d).
Part (f)
least stopping distance = 20.0 × 0.80 + 20.02/(2 × 4.0) = 66.0 m
✓ 1
greatest distance to cross before red = 20.0 × 3.0 = 60.0 m
✓ 1
60.0 m < D < 66.0 m
✓ 1
Award [3] for CNA. Award [2 max] for a correct pair of limits stated the wrong way round.
Part (g)
66.0/20.0 = 3.3 s
✓ 1
Allow ECF from (f).
Part (h)
the reaction distance and the distance covered in the amber time are both proportional to the speed, but the braking distance is proportional to the square of the speed «v2/2a»
✓ 1
so the least stopping distance increases faster with speed than the greatest crossing distance, and the gap between them grows
✓ 1
OWTTE. MP2 only scores if MP1 scores.
Answers: (a) 12.0 m · (d) 45.0 m · (e) 40.1–45.0 m · (f) 60.0–66.0 m · (g) 3.3 s (the remaining parts are explanations — see the table above)
202IBP-A1-N1-08Motion graphs & definitionsPaper 2Hard6 marks
An engineering team tests a rocket-powered sled on a straight horizontal rail. The sled starts from rest at t = 0. Its acceleration–time graph shows three stages: • 0 to 5.0 s: acceleration +12 m s−2 • 5.0 s to 8.0 s: acceleration zero • from 8.0 s: acceleration −20 m s−2 until the sled stops.
(a)
Calculate the maximum speed of the sled.
(1)
(b)
Calculate the time taken for the sled to stop once braking begins.
(1)
(c)
Determine the total distance travelled by the sled.
(2)
(d)
Describe the shape of the velocity–time graph for the whole motion.
(2)
Show mark scheme
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Mark
Notes
Part (a)
v = 12 × 5.0 = 60 m s−1 (area under a–t graph)
✓ 1
Part (b)
t = 60/20 = 3.0 s (so the sled stops at t = 11.0 s); allow ECF from (a)
✓ 1
Part (c)
Stage distances: ½ × 5.0 × 60 = 150 m, 60 × 3.0 = 180 m, ½ × 3.0 × 60 = 90 m
✓ 1
Total = 420 m
✓ 1
Part (d)
Straight line with positive gradient from the origin to 60 m s−1 at 5.0 s, then horizontal until 8.0 s
✓ 1
Straight line with negative gradient, steeper than the first stage, reaching zero at 11.0 s
✓ 1
Exam tip: When given an a–t graph, convert it to a v–t graph first. Constant acceleration means a straight-line v–t section, and zero acceleration means a horizontal one.
203IBP-A1-N1-09Motion graphs & definitionsPaper 2Hard6 marks
A parcel on a warehouse conveyor belt moves along a straight line. Taking its initial direction as positive: • 0 to 15 s: constant velocity +0.80 m s−1 • 15 s to 17 s: velocity decreases uniformly to zero • 17 s to 18 s: velocity changes uniformly from zero to −0.40 m s−1 • 18 s to 28 s: constant velocity −0.40 m s−1
(a)
Determine the displacement of the parcel at t = 17 s.
(2)
(b)
Determine, for the whole 28 s, (i) the total distance travelled and (ii) the final displacement of the parcel.
(2)
(c)
Calculate the average speed and the average velocity of the parcel over the 28 s.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
0.80 × 15 = 12 m and ½ × 2.0 × 0.80 = 0.80 m
✓ 1
Displacement = +12.8 m
✓ 1
Part (b)
Backward distance = ½ × 1.0 × 0.40 + 0.40 × 10 = 0.20 + 4.0 = 4.2 m
✓ 1
(i) distance = 12.8 + 4.2 = 17.0 mand (ii) displacement = 12.8 − 4.2 = +8.6 m
✓ 1
Part (c)
Average speed = 17.0/28 = 0.61 m s−1 (0.607 m s−1)
✓ 1
Average velocity = 8.6/28 = +0.31 m s−1 (0.307 m s−1) in the original direction; allow ECF from (b)
✓ 1
Common error: When the motion reverses, distance and displacement differ. Areas below the time axis of a v–t graph are added for distance but subtracted for displacement.
204IBP-A1-N2-08Equations of motionPaper 2Hard7 marks
A train starts from rest at station P and accelerates uniformly at 0.50 m s−2 until its speed is 20 m s−1. It then travels at 20 m s−1 before decelerating uniformly at 0.80 m s−2, coming to rest at station Q. The track from P to Q is straight and 2.0 km long.
(a)
Calculate the time taken and the distance travelled while the train accelerates.
(2)
(b)
Calculate the time taken and the distance travelled while the train decelerates.
(2)
(c)
Determine the total time for the journey from P to Q.
(2)
(d)
Calculate the average speed of the train between P and Q.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
t = 20 / 0.50 = 40 s
✓ 1
s = ½ × 20 × 40 = 400 m
✓ 1
Part (b)
t = 20 / 0.80 = 25 s
✓ 1
s = ½ × 20 × 25 = 250 m
✓ 1
Part (c)
constant-speed distance = 2000 − 400 − 250 = 1350 m, so time = 1350 / 20 = 67.5 s
✓ 1
total = 40 + 67.5 + 25 = 1.3 × 102 s (132.5 s)
✓ 1
Part (d)
2000 / 132.5 = 15 m s−1 (15.1 m s−1)
✓ 1
Exam tip: Sketching the v–t graph (a trapezium) is a quick check: the total area must equal the 2.0 km between the stations.
205IBP-A1-N2-10Equations of motionPaper 2Hard6 marks
On a calm lake, a jet ski moving in a straight line at a constant 12 m s−1 passes a speedboat that is at rest. At that instant the speedboat sets off after it along the same line with a uniform acceleration of 1.5 m s−2.
(a)
Show that the speedboat draws level with the jet ski 16 s after it starts.
(2)
(b)
Calculate the speed of the speedboat when it draws level with the jet ski.
(1)
(c)
Determine the greatest distance between the jet ski and the speedboat before the speedboat draws level.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
positions equal: 12t = ½ × 1.5 × t2
✓ 1
t = 12 / 0.75 = 16 s (the root t = 0 is the starting instant)
✓ 1
Part (b)
v = 1.5 × 16 = 24 m s−1
✓ 1
Part (c)
the gap is greatest when the two speeds are equal
✓ 1
1.5t = 12, so t = 8.0 s
✓ 1
gap = 12 × 8.0 − ½ × 1.5 × 8.02 = 96 − 48 = 48 m
✓ 1
Exam tip: In a chase, write an expression for each object's position from the same origin and the same starting time. Equal positions give the meeting time, and equal velocities give the largest gap.
A rescue helicopter is rising vertically at a constant speed of 3.0 m s−1. When it is 12 m above a flat field, a crew member releases a supply package from the open door. Air resistance on the package is negligible and g = 9.8 m s−2. Take upwards as positive.
(a)
State the velocity of the package at the instant it is released.
(1)
(b)
Calculate the maximum height above the field reached by the package.
(2)
(c)
Show that the package reaches the field about 1.9 s after release.
(2)
(d)
Determine the speed of the package just before it lands.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
+3.0 m s−1 (3.0 m s−1 upwards)
✓ 1
Part (b)
rise after release = 3.02/(2 × 9.8) = 0.46 m
✓ 1
maximum height = 12 + 0.46 = 12.5 m
✓ 1
Part (c)
−12 = 3.0t − 4.9t2or 4.9t2 − 3.0t − 12 = 0
✓ 1
t = (3.0 + √(9.0 + 235.2))/9.8 = 1.90 s
✓ 1
Part (d)
v = 3.0 − 9.8 × 1.90 orv2 = 3.02 + 2 × 9.8 × 12
✓ 1
speed = 16 m s−1 (15.6 m s−1)
✓ 1
Common error: Using s = +12 m instead of −12 m gives an equation with no real solution. The package ends up below its release point, so its displacement is negative when upwards is positive.
In a physics puzzle, a beanbag is dropped from rest from a window 20 m above a courtyard. At the same instant, a tennis ball is thrown vertically upwards from the courtyard directly below the window, with a speed of 15 m s−1. Both move along the same vertical line. Air resistance is negligible and g = 9.8 m s−2.
Diagram not to scale
(a)
Show that the two objects meet 1.3 s after they are released.
(2)
(b)
Calculate the height above the courtyard at which they meet.
(2)
(c)
Determine the velocity of the tennis ball at the instant they meet, and state whether it is rising or falling.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
height of ball = 15t − 4.9t2 and height of beanbag = 20 − 4.9t2
✓ 1
equating: 15t = 20, so t = 1.33 s ≈ 1.3 s
✓ 1
Part (b)
20 − 4.9 × 1.3332or 15 × 1.333 − 4.9 × 1.3332
✓ 1
= 11 m (11.3 m)
✓ 1
Part (c)
v = 15 − 9.8 × 1.333 = 1.9 m s−1 (1.93 m s−1)
✓ 1
positive value, so the ball is still rising
✓ 1
Exam tip: When two objects share the same acceleration, the ½gt² terms cancel when you equate their positions. Writing both position equations from one common origin makes this visible.
A stone is thrown vertically upwards at 12 m s−1 from the edge of a sea cliff. It rises, then falls past the cliff edge and lands in the sea 26 m below its point of release. Air resistance is negligible and g = 9.8 m s−2. Take upwards as positive.
Diagram not to scale
(a)
State the displacement of the stone from its point of release when it reaches the sea.
(1)
(b)
Calculate the velocity of the stone just before it enters the sea.
(2)
(c)
Calculate the time the stone is in the air.
(2)
(d)
Determine the total distance travelled by the stone.
(2)
Show mark scheme
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Notes
Part (a)
−26 m (26 m downwards)
✓ 1
Part (b)
v2 = 122 + 2 × (−9.8) × (−26) = 654
✓ 1
v = −26 m s−1 (−25.6 m s−1); negative root chosen as the stone moves downwards
Common error: Displacement and distance differ here: the displacement is −26 m, but the stone goes up 7.35 m and back down again before falling the 26 m, so the distance is about 41 m.
209IBP-A1-N5-07Kinematics data & measurementPaper 2Hard7 marks
A steel ball is released from rest by an electromagnet and falls a height h onto a switch pad. An electronic timer records the fall time t. Air resistance is negligible. The results are: h / m: 0.50, 1.00, 1.50, 2.00 t / s: 0.319, 0.452, 0.553, 0.639
Diagram not to scale
(a)
Show that a graph of h against t2 should be a straight line through the origin, and state what its gradient represents.
(2)
(b)
Using the first and last data points, determine a value for g.
(3)
(c)
The student actually measured each height from the bottom of the electromagnet to the pad, so every recorded h is too large by the same amount. Deduce the effect of this on the graph and on the value of g.
(2)
Show mark scheme
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Mark
Notes
Part (a)
from rest, h = ½gt2, which has the form y = mx with no intercept
the line is shifted parallel, so it has a (non-zero) intercept on the h axis
✓ 1
the gradient is unchanged, so the value of g is unaffected
✓ 1
Exam tip: A constant (systematic) offset in a measured quantity shifts a straight-line graph without changing its gradient. Finding the quantity from the gradient removes the effect of the offset.
210IBP-A1-N5-08Kinematics data & measurementPaper 2Hard6 marks
A rover on Mars drops a small sample capsule from rest. The capsule falls (1.21 ± 0.01) m in a time of (0.80 ± 0.02) s. Atmospheric resistance is negligible.
(a)
Calculate the acceleration of free fall on Mars.
(1)
(b)
Determine the percentage uncertainty in your answer to (a).
(2)
(c)
State the value of the acceleration of free fall on Mars with its absolute uncertainty.
(2)
(d)
Suggest which measurement should be improved to reduce the uncertainty in g most effectively. Give a reason.
(1)
Show mark scheme
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Mark
Notes
Part (a)
g = 2h/t2 = 2 × 1.21 / 0.802 = 3.8 m s−2 (3.78 m s−2)
✓ 1
Part (b)
%h = 0.83% and %t = 2.5%
✓ 1
%g = 0.83 + 2 × 2.5 = 5.8%
✓ 1
Part (c)
absolute uncertainty = 0.058 × 3.78 = 0.22 m s−2
✓ 1
(3.8 ± 0.2) m s−2
✓ 1
Part (d)
the time, because its percentage uncertainty is larger and is doubled since t is squared
✓ 1
Exam tip: Quote an absolute uncertainty to one significant figure, then round the value so that its last digit is in the same decimal place as the uncertainty.
211IBP-A1-N5-10Kinematics data & measurementPaper 2Hard8 marks
A video of a beach-volleyball serve is analysed frame by frame. Positions of the ball are measured from its launch point, with x horizontal and y vertically upwards. Air resistance is negligible. t / s: 0, 0.20, 0.40, 0.60 x / m: 0, 1.20, 2.40, 3.60 y / m: 0, 0.804, 1.216, 1.236
Positions of the ball from the video, at t = 0, 0.20 s, 0.40 s and 0.60 s
(a)
Explain what the x data show about the horizontal motion, and calculate the horizontal component of velocity.
(2)
(b)
Calculate the average vertical velocity of the ball between t = 0 and t = 0.20 s, and between t = 0.20 s and t = 0.40 s.
(2)
(c)
Hence determine the acceleration of free fall from the video data.
(2)
(d)
Determine the speed of the ball at the moment it is launched.
(2)
Show mark scheme
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Mark
Notes
Part (a)
equal horizontal distances in equal times, so the horizontal velocity is constant (no horizontal acceleration)
✓ 1
vx = 1.20 / 0.20 = 6.0 m s−1
✓ 1
Part (b)
0.804 / 0.20 = 4.02 m s−1
✓ 1
(1.216 − 0.804) / 0.20 = 2.06 m s−1
✓ 1
Part (c)
these average velocities apply at t = 0.10 s and t = 0.30 s, 0.20 s apart
✓ 1
g = (4.02 − 2.06) / 0.20 = 9.8 m s−2
✓ 1
Part (d)
vertical component at t = 0: 4.02 + 9.8 × 0.10 = 5.0 m s−1
✓ 1
speed = √(6.02 + 5.02) = 7.8 m s−1 (7.81 m s−1)
✓ 1
Exam tip: Treat the horizontal and vertical data separately. Only combine the components with Pythagoras at the end, when a speed (magnitude) is asked for.
212IBP2S-001-05-15Motion graphs & definitionsPaper 2Hard7 marks
A student places a smartphone on the floor of a lift and records the vertical acceleration (upwards positive). The lift starts from rest. The data are summarised below.
0 to 2.0 s: constant acceleration of 1.2 m s−2 2.0 s to 3.0 s: acceleration decreases uniformly from 1.2 m s−2 to 0 3.0 s to 10.0 s: acceleration is 0 10.0 s to 13.0 s: constant acceleration of −1.0 m s−2
(a)
Determine the maximum speed of the lift.
(2)
(b)
Show that the distance travelled between t = 2.0 s and t = 3.0 s lies between 2.7 m and 3.0 m.
(2)
(c)
The distance travelled between t = 2.0 s and t = 3.0 s is 2.8 m. Determine the total height through which the lift rises.
(3)
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Notes
Part (a)
Area under the a–t graph up to 3.0 s: (1.2 × 2.0) + (½ × 1.0 × 1.2)
✓ 1
vmax = 2.4 + 0.6 = 3.0 m s−1
✓ 1
Part (b)
The speed rises from 2.4 to 3.0 m s−1; a straight-line (uniform acceleration) estimate gives ½(2.4 + 3.0) × 1.0 = 2.7 m, and the v–t curve's gradient decreases, so it lies above this line and the distance exceeds 2.7 m
✓ 1
The speed never exceeds 3.0 m s−1, so the distance is less than 3.0 × 1.0 = 3.0 m (ECF from (a))
✓ 1
Part (c)
0–2.0 s: ½ × 1.2 × 2.02 = 2.4 m
✓ 1
3.0–10.0 s: 3.0 × 7.0 = 21 m, and 10.0–13.0 s: ½ × 3.0 × 3.0 = 4.5 m (ECF from (a))
✓ 1
Total = 2.4 + 2.8 + 21 + 4.5 ≈ 31 m (30.7 m)
✓ 1
Exam tip: Split the motion into phases. Work out the speed at each boundary first, then find the distance in each phase using the method that suits it.
213IBP2S-001-05-16Fluid resistance & terminal speedPaper 2Hard8 marks
A medical-supply drone flies horizontally at 12 m s−1 at a height of 45 m above flat ground. It releases a small package.
Diagram not to scale
(a)
Assuming air resistance is negligible, calculate the time taken for the package to reach the ground.
(2)
(b)
Assuming air resistance is negligible, calculate the horizontal distance travelled by the package before it reaches the ground.
(1)
(c)
In practice, air resistance acts on the package. Explain the effect of air resistance on the time of flight and on the horizontal distance travelled.
(3)
(d)
A parachute is fitted so the package reaches a vertical terminal speed of 5.0 m s−1 very soon after release. Estimate the time of fall, and state why your estimate is slightly too small.
(2)
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Notes
Part (a)
Vertical motion starts from rest: 45 = ½ × 9.8 × t2
✓ 1
t = 3.0 s (3.03 s)
✓ 1
Part (b)
x = 12 × 3.03 = 36 m (36.4 m); allow a follow-through from the calculated time of flight
✓ 1
Part (c)
The vertical component of drag opposes the fall, so the downward acceleration is less than g and the time of flight increases
✓ 1
The horizontal component of drag decelerates the package, so its horizontal speed falls below 12 m s−1 during the flight
✓ 1
The net effect is a shorter horizontal distance than the no-drag value: the reduction in horizontal speed outweighs the increase in flight time
✓ 1
Part (d)
t ≈ 45 / 5.0 = 9.0 s
✓ 1
The package starts with zero vertical speed and speeds up to 5.0 m s−1, so its average vertical speed is below 5.0 m s−1 and the true time is slightly longer
✓ 1
Common error: With air resistance, the horizontal velocity is not constant. Drag acts against the velocity, so it has both horizontal and vertical components.
214IBP2S-001-00-15Projectile motionPaper 2Hard8 marks
A firefighter aims a hose so that water enters an open window. The window is 12.0 m above the nozzle and 15.0 m horizontally from it. The water should be moving horizontally as it passes through the window, so that it enters the window at the top of its path. Air resistance is negligible. Take g = 9.8 m s−2.
Diagram not to scale. The water is moving horizontally at the window.
(a)
Show that the vertical component of the water's velocity at the nozzle is about 15 m s−1.
(2)
(b)
Calculate the time the water takes to travel from the nozzle to the window.
(1)
(c)
Determine the speed of the water leaving the nozzle and the angle of the nozzle above the horizontal.
(3)
(d)
In practice, air resistance acts on the water. Explain why the nozzle speed calculated in (c) would not be enough.
(2)
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Notes
Part (a)
Vertical velocity is zero at the window, so 0 = uy2 − 2 × 9.8 × 12.0 . uy = √235 = 15.3 m s−1 .
✓ 2
Part (b)
t = uy / g = 15.3 / 9.8 = 1.56 s (accept 1.57 s) .
✓ 1
Part (c)
ux = 15.0 / 1.56 = 9.6 m s−1, allowing ECF from (b) . u = √(15.32 + 9.62) = 18.1 m s−1 . θ = tan−1(15.3 / 9.6) = 58° above the horizontal; an equivalent method using sin or cos with ECF also earns this mark .
✓ 3
Part (d)
Drag opposes the motion. It reduces the horizontal velocity, which would otherwise stay constant, and it adds to gravity in slowing the upward motion, so the vertical deceleration exceeds g . With the same launch velocity, the water would therefore not rise as high and would fall short of or below the window, so a greater launch speed is needed .
✓ 2
Exam tip: 'Moving horizontally' tells you the vertical velocity is zero at that point. That gives you a final velocity to use in the vertical suvat equations.
215IBP2S-001-00-16Projectile motionPaper 2Hard9 marks
The squirting cucumber plant spreads its seeds by ejecting them. In a model of this, a seed leaves the fruit at 10 m s−1 from ground level and lands 8.0 m away on level ground. Air resistance is initially ignored.
Diagram not to scale
(a)
Determine the two possible launch angles of the seed.
(3)
(b)
Calculate the time of flight for the larger launch angle.
(2)
(c)
Calculate the maximum height reached by the seed for the larger launch angle.
(2)
(d)
Seeds are light, so air resistance is in fact significant. Explain which of the two trajectories would have its range reduced more by air resistance.
(2)
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Notes
Part (a)
Use R = u2 sin2θ / g . sin2θ = 8.0 × 9.8 / 102 = 0.784 . θ = 25.8° or θ = 64.2° .
✓ 3
Part (b)
t = 2u sinθ / g = 2 × 10 × sin64.2° / 9.8 . t = 1.8 s (1.84 s); ECF from (a) .
✓ 2
Part (c)
uy = 10 sin64.2° = 9.0 m s−1 . H = 9.02 / (2 × 9.8) = 4.1 m; ECF .
✓ 2
Part (d)
The 64.2° seed spends about twice as long in the air and travels a longer path, so drag acts on it for longer . Its range is therefore reduced more than that of the 25.8° seed .
✓ 2
Common error: sin2θ = k has two solutions between 0° and 90°: θ and 90° − θ. Give both when the question asks for two angles.
216IBP2S-001-04-14Fluid resistance & terminal speedPaper 2Hard10 marks
In a blood test, red blood cells sink slowly through plasma. Model a cell as a sphere of radius r = 4.0 μm and density ρs = 1100 kg m−3. The plasma has density ρf = 1025 kg m−3 and viscosity η = 1.2 × 10−3 Pa s. The drag force is given by Stokes' law, F = 6πηrv.
(a)
Show that the terminal speed of the cell is vt = 2r2(ρs − ρf)g/(9η).
(3)
(b)
Calculate the terminal speed of the cell.
(2)
(c)
Estimate the time the cell takes to reach terminal speed after being released from rest. Assume its initial acceleration stays constant.
(3)
(d)
Calculate the time for the cell to sink 5.0 mm, and comment on whether it is reasonable to assume the cell moves at terminal speed throughout.
217IBP2S-001-04-16Fluid resistance & terminal speedPaper 2Hard9 marks
An aid agency tests a parachute canopy by dropping it with crates of different total mass m. It records the terminal speed vt for each mass:
m / kg: 2.0, 4.0, 8.0, 16.0 vt / m s−1: 5.0, 7.1, 10.0, 14.1
The drag may be proportional to v or to v2. Take g = 9.8 m s−2.
(a)
Show that the data are consistent with a drag force F = kv2 and not with F = kv.
(3)
(b)
Determine the value of k.
(2)
(c)
A crate of mass 12 kg is released from a hovering helicopter using this canopy. Calculate its acceleration when its speed is 6.0 m s−1.
(2)
(d)
The canopy is replaced by a larger one, for which k is doubled. Calculate the new terminal speed of the 12 kg crate.
(2)
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Notes
Part (a)
At terminal speed the drag equals the weight. For F = kv2, mg = kvt2, so vt2/m should be constant
✓ 1
The values of vt2/m are 12.5, 12.6, 12.5 and 12.4 m2 s−2 kg−1. These are constant within experimental error. Alternatively: each doubling of m multiplies vt by about 1.41 ≈ √2
✓ 1
For F = kv, vt/m would be constant, but it falls from 2.5 to 0.88. Alternatively: vt does not double when m doubles. So linear drag is rejected
✓ 1
Part (b)
k = mg/vt2 = 9.8/12.5, using any row or the mean value
✓ 1
k = 0.78 kg m−1. Accept 0.78 to 0.79
✓ 1
Part (c)
Drag = k × 6.02 ≈ 28 N and weight = 12 × 9.8 = 117.6 N. Allow ECF from the value of k found earlier
✓ 1
a = (117.6 − 28)/12 ≈ 7.5 m s−2. Accept 7.4 to 7.5
✓ 1
Part (d)
vt = √(mg/2k) = √(117.6/(2 × 0.78)). Allow ECF from the value of k found earlier
✓ 1
vt = 8.7 m s−1
✓ 1
Exam tip: To test a proposed power law, work out the quantity the law predicts is constant for every row of data. One calculation per row is stronger evidence than checking a single pair of values.
A model rocket is launched vertically from rest at ground level. Its engine gives it a constant upward acceleration of 20 m s−2 for 3.0 s. The engine then cuts out, and the rocket moves freely under gravity until it hits the ground. Air resistance is negligible. Take g = 9.8 m s−2.
(a)
Calculate the velocity of the rocket when the engine cuts out.
(1)
(b)
Calculate the height of the rocket when the engine cuts out.
(1)
(c)
Determine the maximum height reached by the rocket.
(2)
(d)
Taking upwards as positive, describe the velocity–time graph from launch until the rocket hits the ground.
(3)
(e)
Calculate the total time from launch until the rocket hits the ground.
(2)
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Notes
Part (a)
v = 20 × 3.0 = 60 m s−1 (upwards)
✓ 1
Part (b)
s = ½ × 20 × 3.02 = 90 m
✓ 1
Part (c)
Extra height after cut-out = 602/(2 × 9.8) = 184 m (allow error carried forward from the cut-out velocity)
✓ 1
Maximum height = 90 + 184 = 274 m (accept 273–274 m)
✓ 1
Part (d)
A straight line from the origin with gradient 20 m s−2, reaching 60 m s−1 at 3.0 s
✓ 1
After 3.0 s, a single straight line of gradient −9.8 m s−2 that crosses the time axis at about 9.1 s
✓ 1
The same line continues into negative velocities and ends at about −73 m s−1 at impact
✓ 1
Part (e)
Either: time to rise after cut-out = 60/9.8 = 6.1 s, and time to fall from the maximum height = √(2 × 274/9.8) = 7.5 s; or: solve 0 = 90 + 60t − 4.9t2 to give t = 13.6 s after cut-out (allow error carried forward from earlier values)
✓ 1
Total time = 16.6 s
✓ 1
Exam tip: The motion from engine cut-out to impact is one constant-acceleration phase. Its v–t graph is one unbroken straight line, with no change at the highest point. You can also solve it with a single suvat equation, using displacement −90 m measured from the cut-out point.
219IBP2S-001-03-16Fluid resistance & terminal speedPaper 2Hard8 marks
A wingsuit flyer's vertical velocity is recorded after she leaves a cliff edge. Air resistance is significant.
t / s: 0, 2.0, 4.0, 6.0, 8.0, 10.0 v / m s−1: 0, 17, 29, 36, 40, 42
Data from the table
(a)
Calculate the average acceleration in the first 2.0 s and outline why it is less than g.
(2)
(b)
Explain, with reference to forces, why the gradient of the velocity–time graph decreases with time.
(2)
(c)
Estimate the vertical distance fallen in the first 10.0 s.
(3)
(d)
State whether your estimate of the distance in the previous part is an overestimate or an underestimate, giving a reason.
(1)
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Notes
Part (a)
a = 17/2.0 = 8.5 m s−2
✓ 1
Air resistance acts upwards once she is moving and grows with speed, so over the interval the resultant downward force is less than her weight
✓ 1
Part (b)
Air resistance increases as her speed increases
✓ 1
So the resultant force (weight − drag), and hence the acceleration (the gradient), decreases towards zero as drag approaches her weight
✓ 1
Part (c)
Uses the area under the v–t graph, split into trapezia of width 2.0 s
✓ 1
Areas: 17 + 46 + 65 + 76 + 82 (m)
✓ 1
Distance ≈ 290 m (286 m)
✓ 1
Part (d)
Underestimate, because the graph curves with decreasing gradient, so the straight tops of the trapezia lie below the curve
✓ 1
Common error: Using a single triangle, or the average of the first and last speeds, gives a poor estimate. With a curved v–t graph, split the area into several strips.
220IBP2S-001-02-16Motion graphs & definitionsPaper 2Hard10 marks
A delivery robot moves in a straight line with constant acceleration a. It is already moving with velocity u when it passes a sensor at time t = 0. Its displacement s from the sensor is recorded.
t / s
1.0
2.0
3.0
4.0
5.0
s / m
2.1
5.4
9.9
15.6
22.5
(s/t) / m s−1
2.1
2.7
3.9
(a)
Show that a graph of s/t against t is a straight line, and state what its gradient and intercept represent.
(2)
(b)
Calculate the two missing values of s/t.
(1)
(c)
Determine u and a.
(3)
(d)
Describe the shape of a graph of s against t from t = 0 to 5.0 s. Refer to its starting point and to how its gradient changes.
(2)
(e)
Show that the average velocity over the first 5.0 s equals the instantaneous velocity at t = 2.5 s.
(2)
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Notes
Part (a)
Dividing s = ut + ½at2 by t gives s/t = u + ½at, which has the form y = c + mx
✓ 1
Gradient = a/2 and intercept = u
✓ 1
Part (b)
3.3 m s−1 (at 3.0 s) and 4.5 m s−1 (at 5.0 s)
✓ 1
Part (c)
Gradient = (4.5 − 2.1)/(5.0 − 1.0) = 0.60 m s−2; any pair of points from the completed row, or a best-fit line, gives the same value
✓ 1
a = 2 × gradient = 1.2 m s−2
✓ 1
Intercept: u = 2.1 − 0.60 × 1.0 = 1.5 m s−1
✓ 1
Allow ECF from the values in (b).
—
Part (d)
It is a curve that starts at the origin with a non-zero initial gradient, equal to u = 1.5 m s−1
✓ 1
The gradient increases steadily, so the curve bends upwards as a parabola and reaches 22.5 m at 5.0 s
✓ 1
Part (e)
Average velocity = 22.5/5.0 = 4.5 m s−1
✓ 1
v(2.5) = u + at = 1.5 + 1.2 × 2.5 = 4.5 m s−1, the same value; allow ECF from (c) for this mark
✓ 1
Exam tip: To linearise an equation, rearrange it so that the unknowns appear only as the gradient and the intercept. Then use widely separated points to find the gradient.
221IBP2S-001-01-13Two-body kinematicsPaper 2Hard10 marks
A skier starts from rest at the top of a straight slope and accelerates uniformly down it at 2.0 m s−2. A camera drone flies along a line parallel to the slope at a constant 12 m s−1 in the same direction as the skier. It passes the starting point 4.0 s after the skier sets off. Take t = 0 as the moment the skier starts.
(a)
Calculate the speed of the skier and her distance from the start at t = 4.0 s.
(2)
(b)
Show that the drone never draws level with the skier.
(3)
(c)
Explain why the separation between the drone and the skier is smallest when their speeds are equal, and determine this minimum separation.
(3)
(d)
Determine the minimum constant speed the drone would need in order to just draw level with the skier.
(2)
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Notes
Part (a)
v = 2.0 × 4.0 = 8.0 m s−1 ; s = ½ × 2.0 × 4.02 = 16 m
✓ 2
Part (b)
Skier: x = t2; drone: x = 12(t − 4.0) ; gap = t2 − 12t + 48, or for being level t2 − 12t + 48 = 0 ; discriminant 144 − 192 < 0 so there is no real solution, OR gap = (t − 6)2 + 12 > 0 for all t, so they are never level
✓ 3
Part (c)
While the drone is faster it closes the gap, and once the skier is faster the gap grows, so the gap is smallest when 2.0t = 12, i.e. at t = 6.0 s ; skier position = 6.02 = 36 m and drone position = 12 × 2.0 = 24 m ; minimum separation = 12 m
✓ 3
Part (d)
Condition for just drawing level: t2 − vt + 4.0v = 0 has one repeated root, so v2 − 16v = 0; OR the speeds are equal at the meeting, so v = 2.0t and t2 = v(t − 4.0), giving t = 8.0 s ; v = 16 m s−1, with v = 0 rejected
✓ 2
Exam tip: In a catch-up problem, the gap is smallest, or the chaser just touches, at the moment the two velocities are equal.
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