50 free, high-quality past-paper-style questions on Kinematics for IB Physics HL, each with a full mark scheme. Covers: Equations of uniformly accelerated motion; Distance and displacement; Projectile motion; and more. For when you have finished the past papers, or want to practise one topic at a time.A.1 at Standard 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.215 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1A-1A-02Equations of uniformly accelerated motionPaper 1AEasy1 mark
A car accelerates uniformly from rest. It travels a distance d during the first t seconds.
What distance does it travel during the nextt seconds?
Show mark scheme
Marking point
Mark
Notes
Step 1From rest with constant acceleration, s = ½at², so the distance covered is proportional to the square of the elapsed time.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2After 2t the total distance is (2)² × d = 4d.
—
Step 3The second interval alone covers 4d − d = 3d.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis assumes the speed simply doubles and stays constant in the second interval. The car is still accelerating throughout.
BCorrect: total distance after 2t is 4d, so the second interval covers 4d − d = 3d.
CThis is the total distance after 2t, not the distance in the second interval alone.
DThis would be the total distance after 3t. The question asks only about the second interval.
2A-1A-62Distance and displacementPaper 1AEasy1 mark
A survey drone starts at O and flies 400 m due north in 20 s. It hovers for 10 s and then flies 300 m due east in 20 s, finishing at F, as shown in the plan view.
What is the magnitude of the average velocity of the drone for the whole 50 s?
Plan view of the path of the drone (drawn to scale).Show mark scheme
Marking point
Mark
Notes
Step 1Average velocity = displacement/total time. The displacement is the straight line from O to F.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Displacement = √(400² + 300²) = 500 m.
—
Step 3The total time includes the 10 s hover: average velocity = 500/50 = 10 m s−1.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: displacement 500 m divided by the total time of 50 s.
BThis leaves out the 10 s of hovering: 500/40 = 12.5 m s−1. The average is taken over the whole time interval, including time at rest.
CThis is the average speed, distance/time = 700/50 = 14 m s−1. Average velocity uses the displacement, not the distance travelled.
DThis uses the distance instead of the displacement and also leaves out the hover: 700/40 = 17.5 m s−1.
A ball is thrown from a balcony with a speed of 10.0 m s−1 at 30.0° below the horizontal. It lands on the level ground below 1.20 s later. Air resistance is negligible and g = 9.81 m s−2.
What is the height of the balcony above the ground?
Diagram not to scale
Show mark scheme
Marking point
Mark
Notes
Step 1Initial vertical component = 10.0 sin 30.0° = 5.00 m s−1, directed downwards.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Taking downwards as positive: s = ut + ½at² = 5.00 × 1.20 + ½ × 9.81 × 1.20².
—
Step 3s = 6.00 + 7.06 = 13.1 m.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis treats the initial vertical component as upwards: −6.00 + 7.06 = 1.06 m. The ball is thrown below the horizontal, so it starts with a downward vertical velocity.
BThis ignores the initial vertical component, as if the ball were thrown horizontally: ½ × 9.81 × 1.20² = 7.06 m.
CCorrect: 5.00 × 1.20 + ½ × 9.81 × 1.20² = 13.1 m.
DThis uses the horizontal component 10.0 cos 30.0° = 8.66 m s−1 as the vertical one: 8.66 × 1.20 + 7.06 = 17.5 m.
A delivery drone rises vertically from the ground. Its displacement–time graph is a smooth curve that reaches a maximum at t = 4.0 s and then decreases. Which statement about the drone at t = 4.0 s is correct?
Show mark scheme
Marking point
Mark
Notes
Velocity is the gradient of a displacement–time graph. At a maximum the tangent is horizontal, so the gradient is zero. The drone is therefore momentarily at rest at its greatest displacement. The acceleration need not be zero, because the velocity is changing from positive to negative. A maximum in displacement is not a maximum in velocity. The displacement here is at its largest value, not zero.
✓ 1
Answer A
Answer: A
Common error: Zero velocity does not mean zero acceleration. At a turning point of an s–t graph the velocity is changing sign, so the object is still accelerating.
5IBP2H-002-01-02Relative velocity and vectorsPaper 1AEasy1 mark
A cyclist rides due east at 6.0 m s−1. Rain falls vertically relative to the ground at 8.0 m s−1. What is the speed of the rain relative to the cyclist?
Velocities relative to the ground (side view)
Show mark scheme
Marking point
Mark
Notes
The velocity of the rain relative to the cyclist is vrain − vcyclist. This has a vertical component of 8.0 m s−1 downwards and a horizontal component of 6.0 m s−1 towards the west. The two components are perpendicular, so the relative speed is √(6.0² + 8.0²) = 10 m s−1. Option 2.0 comes from subtracting the speeds and 14 from adding them, both as if the velocities were collinear. Option 7.0 is simply the mean of the two speeds.
✓ 1
Answer C
Answer: C
Exam tip: Relative velocity is a vector subtraction. When the two velocities are perpendicular, use Pythagoras rather than adding or subtracting the speeds.
6IBP2H-002-00-01Instantaneous and average valuesPaper 1AEasy1 mark
An athlete warming up runs exactly 1.5 laps of a circular track of radius 50 m, starting and finishing on the track. What are the distance travelled and the magnitude of the displacement?
Show mark scheme
Marking point
Mark
Notes
Distance is the total path length: 1.5 × 2π × 50 = 471 m. After 1.5 laps the athlete is diametrically opposite the start, so the displacement magnitude is the diameter, 100 m. Option B treats 1.5 laps as a whole number of laps. Option C uses only one lap for the distance. Option D wrongly makes displacement equal to distance.
✓ 1
Answer A
Answer: A
Common error: Displacement depends only on the start and end positions. Half a lap leaves you a full diameter from the start, not back where you began.
A wearable fitness tracker records the forward acceleration of a patient's walking frame against time during a physiotherapy session. What does the area between the acceleration–time graph and the time axis represent?
Show mark scheme
Marking point
Mark
Notes
Acceleration is the rate of change of velocity, so acceleration × time gives the change in velocity. The area under an a–t graph is therefore Δv. Option A confuses this with the area under a v–t graph. Option C is the gradient of the a–t graph, not the area. Option D would need the displacement divided by the time.
✓ 1
Answer B
Answer: B
Common error: An area under a graph gives the change in velocity, not the velocity itself. You need the initial velocity to find the actual velocity.
A projectile is launched with speed u at 60° above the horizontal. Air resistance is negligible.
What is the ratio (kinetic energy at maximum height) / (initial kinetic energy)?
Show mark scheme
Marking point
Mark
Notes
Step 1At the highest point the vertical component of velocity is zero, so the only velocity left is the horizontal component u cos 60°, which never changes.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Kinetic energy is proportional to speed squared, so the ratio is (u cos 60°)²/u² = cos²60°.
—
Step 3cos 60° = 0.5, so cos²60° = 0.25 = 1/4.
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: only the horizontal component u cos 60° = 0.5u survives at the top, and (0.5)² = 1/4.
BThis is cos 60°, the ratio of the speeds. Kinetic energy depends on the square of the speed.
CThis is sin²60°, the fraction of the kinetic energy that has been converted to potential energy, not the fraction that remains.
DThis is sin 60°, the fraction of the velocity that is vertical at launch — it has all gone at the top.
Two identical balls P and Q are projected from the same point on level ground with the same speed. P is projected at 30° above the horizontal and Q at 60° above the horizontal. Air resistance is negligible.
Which statements are correct? I. P and Q land at the same distance from the point of projection. II. P and Q spend the same time in the air. III. The maximum height reached by Q is three times the maximum height reached by P.
Show mark scheme
Marking point
Mark
Notes
Step 1Range = (u cos θ)(2u sin θ/g) ∝ sin θ cos θ. sin 30° cos 30° = sin 60° cos 60°, so the ranges are equal: I is correct.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Time of flight = 2u sin θ/g ∝ sin θ: the ratio is sin 60°/sin 30° = √3, so the times are different: II is wrong.
—
Step 3Maximum height = (u sin θ)²/2g ∝ sin²θ: (√3/2)²/(1/2)² = (3/4)/(1/4) = 3, so III is correct.
✓ 1
Answer B
Answer: B · 3 stages of work, one mark
Every option, and why
AThis assumes that equal ranges need equal times of flight. Q has a smaller horizontal component and stays in the air √3 times longer.
BCorrect: complementary angles give equal ranges, and the heights are in the ratio sin²60° : sin²30° = 3 : 1.
CThis rejects I, perhaps believing the steeper launch goes further; the range depends on sin 2θ, which is the same for 30° and 60°.
DII is wrong: the times of flight are in the ratio sin 60° : sin 30° = √3 : 1.
A motorised camera dolly moves from rest along a straight rail. The graph shows its position x against time t. The tangent to the curve at P, where t = 6.0 s, is drawn.
Which row gives the instantaneous speed of the dolly at t = 6.0 s and its average speed for the first 6.0 s?
Position x of the dolly against time t. The dashed line is the tangent to the curve at P.
Instantaneous speed at 6.0 s / m s−1Average speed 0–6.0 s / m s−1
Show mark scheme
Marking point
Mark
Notes
Step 1Instantaneous speed = gradient of the tangent at P = (10.8 − 0)/(8.0 − 4.0) = 2.7 m s−1.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Average speed = total distance/total time = 5.4/6.0 = 0.90 m s−1 (the gradient of the chord from the origin to P).
—
Step 3The two differ because the acceleration is not uniform — the curve gets steeper and steeper.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis reads the coordinates of P and divides them (5.4/6.0) for the instantaneous speed — that is the chord gradient, not the tangent gradient.
BThis swaps the two: the tangent gives the instantaneous value (2.7 m s−1), the chord from the origin gives the average (0.90 m s−1).
CCorrect: tangent gradient 2.7 m s−1; chord gradient 0.90 m s−1.
DThis takes the average speed as half the final speed (2.7/2 = 1.35 m s−1), which is only true for uniform acceleration from rest.
An electric scooter starts from rest and moves in a straight line. The graph shows how its acceleration a varies with time t.
Which row gives the speed of the scooter at t = 4.0 s and the time at which its speed is greatest?
Acceleration a of the scooter against time t.
Speed at 4.0 s / m s−1Speed is greatest
Show mark scheme
Marking point
Mark
Notes
Step 1The change in velocity is the area under the acceleration–time graph: ½ × 3.0 × 4.0 = 6.0 m s−1. Starting from rest, the speed at 4.0 s is 6.0 m s−1.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2The acceleration is positive (though decreasing) all the way to 4.0 s, so the speed keeps increasing until then.
—
Step 3After 4.0 s the acceleration is zero, so the speed stays constant at its maximum value of 6.0 m s−1.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
A0.75 is the gradient of the a–t graph (3.0/4.0), which has no useful meaning here; and the speed is zero, not greatest, at t = 0.
BThe area is right, but this confuses the greatest acceleration (at t = 0) with the greatest speed.
CCorrect: area = 6.0 m s−1; the speed rises while a > 0 and is then constant.
DThis uses v = at with the initial acceleration (3.0 × 4.0 = 12 m s−1), treating the non-uniform acceleration as uniform.
A crew member throws a lifebuoy horizontally from the deck of a ship, 15.0 m above the sea, with a speed of 8.0 m s−1. Air resistance is negligible and g = 9.81 m s−2.
What is the angle between the velocity of the lifebuoy and the horizontal just before it reaches the sea?
Diagram not to scale
Show mark scheme
Marking point
Mark
Notes
Step 1The horizontal component stays at 8.0 m s−1 because no horizontal force acts.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Vertically from rest: vy² = 2gs = 2 × 9.81 × 15.0, so vy = 17.2 m s−1 downwards.
—
Step 3tan θ = 17.2/8.0, so θ = 65° below the horizontal.
✓ 1
Answer C
Answer: C · 3 stages of work, one mark
Every option, and why
AThis is the angle between the velocity and the vertical: tan−1(8.0/17.2) = 25°.
BThis is the direction of the displacement, not the velocity: the lifebuoy lands 14.0 m out, and tan−1(15.0/14.0) = 47°.
CCorrect: tan−1(vy/vx) = tan−1(17.2/8.0) = 65°.
DThis forgets the square root and uses vy² = 294 in place of vy: tan−1(294/8.0) = 88°.
In a brake test, an electric scooter moving in a straight line decelerates uniformly to rest. The graph shows how the square of its speed, v², varies with its displacement s from the point where the brakes are applied.
Which row gives the magnitude of the deceleration and the time taken to stop?
Square of the speed against displacement for the scooter (drawn to scale).
deceleration / m s−2time to stop / s
Show mark scheme
Marking point
Mark
Notes
Step 1From v² = u² + 2as, a graph of v² against s is a straight line of gradient 2a.
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Gradient = −100/20 = −5.0 m s−2, so a = −2.5 m s−2: the deceleration is 2.5 m s−2.
—
Step 3The intercept gives u² = 100 m² s−2, so u = 10 m s−1, and t = u/a = 10/2.5 = 4.0 s. (Check: s = ½(u + v)t = ½ × 10 × 4.0 = 20 m.)
✓ 1
Answer A
Answer: A · 3 stages of work, one mark
Every option, and why
ACorrect: the gradient is 2a, so a = 2.5 m s−2, and t = 10/2.5 = 4.0 s.
BThe deceleration is right, but the time is found as distance/initial speed, 20/10 = 2.0 s, as if the scooter kept its initial speed. Its average speed is only 5.0 m s−1.
CThe time is right, but the gradient of the v²–s graph has been taken as the acceleration itself; it is 2a.
DThis takes the gradient as the acceleration (5.0 m s−2) and then uses t = u/a = 10/5.0 = 2.0 s consistently with that error.
16IBP2H-002-01-03Motion graphsPaper 1AMedium1 mark
A sprinter starts from rest. Her acceleration decreases uniformly from 8.0 m s−2 at t = 0 to zero at t = 2.5 s. What is her speed at t = 2.5 s?
Acceleration of the sprinter against time
Show mark scheme
Marking point
Mark
Notes
The area under an acceleration–time graph equals the change in velocity. Here the area is a triangle: ½ × 8.0 × 2.5 = 10 m s−1. She started from rest, so her speed is 10 m s−1. Option 20 m s−1 uses a rectangle instead of a triangle. Option 3.2 is the gradient of the graph, not its area. Option 0 confuses zero acceleration with being at rest.
✓ 1
Answer C
Answer: C
Common error: Zero acceleration means constant velocity, not zero velocity. At t = 2.5 s the sprinter has stopped speeding up, but she is still moving fast.
17IBP2H-002-01-04Fluid resistance and projectilesPaper 1AMedium1 mark
A javelin is launched with the same speed and angle twice: once in a vacuum and once in air, where fluid resistance cannot be neglected. Which statement is correct for the flight in air?
Show mark scheme
Marking point
Mark
Notes
On the way up, drag and weight both act downwards, so the javelin decelerates faster than g and reaches its highest point quickly. On the way down, drag opposes the weight, so the downward acceleration is less than g and the descent takes longer. Drag has a horizontal component, so the horizontal velocity decreases. At the top the velocity is horizontal, so drag acts horizontally backwards and the total acceleration is not g vertically. Because the motion up and down differs, the path is asymmetric.
✓ 1
Answer A
Answer: A
Exam tip: With drag, the descent is steeper and slower than the ascent. To justify this, compare the direction of the drag force on the way up with its direction on the way down.
18IBP2H-002-01-05Stroboscopic analysis of a projectilePaper 1AMedium1 mark
A stroboscope flashing at 20 Hz photographs a ball in projectile motion. Six consecutive images of the ball are recorded. The horizontal distance between the first and sixth images is 0.75 m. What is the horizontal component of the ball's velocity?
Show mark scheme
Marking point
Mark
Notes
Six images span five time intervals. Each interval is 1/20 = 0.050 s, so the total time is 5 × 0.050 = 0.25 s. The horizontal velocity is 0.75 / 0.25 = 3.0 m s−1. Option 2.5 m s−1 counts six intervals instead of five. Option 0.0375 m s−1 divides the distance by the frequency. Option 15 m s−1 multiplies the full distance by the frequency, treating it as the distance covered in a single interval.
✓ 1
Answer C
Answer: C
Common error: n images span (n − 1) time intervals. Count the gaps between images, not the images themselves.
19IBP2H-002-01-06Relative velocity and vectorsPaper 1AMedium1 mark
An aircraft flies with an airspeed of 250 m s−1 and its nose points due north. A wind blows from the west at 50 m s−1. What is the direction of the aircraft's velocity relative to the ground?
Plan view; the wind blows from the west
Show mark scheme
Marking point
Mark
Notes
A wind blowing from the west pushes the aircraft towards the east. The ground velocity is the vector sum of 250 m s−1 north and 50 m s−1 east. Its angle from north is tan−1(50/250) = 11° east of north. Option 'due north' ignores the wind altogether. Option '11° west of north' misreads the wind direction. Option 79° inverts the ratio in the tangent and so measures the angle from east, not from north.
✓ 1
Answer C
Answer: C
Common error: Winds are named by where they come from. A westerly wind blows towards the east.
20IBP2H-002-00-02Motion graphsPaper 1AMedium1 mark
A test vehicle starts from rest. Its velocity is proportional to the square of the time, so that v = kt2, where k is a constant. What is the ratio
(average velocity from t = 0 to t = T) ÷ (instantaneous velocity at t = T)?
Show mark scheme
Marking point
Mark
Notes
The displacement is the area under the v–t curve: s = kT3/3. The average velocity is s/T = kT2/3. The instantaneous velocity at T is kT2, so the ratio is 1/3. Option A (1/2) holds only for uniform acceleration, where v is proportional to t. Option B (2/3) is the fraction of the enclosing rectangle that lies under a parabola drawn up to its vertex, which is a different curve. Option D treats average and instantaneous values as the same.
✓ 1
Answer C
Answer: C
Exam tip: The rule 'average velocity = (u + v)/2' applies only when the acceleration is uniform. When it is not, go back to total displacement ÷ total time.
21IBP2H-002-00-03Fluid resistance and projectilesPaper 1AMedium1 mark
A water jet from a fire hose follows a parabolic path. Air resistance is negligible. Consider a small volume of water at the highest point of its path. Which statement is correct?
Show mark scheme
Marking point
Mark
Notes
With no air resistance, the only force on the water is its weight, so the acceleration is g downwards at every point, including the top. At the highest point only the vertical component of velocity is zero. The horizontal component is unchanged, so the velocity is horizontal. Options A and C forget the horizontal velocity. Options A and B wrongly assume that the acceleration vanishes where the vertical velocity is zero.
✓ 1
Answer D
Answer: D
Common error: A zero vertical velocity does not mean a zero acceleration. The velocity is changing at that instant, so the acceleration cannot be zero.
22IBP2H-002-00-04Fluid resistance and projectilesPaper 1AMedium1 mark
A badminton shuttlecock is hit upwards at an angle and experiences significant air resistance. It lands at the height from which it was hit. Which statement about its flight is correct?
Show mark scheme
Marking point
Mark
Notes
On the way up, the weight and the drag both act downwards, so the deceleration is greater than g. On the way down, the drag opposes the weight, so the downward acceleration is less than g. The vertical distance is the same both ways, so the descent takes longer. Option A reverses this. Option C ignores the asymmetry that drag causes: the descent is steeper and the range is shorter. Option D is wrong because drag has a horizontal component that reduces the horizontal velocity.
✓ 1
Answer B
Answer: B
Exam tip: Compare the net vertical acceleration on the way up with that on the way down. The equal vertical distance then tells you which stage takes longer.
23IBP2H-002-00-06Instantaneous and average valuesPaper 1AMedium1 mark
An astronaut on the surface of an asteroid with no atmosphere launches a small ball vertically upwards at 2.0 m s−1. The ball returns to the launch point 8.0 s later. What is the acceleration of free fall on the asteroid?
Show mark scheme
Marking point
Mark
Notes
The velocity changes from +2.0 m s−1 to −2.0 m s−1, a change of magnitude 4.0 m s−1, in 8.0 s. So g = 4.0/8.0 = 0.50 m s−2. Equivalently, the rise takes 4.0 s and g = 2.0/4.0. Option A divides the launch speed by the whole flight time. Option C uses the change in velocity divided by the rise time only. Option D inverts the ratio.
✓ 1
Answer B
Answer: B
Exam tip: For a symmetrical vertical flight, the total change in velocity is 2u. Alternatively, use only the rising half of the motion.
25A-1A-36Fluid resistance and projectilesPaper 1AHard1 mark
A badminton shuttlecock is hit upwards at an angle to the horizontal. Air resistance on the shuttlecock is significant.
What is the acceleration of the shuttlecock at the highest point of its path?
Show mark scheme
Marking point
Mark
Notes
Step 1At the highest point the velocity is horizontal (vertical component zero, horizontal component not zero).
—
All 3 steps must be completed — there is no mark for a part-answer.
Step 2Two forces act: the weight mg vertically downwards and the air resistance horizontally backwards, opposite to the velocity.
—
Step 3The resultant is the vector sum, so the acceleration has a vertical component g and a horizontal component backwards: its magnitude is √(g² + (Fd/m)²) > g and it points downwards and backwards.
✓ 1
Answer D
Answer: D · 3 stages of work, one mark
Every option, and why
AOnly the vertical component of the velocity is zero at the top; the weight still acts, so the acceleration cannot be zero.
BAt the top the air resistance is horizontal, so it cannot reduce the vertical component of the acceleration below g.
CThis would be true without air resistance; here the horizontal drag adds a backwards component to the acceleration.
DCorrect: weight (down) plus drag (backwards) give a resultant larger than mg, directed downwards and backwards.
26IBP2H-002-01-07Fluid resistance and projectilesPaper 1AHard1 mark
A skydiver falls vertically. The drag force on her is proportional to the square of her speed. Her terminal speed is vT. What is the magnitude of her acceleration when her speed is 0.5vT?
Show mark scheme
Marking point
Mark
Notes
At terminal speed the drag equals the weight: kvT2 = mg. At speed 0.5vT, the drag is k(0.5vT)2 = 0.25mg. The resultant downward force is mg − 0.25mg = 0.75mg, so a = 0.75g. Option 0.50g assumes drag is proportional to speed, not speed squared. Option 0.25g gives the drag as a fraction of the weight instead of the resultant. Option g ignores drag.
✓ 1
Answer C
Answer: C
Exam tip: Use the terminal-speed condition to express the drag as a fraction of the weight. Then subtract that drag from the weight to find the resultant force.
27IBP2H-002-01-08Relative velocity and vectorsPaper 1AHard1 mark
Two stones, X and Y, are launched at the same instant from the same point with different velocities. Air resistance is negligible. How does stone Y move relative to stone X while both are in flight?
Show mark scheme
Marking point
Mark
Notes
Both stones have the same acceleration, g downwards. The relative acceleration aY − aX is therefore zero. The relative velocity stays equal to the difference in their launch velocities, which is constant and non-zero. So, relative to X, stone Y moves in a straight line at constant velocity. It moves along a parabola only relative to the ground. Its relative acceleration is zero, not g. It is not at rest, because the launch velocities differ.
✓ 1
Answer C
Answer: C
Exam tip: When two objects share the same acceleration, their relative motion is uniform. This makes many interception problems much simpler.
28IBP2H-002-00-05Equations of uniformly accelerated motionPaper 1AHard1 mark
A sensor package is released from rest above the surface of a planet with no atmosphere. It falls a distance d during the first second. What distance does it fall during the third second of its motion?
Show mark scheme
Marking point
Mark
Notes
From rest, s = ½at2, so the total distances after 1 s, 2 s and 3 s are d, 4d and 9d. The distance fallen during the third second is 9d − 4d = 5d. The distances in successive equal time intervals follow the ratio 1 : 3 : 5. Option D is the total distance after 3 s. Option B is the total distance after 2 s. Option A is the distance fallen in the second second.
✓ 1
Answer C
Answer: C
Common error: 'During the third second' means the interval from t = 2 s to t = 3 s. It does not mean the total time up to t = 3 s.
A cyclist stops pedalling on level ground. The cyclist then decelerates with a deceleration whose magnitude is proportional to the cyclist's speed. Which statement describes the subsequent motion?
Show mark scheme
Marking point
Mark
Notes
The gradient of a v–t graph is the acceleration. The deceleration is proportional to the speed, and the speed is falling, so the gradient becomes less steep as time passes. The curve flattens towards v = 0. Option A would need a constant deceleration. Option C is wrong because the s–t gradient is the speed, which decreases. Option D would need a constant speed.
✓ 1
Answer B
Answer: B
Exam tip: For non-uniform motion, translate each graph feature into a quantity: the v–t gradient is acceleration and the s–t gradient is velocity. Then ask how that quantity changes.
30A-1B-10Stroboscopic analysis of a projectilePaper 1BEasy7 marks
A small steel ball rolls off the edge of a horizontal bench and is photographed against a square grid using a stroboscope that flashes every 0.10 s. The first image (gold) is at the instant the ball leaves the bench. The horizontal position x and the vertical distance fallen y of each image are read from the photograph, each with an uncertainty of ±0.01 m. Air resistance is negligible.
The figure shows the photograph and the table gives the measurements.
t / s
0
0.10
0.20
0.30
0.40
0.50
x / m
0
0.20
0.40
0.60
0.80
1.00
y / m
0
0.05
0.20
0.44
0.78
1.23
Successive images of the ball at intervals of 0.10 s; the first image (gold) is at the moment of launch.
(a)
(i)
Determine the initial speed of the ball, using the full range of the data.
(1)
(b)
(i)
Test, using at least three images, the hypothesis that the vertical motion is uniformly accelerated from rest.
(2)
(ii)
Hence determine the acceleration of free fall.
(1)
(iii)
Determine the speed of the ball at t = 0.50 s and the angle that its velocity makes with the horizontal.
(2)
(iv)
Explain why the image at 0.50 s gives a more reliable value of the acceleration than the image at 0.10 s.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
u = 1.00/0.50 = 2.0 m s−1 (constant horizontal velocity: equal horizontal spacing)
✓ 1
Using a single 0.10 s interval (0.20/0.10) also gives 2.0 m s−1 but must be justified by the equal spacing.
Part (b)(i)
For uniform acceleration from rest y = ½at², so y/t² should be constant; calculates at least three values, e.g. 5.00, 4.89, 4.88, 4.92 m s−2
✓ 1
Equivalent: successive differences in y increase by a constant 0.10 m. Two points only scores [1 max].
The values are the same (≈ 4.9 m s−2) within the reading uncertainty, so the hypothesis is supported
✓ 1
Conclusion must be consistent with the values.
Part (b)(ii)
g = 2 × y/t², e.g. 2 × 1.23/0.50² = 9.8 m s−2
✓ 1
Allow ECF from (b)(i). Accept 9.6–10.0 m s−2.
Part (b)(iii)
vy = gt = 9.8 × 0.50 = 4.9 m s−1; speed = √(2.0² + 4.9²) = 5.3 m s−1
✓ 1
Allow ECF from (a)(i) and (b)(ii).
Angle = tan−1(4.9/2.0) = 68° below the horizontal
✓ 1
Allow ECF. Accept 67–69°.
Part (b)(iv)
The reading uncertainty ±0.01 m is the same for every image: it is 20 % of 0.05 m but under 1 % of 1.23 m (and the timing is equally precise), so the fractional uncertainty in g is much smaller for the larger distance
✓ 1
Allow ECF from (b)(ii). Must compare fractional (percentage) uncertainties.
Answers: (a)(i) 2.0 m s−1 · (b)(i) y/t² ≈ 4.9 m s−2 · (b)(ii) 9.8 m s−2 · (b)(iii) 5.3 m s−1 at 68° below the horizontal (the remaining parts are explanations — see the table above)
31A-1B-24Equations of uniformly accelerated motionPaper 1BEasy7 marks
A steel ball is held by an electromagnet above a trapdoor switch. Switching off the electromagnet starts an electronic timer; the timer stops when the ball hits the trapdoor. The height h fallen is measured with a metre rule to ±0.002 m, and each time t is the mean of five drops with an uncertainty of ±0.003 s. Air resistance is negligible.
The magnetism of the electromagnet takes a short, constant time t0 to decay, so the ball is released slightly after the timer starts. The time recorded is then t = √(2/g) √h + t0. The graph shows t against √h.
h / m
√h / m½
t / s
0.200
0.447
0.242
0.400
0.632
0.326
0.600
0.775
0.389
0.800
0.443
1.000
1.000
0.491
1.200
1.095
0.533
Graph drawn to scale. The line of best fit has been extended to √h = 0.
(a)
(i)
Calculate √h for h = 0.800 m, together with its absolute uncertainty.
(2)
(ii)
Determine the gradient of the line of best fit. Give its unit.
(2)
(b)
(i)
Hence determine g.
(2)
(ii)
Determine t0, and state why it does not affect your answer to (b)(i).
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
√0.800 = 0.894 m½
✓ 1
Percentage uncertainty halves for a square root: ½ × (0.002/0.800) × 100 % = 0.125 %, so Δ√h = ±0.001 m½
✓ 1
Accept ±0.0011. ±0.002 m½ (uncertainty of h copied unchanged) scores 0 for this mark.
Part (a)(ii)
Gradient using the full range of the line, e.g. (0.533 − 0.242)/(1.095 − 0.447) = 0.449
✓ 1
Accept 0.44–0.46. A triangle spanning less than half the line scores [1 max].
Unit: s m−½
✓ 1
Accept s/m½ or s m−0.5.
Part (b)(i)
Gradient = √(2/g), so g = 2/gradient²
✓ 1
Allow ECF from (a)(ii).
g = 2/0.449² = 9.9 m s−2
✓ 1
Accept 9.4–10.4 m s−2 (from a gradient in the range 0.44–0.46). g = 1/gradient² or 2 × gradient scores [1 max].
Part (b)(ii)
Intercept on the t axis: t0 = 0.04 s (accept 0.035–0.045 s); it adds the same time to every reading, so it shifts the line without changing its gradient
✓ 1
Allow ECF from (a)(ii) for an intercept calculated from the candidate's gradient and one data point. Both the value and the reason are needed.
Answers: (a)(i) 0.894 ± 0.001 m½ · (a)(ii) 0.449 s m−½ · (b)(i) 9.9 m s−2 · (b)(ii) t0 ≈ 0.04 s (the remaining parts are explanations — see the table above)
32A-1B-15Non-uniform accelerationPaper 1BHard7 marks
An electric racing kart and its driver, of total mass 180 kg, start from rest on a straight level track. A radar gun records the speed v at times t; each speed is ±0.2 m s−1. The motor delivers a constant mechanical power P and resistive forces may be neglected, so PΔt = Δ(½mv²) and a graph of v² against t has gradient 2P/m.
During the run the battery supplies a current of 160 A at a terminal potential difference of 72.0 V.
t / s
1.0
1.5
2.0
2.5
3.0
3.5
v / m s−1
7.7
10.5
12.7
14.5
16.2
17.5
Graph drawn to scale. The point for t = 3.5 s has not been plotted.
(a)
(i)
Calculate v² at t = 3.5 s together with its absolute uncertainty.
(2)
(ii)
Determine P, using the full range of the data.
(2)
(iii)
Determine the efficiency with which the motor transfers energy from the battery to the kinetic energy of the kart.
(1)
(iv)
The line of best fit crosses the time axis at about 0.4 s instead of passing through the origin. The maximum forward force that friction between the tyres and the track can provide is 1.4 × 103 N. Explain the intercept, using your answer to (a)(ii).
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
v² = 17.5² = 306 m² s−2
✓ 1
Δ(v²) = 2 × (0.2/17.5) × 306 = ±7 m² s−2
✓ 1
Absolute uncertainty to 1 s.f. ±3.5 (not doubled) scores 0 for this mark.
Part (a)(ii)
Gradient of the line from (1.0, 59) to (3.5, 306): (306 − 59)/2.5 = 99 m² s−3
Allow ECF from (a)(ii). Accept 0.75–0.82 (or 75–82 %).
Part (a)(iv)
Delivering power P needs a driving force F = P/v, which exceeds 1.4 × 103 N below v = 8.9 × 103/1.4 × 103 = 6.4 m s−1
✓ 1
Allow ECF from (a)(ii). Accept 6.1–6.7 m s−1.
So at first the force is limited (wheels would spin) and the power delivered (Fv) is less than P: less kinetic energy is gained than the model predicts, so the early v² values are low and the line is displaced to later times
✓ 1
Do not accept "reaction time" or "random error".
Answers: (a)(i) 306 ± 7 m² s−2 · (a)(ii) 8.9 × 103 W · (a)(iii) 0.77 · (a)(iv) 6.4 m s−1(the remaining parts are explanations — see the table above)
A lift in an office building starts from rest and rises vertically. It accelerates uniformly at 1.2 m s−2 for 2.5 s. It then moves at constant speed for 8.0 s, and finally decelerates uniformly to rest in 3.0 s.
(a)
Calculate the maximum speed of the lift.
(1)
(b)
Determine the total distance travelled by the lift.
(3)
(c)
Calculate the average speed of the lift for the whole journey.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
v = u + at = 0 + 1.2 × 2.5 = 3.0 m s−1
✓ 1
Part (b)
Accelerating stage: ½ × 3.0 × 2.5 = 3.75 m
✓ 1
Constant speed stage: 3.0 × 8.0 = 24 m, and decelerating stage: ½ × 3.0 × 3.0 = 4.5 m
✓ 1
Total = 32.25 m ≈ 32 m
✓ 1
ECF from (a).
—
Part (c)
Total time = 2.5 + 8.0 + 3.0 = 13.5 s
✓ 1
Average speed = 32.25/13.5 = 2.4 m s−1
✓ 1
ECF from (b).
—
Exam tip: Divide multi-stage journeys into stages. The area under each stage of the v–t graph gives the distance for that stage.
An astronaut strikes a golf ball on the Moon. The ball leaves the level ground at 28.0 m s−1 at 35.0° to the horizontal. The Moon has mass 7.35 × 1022 kg and radius 1.74 × 106 m and no atmosphere.
Height against horizontal distance for the same strike on the Earth (drawn to scale).
(a)
(i)
Show that the gravitational field strength at the surface of the Moon is about 1.6 N kg−1.
(2)
(ii)
Calculate the time of flight of the ball.
(2)
(iii)
Calculate the horizontal distance travelled by the ball.
(1)
(b)
(i)
Show that, for the same launch velocity and no air resistance, the ratio of the range on the Moon to the range on the Earth is gEarth/gMoon.
(2)
(ii)
The graph shows the path of the same strike on the Earth with and without air resistance. Use the graph to determine the percentage reduction in range caused by air resistance.
(2)
(iii)
Outline two ways, other than the range, in which the path with air resistance differs from the path without it, and give a reason for each.
(2)
(c)
(i)
Explain why the ball cannot escape from the Moon, however hard a golf ball could be struck by an astronaut.
The graph shows how the velocity of a sprinter of mass 70 kg varies with time during the first 6.0 s of a race along a straight, level track.
Velocity v of the sprinter against time t (drawn to scale).
(a)
(i)
Determine the instantaneous acceleration of the sprinter at t = 1.0 s.
(2)
(ii)
Estimate the distance run in the first 4.0 s.
(2)
(iii)
Hence calculate the average speed over the first 4.0 s.
(1)
(b)
(i)
Calculate the resultant horizontal force on the sprinter at t = 1.0 s.
(1)
(ii)
Explain, with reference to Newton's third law, how the sprinter is able to accelerate forwards.
(2)
(iii)
Use the graph to determine the minimum coefficient of static friction between the sprinter's shoes and the track at the start. Assume that the vertical force from the track equals the sprinter's weight.
(2)
(c)
(i)
Determine the kinetic energy of the sprinter at t = 6.0 s and the minimum average power developed during the first 6.0 s.
(2)
(ii)
The sprinter's leg muscles transfer chemical energy to the kinetic energy of the sprinter with an efficiency of 25 %. Estimate the rise in temperature of the 30 kg of leg muscle during the 6.0 s. The specific heat capacity of muscle is 3.5 × 103 J kg−1 K−1.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Tangent drawn at t = 1.0 s and its gradient found from a large triangle
✓ 1
a = 3.7 m s−2
✓ 1
Accept 3.2–4.2 m s−2.
Part (a)(ii)
Area under the graph from 0 to 4.0 s, by counting squares or trapezia
✓ 1
Distance ≈ 28 m
✓ 1
Accept 26–30 m.
Part (a)(iii)
28/4.0 = 7.0 m s−1
✓ 1
Allow ECF from (a)(ii).
Part (b)(i)
F = 70 × 3.7 = 2.6 × 102 N
✓ 1
Allow ECF from (a)(i). Accept 220–290 N.
Part (b)(ii)
The sprinter's foot pushes backwards on the track (a frictional force)
✓ 1
By Newton's third law the track pushes forwards on the sprinter with an equal force; this external force on the sprinter causes the acceleration
✓ 1
A reference to the forces acting on different bodies is needed.
Part (b)(iii)
Initial acceleration = gradient of the tangent at t = 0 ≈ 7.5 m s−2
✓ 1
Accept 6.8–8.0 m s−2.
Friction provides ma and Ff ≤ μmg, so μ ≥ a/g = 7.5/9.81 = 0.76
✓ 1
Accept 0.69–0.82, consistent with the gradient read.
Part (c)(i)
Ek = ½ × 70 × 10.4² = 3.8 × 103 J
✓ 1
Accept 10.3–10.5 m s−1 read from the graph (3.7–3.9 × 103 J).
Average power ≥ 3.79 × 103/6.0 = 6.3 × 102 W
✓ 1
Allow ECF from the kinetic energy. Minimum because work against air resistance is ignored.
Part (c)(ii)
Thermal energy = 3 × 3.79 × 103 = 1.1 × 104 J (75 % of the input, three times the useful output)
✓ 1
Allow ECF from (c)(i). Using 75 % of the kinetic energy instead of 3 × scores [1 max].
ΔT = 1.14 × 104/(30 × 3.5 × 103) = 0.11 K
✓ 1
Accept 0.10–0.12 K.
Answers: (a)(i) 3.7 m s−2 · (a)(ii) 28 m · (a)(iii) 7.0 m s−1 · (b)(i) 2.6 × 102 N · (b)(iii) 0.76 · (c)(i) 3.8 × 103 J; 6.3 × 102 W · (c)(ii) 0.11 K (the remaining parts are explanations — see the table above)
In an ink-jet printer, a negatively charged ink droplet of mass 1.2 × 10−10 kg and charge −1.8 × 10−13 C enters the region between two horizontal parallel plates, midway between them, moving horizontally at 20 m s−1. The plates are 16 mm long and 2.0 mm apart, and the potential difference between them is 1.5 kV, with the upper plate positive. The paper is 12 mm beyond the end of the plates, as shown. The field between the plates is uniform and there is no field outside them. Air resistance is negligible.
Droplet deflected between the charged plates (not to scale).
(a)
(i)
Show that the electric force on the droplet is about 1.4 × 10−7 N.
(2)
(ii)
Show that the effect of gravity on the motion of the droplet can be ignored.
(1)
(b)
(i)
Calculate the time the droplet spends between the plates and its acceleration there.
(2)
(ii)
Determine the vertical displacement of the droplet as it leaves the plates.
(2)
(iii)
Calculate the vertical component of the velocity of the droplet as it leaves the plates.
(1)
(iv)
Determine the total vertical displacement of the droplet, from its line of entry, when it reaches the paper.
(3)
(c)
(i)
Show that the vertical displacement of a droplet as it leaves the plates is y = qVL²/(2mdu²), where L is the length of the plates and u the speed of entry. Hence state the effect on this displacement of doubling the speed of entry.
(2)
(ii)
Calculate the gain in kinetic energy of the droplet while it is between the plates.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
E = V/d = 1.5 × 103/2.0 × 10−3 = 7.5 × 105 V m−1
✓ 1
Field strength must be calculated with d in m.
F = qE = 1.8 × 10−13 × 7.5 × 105 = 1.35 × 10−7 N
✓ 1
The answer must be given to at least 3 s.f., or the full substitution shown.
Part (a)(ii)
Weight = 1.2 × 10−10 × 9.81 = 1.2 × 10−9 N, about 1 % of (or about 115 times smaller than) the electric force
✓ 1
Allow ECF from (a)(i). A comparison is needed.
Part (b)(i)
t = 16 × 10−3/20 = 8.0 × 10−4 s: the horizontal velocity is constant because the force is vertical
✓ 1
a = F/m = 1.35 × 10−7/1.2 × 10−10 = 1.1 × 103 m s−2 (upwards)
✓ 1
Allow ECF from (a)(i). Accept 1125 m s−2; 1.17 × 103 m s−2 from 1.4 × 10−7 N.
Part (b)(ii)
The droplet starts with zero vertical velocity: y = ½at²
✓ 1
Allow ECF from (b)(i).
y = ½ × 1125 × (8.0 × 10−4)² = 3.6 × 10−4 m (0.36 mm), towards the upper plate
✓ 1
Accept 3.6–3.8 × 10−4 m.
Part (b)(iii)
vy = at = 1125 × 8.0 × 10−4 = 0.90 m s−1
✓ 1
Allow ECF from (b)(i). Accept 0.93–0.94 m s−1 from a = 1.17 × 103 m s−2.
Part (b)(iv)
Beyond the plates no force acts, so the droplet moves in a straight line at constant velocity (Newton's first law); time to the paper = 12 × 10−3/20 = 6.0 × 10−4 s
✓ 1
Award this mark for the time with any valid reason for straight-line motion, or for a similar-triangles method.
Extra displacement = 0.90 × 6.0 × 10−4 = 5.4 × 10−4 m
✓ 1
Allow ECF from (b)(iii).
Total = 3.6 × 10−4 + 5.4 × 10−4 = 9.0 × 10−4 m (0.90 mm)
✓ 1
Allow ECF from (b)(ii). Accept 9.0–9.4 × 10−4 m. Continuing with acceleration beyond the plates scores [2 max].
Part (c)(i)
a = qV/md and t = L/u, so y = ½(qV/md)(L/u)² = qVL²/(2mdu²)
✓ 1
Both substitutions must be seen.
y ∝ 1/u², so doubling the speed reduces the displacement to a quarter
✓ 1
Accept "divided by 4".
Part (c)(ii)
Gain = ½mvy² = ½ × 1.2 × 10−10 × 0.90² = 4.9 × 10−11 J
✓ 1
Allow ECF from (b)(iii). Accept the work done by the electric force, Fy = 1.35 × 10−7 × 3.6 × 10−4 = 4.9 × 10−11 J, Allow ECF from (b)(ii).
Answers: (a)(i) 1.35 × 10−7 N · (b)(i) 8.0 × 10−4 s; 1.1 × 103 m s−2 · (b)(ii) 3.6 × 10−4 m · (b)(iii) 0.90 m s−1 · (b)(iv) 9.0 × 10−4 m · (c)(ii) 4.9 × 10−11 J (the remaining parts are explanations — see the table above)
37IBP2H-002-01-10Relative velocity and vectorsPaper 2Medium6 marks
A medical-supply drone flies at an airspeed of 18 m s−1. It must deliver blood samples to a hospital 2.4 km due east of its base. A steady wind blows towards the south at 6.0 m s−1.
Plan view, not to scale
(a)
Determine the direction in which the drone must point so that it travels due east relative to the ground.
(2)
(b)
Calculate the speed of the drone relative to the ground.
(2)
(c)
Calculate the time taken to reach the hospital.
(1)
(d)
Outline why the journey takes longer than it would in still air.
(1)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The northward component of the airspeed must cancel the wind: 18 sin θ = 6.0
✓ 1
θ = 19.5° ≈ 19° north of east
✓ 1
Part (b)
v = √(18² − 6.0²) or v = 18 cos 19.5°
✓ 1
v = 17 m s−1 (16.97 m s−1)
✓ 1
Part (c)
t = 2400 / 16.97 = 141 s ≈ 140 s ECF from (b)
✓ 1
Part (d)
Part of the drone's velocity relative to the air is used to cancel the wind, so its eastward component is less than 18 m s−1
✓ 1
Common error: Do not add the wind speed to the airspeed using Pythagoras. Here the airspeed is the hypotenuse of the vector triangle, so the ground speed is √(18² − 6²).
38IBP2H-002-01-11Stroboscopic analysis of a projectilePaper 2Medium8 marks
A ball rolls horizontally off a bench and is photographed with a stroboscope flashing at 25 Hz. Image 0 is taken at the instant the ball leaves the bench. The table gives the horizontal distance x and the vertical drop y for each image.
Show that the horizontal velocity of the ball is constant and determine its value.
(2)
(b)
Use the data to show that the vertical acceleration is uniform and determine its value.
(3)
(c)
Determine the velocity of the ball at image 5.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The x-values increase by the same 6.0 cm in each interval, so the horizontal velocity is constant
✓ 1
Time interval = 1/25 = 0.040 s, so vx = 0.060 / 0.040 = 1.5 m s−1
✓ 1
Part (b)
Successive vertical displacements are 0.78, 2.36, 3.92, 5.48 and 7.06 cm
✓ 1
Their differences (1.58, 1.56, 1.56 and 1.58 cm) are constant within experimental uncertainty, so the acceleration is uniform
✓ 1
a = Δ(Δy) / T² = 0.0157 / 0.040² ≈ 9.8 m s−2, using the mean difference of 1.57 cm
✓ 1
Part (c)
t = 0.20 s and vy = at = 9.8 × 0.20 = 1.96 m s−1 (accept vy = 2y/t = 2 × 0.196 / 0.20; allow error carried forward from the acceleration found earlier)
✓ 1
v = √(1.5² + 1.96²) ≈ 2.5 m s−1
✓ 1
Direction: tan θ = 1.96/1.5, so θ ≈ 53° below the horizontal
✓ 1
Exam tip: For equal time intervals T, a constant second difference in displacement shows uniform acceleration, with a = Δ(Δy)/T². Always convert centimetres to metres before calculating, and give both the magnitude and the direction when asked for a velocity.
39IBP2H-002-01-13Projectile motionPaper 2Medium8 marks
An astronaut on the Moon strikes a golf ball. It leaves the ground at 40 m s−1 at 35° above the horizontal. The Moon has no atmosphere. The acceleration of free fall on the Moon is 1.62 m s−2. The ground is level.
Golf ball struck on level ground on the Moon (not to scale)
(a)
Calculate the time of flight of the ball.
(2)
(b)
Calculate the maximum height reached by the ball.
(2)
(c)
Determine the horizontal range of the ball.
(2)
(d)
State and explain the launch angle that would give the greatest range on the Moon for the same launch speed.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Initial vertical velocity uy = 40 sin 35° = 22.9 m s−1
✓ 1
t = 2uy / g = 2 × 22.9 / 1.62 = 28 s (28.3 s); any valid equivalent method, e.g. 0 = uyt − ½gt², is accepted
✓ 1
Part (b)
H = uy2 / 2g = 22.92 / (2 × 1.62), or an equivalent method such as H = ½uy(t/2)
✓ 1
H = 160 m (162 m); allow ECF for the value of uy from (a)
✓ 1
Part (c)
Horizontal velocity ux = 40 cos 35° = 32.8 m s−1, which stays constant because there is no horizontal force
✓ 1
R = 32.8 × 28.3 = 930 m (928 m); allow ECF for the time of flight from (a)
✓ 1
Part (d)
R = u2 sin 2θ / g, valid for level ground with no air resistance
✓ 1
R is greatest when sin 2θ = 1, so θ = 45°. The angle does not depend on g, because g only scales the range
✓ 1
Common error: Use the value of g given for the Moon, not 9.8 m s⁻², in every part.
40IBP2H-002-01-16Fluid resistance and projectilesPaper 2Medium9 marks
In a falling-sphere viscometer, a small solid plastic sphere is released from rest in a tall tube of glycerol. The plastic is only slightly denser than glycerol. A video records the distance fallen x at time t.
t / s: 0.0, 0.2, 0.4, 0.6, 0.8, 1.0 x / cm: 0.0, 1.6, 4.6, 7.8, 11.0, 14.2
(a)
Outline how the data show that the sphere reaches terminal speed.
(2)
(b)
Determine the terminal speed of the sphere.
(2)
(c)
Explain, in terms of the forces acting, why the speed of the sphere becomes constant.
(3)
(d)
The sphere is replaced by a sphere of the same plastic but with a larger radius. Explain the effect on the terminal speed.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The distances fallen in successive 0.2 s intervals are 1.6, 3.0, 3.2, 3.2 and 3.2 cm
✓ 1
From t = 0.4 s onwards, equal distances are covered in equal times, so the speed is constant
✓ 1
Part (b)
v = 3.2 cm ÷ 0.2 s, or (14.2 − 4.6) cm ÷ 0.6 s
✓ 1
v = 0.16 m s−1
✓ 1
Part (c)
The weight acts downwards; the buoyancy (upthrust) and the viscous drag act upwards
✓ 1
The drag increases as the speed increases, so the resultant force and therefore the acceleration decrease
✓ 1
When drag + buoyancy = weight, the resultant force is zero, so the acceleration is zero and the speed stays constant
✓ 1
Part (d)
Weight minus buoyancy is proportional to r³, but the drag (F = 6πηrv) is proportional to r only, for a given speed; or: the downward force increases more than the drag does at the same speed
✓ 1
So a larger speed is needed for the drag to balance the forces, and the terminal speed increases (v ∝ r²)
✓ 1
Common error: Do not divide the total distance by the total time to find the terminal speed. Use only the interval in which equal distances are covered in equal times.
41IBP2H-002-00-10Motion graphsPaper 2Medium10 marks
A sports scientist uses a radar gun to measure the speed of a sprinter during the first 6.0 s of a race. The sprinter starts from rest.
t / s : 0, 1.0, 2.0, 3.0, 4.0, 5.0, 6.0 v / m s−1 : 0, 4.2, 7.0, 8.8, 9.8, 10.3, 10.5
Speed of the sprinter against time
(a)
Calculate the average acceleration of the sprinter between t = 0 and t = 2.0 s.
(1)
(b)
Explain how the data show that the acceleration of the sprinter is not uniform.
(2)
(c)
Estimate the distance run in the first 6.0 s.
(3)
(d)
Estimate the instantaneous acceleration of the sprinter at t = 3.0 s.
(2)
(e)
State and explain whether your estimate in (c) is likely to be greater or smaller than the true distance.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
a = (7.0 − 0)/2.0 = 3.5 m s−2
✓ 1
Part (b)
The increase in speed in each successive 1.0 s interval is 4.2, 2.8, 1.8, 1.0, 0.5 and 0.2 m s−1
✓ 1
The change in velocity per equal time interval decreases, so the acceleration decreases with time
✓ 1
Part (c)
The distance is found from the area under the v–t data, using trapezia of width 1.0 s (or an equivalent method)
✓ 1
Area = 2.1 + 5.6 + 7.9 + 9.3 + 10.05 + 10.4
✓ 1
Distance ≈ 45 m
✓ 1
Part (d)
Use the change in velocity across the interval centred on 3.0 s: (9.8 − 7.0)/(4.0 − 2.0)
✓ 1
a ≈ 1.4 m s−2
✓ 1
Accept values from 1.3 to 1.5 m s−2 from a valid tangent or difference method.
—
Part (e)
The estimate is smaller than the true distance (an underestimate)
✓ 1
The v–t curve has a decreasing gradient and bulges upwards, so the straight tops of the trapezia lie below the curve and miss some area
✓ 1
Exam tip: For an instantaneous value from tabulated data, use a small interval centred on the time asked for. Do not use an interval that starts at zero.
42IBP2H-002-00-11Projectile motionPaper 2Medium8 marks
A basketball player releases a ball at a height of 2.10 m above the floor. The ball leaves with a speed of 7.60 m s−1 at 52.0° above the horizontal. The rim of the hoop is 3.05 m above the floor and 4.60 m horizontally from the release point. Air resistance is negligible. Take g = 9.8 m s−2.
Set-up for the throw (not to scale)
(a)
Calculate the horizontal and vertical components of the initial velocity.
(2)
(b)
Calculate the time taken for the ball to reach its highest point.
(1)
(c)
Determine the maximum height of the ball above the floor.
(2)
(d)
Determine the height of the ball above the floor at the instant it reaches the horizontal position of the rim, 4.60 m from the release point. Hence state whether the ball is rising or falling at that instant.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Horizontal: 7.60 cos 52.0° = 4.68 m s−1
✓ 1
Vertical: 7.60 sin 52.0° = 5.99 m s−1
✓ 1
Part (b)
t = 5.99/9.8 = 0.611 s
✓ 1
ECF from (a).
—
Part (c)
Rise above release = (5.99)2/(2 × 9.8) = 1.83 m
✓ 1
Maximum height = 2.10 + 1.83 = 3.93 m
✓ 1
ECF from (a).
—
Part (d)
Time to travel 4.60 m horizontally: t = 4.60/4.68 = 0.983 s
✓ 1
Height = 2.10 + 5.99 × 0.983 − ½ × 9.8 × 0.9832 = 3.25 m
✓ 1
Falling, because 0.983 s is later than the 0.611 s needed to reach the highest point. Equivalent valid reasoning also earns this mark, for example that the vertical velocity 5.99 − 9.8 × 0.983 is negative
✓ 1
Accept answers obtained with g = 9.81 m s−2. ECF from (a) and (b).
—
Exam tip: Find the time from the horizontal motion first, because the horizontal velocity is constant. Then use that time in the vertical equation, and remember to add the release height.
43IBP2H-002-00-15Motion graphsPaper 2Medium8 marks
Trackside sensors A, B and C monitor an electric train that accelerates uniformly along a straight track. A is 120 m from B, and B is a further 120 m from C. The front of the train takes 6.0 s to travel from A to B and 4.0 s to travel from B to C.
Train approaching sensors A, B and C (not to scale)
(a)
Let u be the speed of the train at A and a be its acceleration. Write two equations linking u and a, one for the motion from A to B and one for the motion from A to C.
(2)
(b)
Determine u and a.
(3)
(c)
Calculate the speed of the train at C.
(1)
(d)
Calculate the average velocity between A and C. Show that it is consistent with uniform acceleration.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
A to B: 120 = 6.0u + ½a(6.0)2, so 120 = 6u + 18a
✓ 1
A to C: 240 = 10u + ½a(10)2, so 240 = 10u + 50a
✓ 1
Part (b)
From the first equation, u = 20 − 3a. Substituting into the second gives 240 = 200 − 30a + 50a
✓ 1
a = 2.0 m s−2
✓ 1
u = 14 m s−1
✓ 1
ECF from (a).
—
Part (c)
v = 14 + 2.0 × 10 = 34 m s−1
✓ 1
ECF from (b).
—
Part (d)
Average velocity = 240/10 = 24 m s−1
✓ 1
For uniform acceleration, the average velocity is (u + v)/2 = (14 + 34)/2 = 24 m s−1, which agrees
✓ 1
ECF from (c).
—
Common error: The train is not at rest at B. Measure the time for each equation from A, or else use the speed at B as the initial speed for the B to C stage.
44IBP2H-002-00-16Projectile motionPaper 2Medium8 marks
A student on a balcony throws a small rubber ball vertically upwards at 12 m s−1. The ball leaves the hand 18 m above the ground. It rises, then falls past the balcony and lands on the ground. Air resistance is negligible.
Ball thrown vertically upwards from the balcony (not to scale)
(a)
Calculate the maximum height reached by the ball above the point of release.
(2)
(b)
Calculate the time from release until the ball hits the ground.
(2)
(c)
Determine the total distance travelled by the ball. Determine also the magnitude and direction of its displacement when it hits the ground.
(2)
(d)
Calculate the average speed and the average velocity of the ball over the whole flight.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
0 = 122 − 2 × 9.8 × h or h = v2/2g
✓ 1
h = 7.3 m (7.35 m)
✓ 1
Part (b)
Taking upwards as positive: −18 = 12t − 4.9t2, or an equivalent method such as time up plus time down
✓ 1
t = [12 + √(144 + 352.8)]/9.8 = 3.5 s (3.50 s)
✓ 1
Part (c)
Distance = 2 × 7.35 + 18 = 32.7 m ≈ 33 m
✓ 1
Displacement = 18 m downwards
✓ 1
Allow ECF from (a).
—
Part (d)
Average speed = total distance ÷ time = 32.7/3.50 = 9.3 m s−1. Accept 9.3–9.4 m s−1 if rounded values from (c) are used.
✓ 1
Average velocity = 18/3.50 = 5.1 m s−1 downwards
✓ 1
Allow ECF from (b) and (c).
—
Common error: Average speed uses the total distance, but average velocity uses the displacement. The two give different answers whenever an object changes direction.
In a compressed-air launcher, a projectile of mass 0.25 kg fits closely inside a barrel. Air behind the projectile, initially at a pressure of 6.0 × 105 Pa and volume 0.50 × 10−3 m³, pushes it along the barrel. The projectile leaves the barrel when the volume of air behind it is 2.0 × 10−3 m³. The air in front of the projectile stays at atmospheric pressure, 1.0 × 105 Pa. The graph shows how the pressure of the air behind the projectile varies with its volume.
Pressure p of the air behind the projectile against its volume V (drawn to scale).
(a)
(i)
Estimate the work done by the air behind the projectile.
(2)
(ii)
Show that the net work done on the projectile by the air on both sides of it is about 270 J.
(2)
(b)
(i)
Friction in the barrel dissipates 30 % of the net work. Calculate the speed of the projectile as it leaves the barrel.
(2)
(ii)
The launcher fires the projectile at 30° above the horizontal towards a vertical wall that is 75 m away horizontally. Determine the height above the launch point at which the projectile hits the wall. Air resistance is negligible.
(2)
(iii)
Deduce whether the projectile is rising or falling when it hits the wall.
(1)
(c)
(i)
The graph was drawn assuming the air behind the projectile stays at constant temperature. In practice the expansion is very rapid. Explain why the projectile leaves the barrel with a lower speed than predicted.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)(i)
Work = area under the p–V graph between 0.50 and 2.0 × 10−3 m³ (counting squares or trapezia)
✓ 1
Work ≈ 4.2 × 102 J
✓ 1
Accept 390–440 J. Powers of ten must be handled: one large square = 50 J.
Part (a)(ii)
The atmosphere in front does negative work p0ΔV = 1.0 × 105 × 1.5 × 10−3 = 150 J
✓ 1
The hidden step: the projectile pushes the outside air out of the barrel.
Net work = 416 − 150 = 266 J
✓ 1
Allow ECF from (a)(i). Must see a value to 3 s.f. or full working.
Part (b)(i)
Ek = 0.70 × 266 = 186 J
✓ 1
Allow ECF from (a)(ii); use 270 J if (a)(ii) was not attempted.
v = √(2 × 186/0.25) = 38.6 m s−1
✓ 1
Accept 38–39 m s−1.
Part (b)(ii)
Horizontal velocity = 38.6 cos 30° = 33.4 m s−1, constant, so the time to reach the wall = 75/33.4 = 2.24 s
✓ 1
Allow ECF from (b)(i).
Height = 19.3 × 2.24 − ½ × 9.81 × 2.24² = 19 m
✓ 1
Accept 18–19 m (17.8–19.1 m for a launch speed of 38–39 m s−1). Using the full launch speed for the vertical motion scores [1 max].
Part (b)(iii)
Vertical velocity at the wall = 19.3 − 9.81 × 2.24 = −2.7 m s−1, so it is falling (it passed its highest point at 19.3/9.81 = 1.97 s)
✓ 1
Allow ECF from (b)(ii). The conclusion must follow from a calculated vertical velocity or a comparison of times.
Part (c)(i)
The expansion is (nearly) adiabatic: there is no time for thermal energy to flow into the air
✓ 1
The air does work on the projectile at the expense of its internal energy, so its temperature falls
✓ 1
So the pressure at each volume is lower than on the isothermal curve, the area under the curve (work done) is smaller, and less kinetic energy is given to the projectile
✓ 1
Allow ECF from (a)(i): reference to a smaller area or less work needed.
Answers: (a)(i) 4.2 × 102 J · (b)(i) 38.6 m s−1 · (b)(ii) 19 m · (b)(iii) falling (the remaining parts are explanations — see the table above)
46IBP2H-002-01-12Fluid resistance and projectilesPaper 2Hard9 marks
A badminton shuttlecock has a mass of 5.0 g and a terminal speed of 6.8 m s−1. Assume that the drag force F on the shuttlecock is given by F = kv2, where k is a constant.
(a)
Determine k.
(2)
(b)
A player smashes the shuttlecock so that it leaves the racket horizontally at 60 m s−1. Calculate the magnitude of the deceleration caused by drag immediately after impact.
(2)
(c)
Explain why the trajectory of a high, lofted shuttlecock shot is strongly asymmetric, with a very steep descent.
(3)
(d)
Suggest why the shuttlecock reaches the ground at a speed close to 6.8 m s−1 even though it was hit at 60 m s−1.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
At terminal speed, the drag equals the weight: mg = kvT2
✓ 1
k = (5.0×10−3 × 9.8) / 6.82 = 1.1×10−3 kg m−1 (1.06×10−3)
✓ 1
Part (b)
F = kv2 = 1.06×10−3 × 602 = 3.8 N (allow ECF from the value of k)
✓ 1
a = F/m = 3.8 / 5.0×10−3 ≈ 7.6×102 m s−2, about 78g. Accept 760–790 m s−2, depending on how k was rounded
✓ 1
Part (c)
At high speed, drag is large compared with the weight, so the horizontal component of velocity decreases rapidly
✓ 1
As the horizontal velocity falls, the velocity becomes increasingly vertical and the path bends sharply downwards
✓ 1
On the way down, the shuttlecock falls almost vertically at close to its terminal speed, so the descent is much steeper than the ascent
✓ 1
Part (d)
At any speed above vT, drag exceeds the weight, so there is a resultant force opposing the motion and the shuttlecock decelerates
✓ 1
The deceleration is very large (tens of g at high speed), so the speed falls to about vT within a few metres, much less than the length of the flight. It then stays close to vT because drag ≈ weight
✓ 1
Exam tip: The terminal speed condition, mg = kvT², gives the drag constant directly. Keep the unrounded value of k for later parts to avoid rounding drift.
In a crash test, a car travelling at 15 m s−1 hits a rigid barrier and comes to rest in 0.100 s. The magnitude of its deceleration increases uniformly from zero to a peak value amax at t = 0.050 s. It then decreases uniformly to zero at t = 0.100 s.
Shape of the deceleration–time graph (not to scale)
(a)
Determine amax.
(2)
(b)
Express amax as a multiple of g.
(1)
(c)
Describe the shape of the velocity–time graph for the collision. Refer to the gradient at key times.
(3)
(d)
Determine the distance moved by the car during the collision.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
Change in velocity = area under the a–t graph: ½ × 0.100 × amax = 15
✓ 1
amax = 300 m s−2
✓ 1
Part (b)
300 / 9.81 ≈ 31g (accept 30.6g; ECF from (a))
✓ 1
Part (c)
Starts at 15 m s−1 at t = 0 with zero gradient and ends at 0 at t = 0.100 s with zero gradient
✓ 1
The gradient becomes increasingly negative up to t = 0.050 s, where v = 7.5 m s−1. The graph is steepest there, with gradient −300 m s−2 (point of inflection)
✓ 1
After t = 0.050 s the curve becomes less steep. It is an S-shape with rotational symmetry about (0.050 s, 7.5 m s−1)
✓ 1
Part (d)
Mean velocity is 7.5 m s−1, by the symmetry of the v–t curve about (0.050 s, 7.5 m s−1). Alternatively, by integration: 0.625 m in the first half and 0.125 m in the second half
✓ 1
d = 7.5 × 0.100 = 0.75 m
✓ 1
Exam tip: The area under an a–t graph gives Δv, and the gradient of a v–t graph gives a. Use these links to work out the shape of one graph from the other, including where the curve is steepest.
48IBP2H-002-00-12Projectile motionPaper 2Hard9 marks
A delivery drone is climbing with a velocity of 6.0 m s−1 at 30° above the horizontal. When it is 45 m above level ground, it releases a parcel. Air resistance on the parcel is negligible.
Parcel released from the drone (not to scale)
(a)
Explain why the parcel initially continues to move upwards after it is released.
(2)
(b)
Show that the parcel reaches the ground about 3.4 s after release.
(3)
(c)
Calculate the horizontal distance between the release point and the landing point.
(2)
(d)
Determine the speed of the parcel just before it hits the ground.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
At release, the parcel has the same velocity as the drone, which includes an upward component of 6.0 sin 30° = 3.0 m s−1
✓ 1
The only force is its weight, which causes a downward acceleration of 9.8 m s−2. This reduces the upward velocity gradually, so the parcel keeps rising until its vertical velocity reaches zero
✓ 1
Part (b)
Take upwards as positive: −45 = 3.0t − 4.9t2
✓ 1
Rearranged: 4.9t2 − 3.0t − 45 = 0, so t = [3.0 + √(9.0 + 882)]/9.8
Allow ECF from (b) for the vertical velocity and from the horizontal velocity found in (c).
—
Common error: An object released from a moving vehicle shares the vehicle's velocity at release. Do not treat the parcel as starting from rest, or as moving purely horizontally.
A rocket sled on a straight test track starts from rest. Its acceleration decreases linearly with time, from 20 m s−2 at t = 0 to zero at t = 4.0 s. After t = 4.0 s the acceleration remains zero.
Acceleration of the sled against time
(a)
Determine the velocity of the sled at t = 4.0 s.
(2)
(b)
Describe and explain the shape of the velocity–time graph between t = 0 and t = 4.0 s.
(2)
(c)
Explain why the displacement of the sled in the first 4.0 s is greater than 80 m.
(2)
(d)
The v–t curve between t = 0 and t = 4.0 s is a parabola with its vertex at t = 4.0 s. The area under such a curve is two-thirds of the area of the rectangle that encloses it. Determine the average velocity of the sled over the first 4.0 s.
(2)
Show mark scheme
Marking point
Mark
Notes
Part (a)
The change in velocity equals the area under the a–t graph, which is a triangle: ½ × 20 × 4.0
✓ 1
v = 40 m s−1
✓ 1
Part (b)
The graph is a curve rising from the origin, with a gradient that decreases until it is horizontal at t = 4.0 s
✓ 1
The gradient of a v–t graph is the acceleration, which decreases steadily to zero
✓ 1
Part (c)
A uniform acceleration to 40 m s−1 in 4.0 s would give a displacement of ½ × 40 × 4.0 = 80 m
✓ 1
Here the velocity rises more quickly at first, so the curve lies above the straight line at all times between 0 and 4.0 s and the area under it is greater
✓ 1
Part (d)
Displacement = ⅔ × 40 × 4.0 = 107 m
✓ 1
Average velocity = 107/4.0 = 27 m s−1
✓ 1
ECF from (a).
—
Exam tip: With non-uniform acceleration, do not use the suvat equations. Use the areas and gradients of graphs instead: the area under a–t gives Δv, and the area under v–t gives displacement.
50IBP2H-002-00-14Projectile motionPaper 2Hard8 marks
A ski jumper leaves the end of a take-off ramp horizontally at 24 m s−1. The jumper lands on a straight landing slope that starts at the end of the ramp and descends at 35° below the horizontal. Air resistance and lift are negligible.
Ski jumper leaving the ramp (not to scale)
(a)
Show that the time of flight t is given by t = 2u tan θ / g, where u is the take-off speed and θ is the angle of the slope.
(2)
(b)
Calculate the time of flight.
(1)
(c)
Calculate the distance from take-off to landing, measured along the slope.
(2)
(d)
Show that the angle between the jumper's velocity and the horizontal at landing does not depend on the take-off speed, and calculate this angle.
(3)
Show mark scheme
Marking point
Mark
Notes
Part (a)
At landing, the horizontal displacement is ut and the vertical drop is ½gt2
✓ 1
The landing point is on the slope, so tan θ = ½gt2/(ut), which rearranges to t = 2u tan θ / g
✓ 1
Part (b)
t = 2 × 24 × tan 35°/9.8 = 3.4 s
✓ 1
Part (c)
Horizontal distance = 24 × 3.43 = 82.3 m
✓ 1
Distance along slope = 82.3/cos 35° = 100 m
✓ 1
ECF from (b).
—
Part (d)
At landing, tan φ = vy/vx = gt/u
✓ 1
Substituting t from (a) gives tan φ = 2 tan θ, which does not contain u
✓ 1
φ = tan−1(2 tan 35°) = 54° (54.5°)
✓ 1
Exam tip: For a landing on an incline, the condition 'on the slope' gives an extra equation: the ratio of the vertical drop to the horizontal distance equals tan θ.
We use essential cookies to make this site work. With your permission we'd also use analytics and advertising cookies to understand which resources help students most and how families find us. Nothing optional is set unless you accept. Privacy policy
EssentialNeeded for the site and your cookie choice to work.
Always on
AnalyticsAnonymous visit statistics that help improve the free resources.
MarketingMeasures whether ads reach the right families. Off unless you turn it on.