ExaminerPrep › Question bank › IAL Physics › 1.1 Motion, vectors and projectiles
Edexcel International A Level Physics · Unit 1 · Mechanics and Materials
40 free, high-quality past-paper-style questions on Motion, vectors and projectiles for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
Which list contains only vector quantities?
| Answer | Mark | Notes |
|---|---|---|
| Displacement, momentum and force all have a direction as well as a magnitude, so they are vectors. Mass, energy, power and speed are scalars, so every other list contains at least one scalar. | ✓ 1 | Answer A |
Answer: A
Common error: Weight is a vector (a force acting downwards) but mass is a scalar. Keep the two apart.
A rope pulls a crate with a force of 60 N at 35° above the horizontal.
What is the vertical component of this force?
| Answer | Mark | Notes |
|---|---|---|
| The vertical component is opposite the 35° angle: 60 sin 35° = 34 N (34.4 N). | ✓ 1 | Answer A |
Answer: A
Exam tip: Mark the angle on a sketch first. The component next to the angle uses cos; the one opposite the angle uses sin.
A ferry heads straight across a river at 2.4 m s−1 relative to the water. The river flows at 0.70 m s−1 at right angles to the ferry's heading.
What is the speed of the ferry relative to the river bank?
| Answer | Mark | Notes |
|---|---|---|
| The velocities are perpendicular, so add them by Pythagoras: √(2.42 + 0.702) = 2.5 m s−1. | ✓ 1 | Answer B |
Answer: B
Common error: Adding or subtracting the magnitudes of two vectors only works when they act along the same line.
An aircraft flies horizontally at constant velocity and releases a supply crate. Air resistance on the crate is negligible.
Where is the aircraft at the instant the crate reaches the ground?
| Answer | Mark | Notes |
|---|---|---|
| With no air resistance the crate keeps the horizontal velocity it had on release, which is the velocity of the aircraft. Both cover the same horizontal distance in the same time, so the aircraft is directly above the crate. | ✓ 1 | Answer D |
Answer: D
Exam tip: The horizontal and vertical motions of a projectile are independent: gravity changes only the vertical velocity.
A kite is held by a string. The tension in the string is 18 N and the string makes an angle of 55° with the horizontal ground.
Calculate the horizontal component of the tension.
Calculate the vertical component of the tension.
The wind strengthens and the string rises to 70° above the horizontal. The tension stays at 18 N. State and explain what happens to the horizontal component of the tension.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 18 cos 55° = 10 N (10.3 N) | ✓ 1 | |
| Part (b) | ||
| 18 sin 55° = 15 N (14.7 N) | ✓ 1 | |
| Part (c) | ||
| It decreases | ✓ 1 | |
| cos 70° is smaller than cos 55° / the string is closer to vertical, so less of the tension acts horizontally (6.2 N) | ✓ 1 | |
Exam tip: In a “state and explain” question the first mark is for the change and the second is for the physics reason. A reason without a clear statement of the change may not score.
The velocity–time graph shows the motion of a lift as it travels upwards from one floor to another.
What is the distance travelled by the lift?
| Answer | Mark | Notes |
|---|---|---|
| Distance = area under the graph = ½ × 4 × 2.0 + 6 × 2.0 + ½ × 4 × 2.0 = 4 + 12 + 4 = 20 m. | ✓ 1 | Answer D |
Answer: D
Exam tip: Split the area under a velocity–time graph into triangles and rectangles, and read each time interval from the axis: here 4 s, 6 s and 4 s.
The diagrams show the same two vectors P and Q in each case, together with a third vector R.
In which diagram is R = P + Q?
| Answer | Mark | Notes |
|---|---|---|
| To add vectors, draw them tip to tail; the resultant runs from the tail of the first to the tip of the last. Only diagram 4 does this. In diagram 1 the arrow on R is reversed, so P + Q + R = 0. Diagrams 2 and 3 show the differences Q − P and P − Q. | ✓ 1 | Answer D |
Answer: D
Common error: Joining the tips of two vectors drawn from the same point gives their difference, not their sum.
The diagram shows a force of 40 N pulling a wheeled case along level ground. The angle is measured from the vertical.
What is the horizontal component of the 40 N force?
| Answer | Mark | Notes |
|---|---|---|
| The horizontal component is opposite the 50° angle, which is measured from the vertical: 40 sin 50° = 31 N (30.6 N). | ✓ 1 | Answer B |
Answer: B
Common error: Using cos automatically. Here the angle is measured from the vertical, so the horizontal component is F sin 50°.
The diagram shows the path of a ball thrown over level ground. Air resistance is negligible. X is a point on the path after the ball has passed its highest point.
Which arrow shows the direction of the velocity of the ball at X?
| Answer | Mark | Notes |
|---|---|---|
| The velocity of a projectile is always along the tangent to its path. After the highest point the vertical component is downwards, so the velocity at X points forwards and downwards along the path: arrow 1. | ✓ 1 | Answer A |
Answer: A
Common error: Confusing velocity with acceleration. For a projectile the acceleration is always vertically downwards (arrow 2), but the velocity is along the path.
A gardener pushes a heavy roller along a level lawn. The push of 150 N acts along the handle, as shown in the diagram.
Calculate the horizontal component of the push.
Calculate the vertical component of the push and state its direction.
The gardener now pulls the roller with the same force along the handle, at the same angle. State what happens to the vertical component of the force.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 150 cos 35° = 123 N | ✓ 1 | |
| Part (b) | ||
| 150 sin 35° = 86 N (86.0 N) | ✓ 1 | |
| downwards | ✓ 1 | |
| Part (c) | ||
| same size (86 N) but now upwards | ✓ 1 | |
Exam tip: Read the angle from the diagram before choosing sin or cos: the 35° is between the handle and the horizontal, so the horizontal component uses cos 35°.
A trolley rolls down a runway pulling a paper tape through a ticker-timer. The tape is cut into strips, each covering a time of 0.10 s, and the strips are placed side by side to make the chart shown.
Calculate the mean speed of the trolley during strip 1.
Show that the acceleration of the trolley is about 1.2 m s−2.
Calculate the total distance travelled during the 0.50 s shown.
Explain how the chart shows that the acceleration is uniform.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 0.020 ÷ 0.10 = 0.20 m s−1 | ✓ 1 | |
| Part (b) | ||
| mean speed during strip 5 = 0.068 ÷ 0.10 = 0.68 m s−1; the middles of strips 1 and 5 are 0.40 s apart | ✓ 1 | |
| a = (0.68 − 0.20) ÷ 0.40 = 1.20 m s−2 | ✓ 1 | |
| allow: each strip is 1.2 cm longer than the last, so a = Δs ÷ T2 = 0.012 ÷ 0.102 = 1.2 m s−2 | — | |
| Part (c) | ||
| 2.0 + 3.2 + 4.4 + 5.6 + 6.8 = 22.0 cm (0.220 m) | ✓ 1 | |
| Part (d) | ||
| each strip is longer than the one before by the same amount (1.2 cm), so the velocity increases by equal amounts in equal times | ✓ 1 | |
Exam tip: The mean speed during a strip is the speed at the middle of its time interval. Measure the time between strip middles, not from the start of the tape.
The acceleration–time graph for an electric scooter is a horizontal line at 2.5 m s−2 from t = 0 to t = 4.0 s. At t = 0 the velocity of the scooter is 3.0 m s−1.
What is the velocity of the scooter at t = 4.0 s?
| Answer | Mark | Notes |
|---|---|---|
| The area under the acceleration–time graph is the change in velocity: 2.5 × 4.0 = 10 m s−1. Final velocity = 3.0 + 10 = 13 m s−1. | ✓ 1 | Answer C |
Answer: C
Common error: The area under an acceleration–time graph is the change in velocity. Add it to the initial velocity.
The displacement–time graph for a robot vacuum cleaner is a straight line from (0, 12 m) to (8.0 s, 60 m).
What is the velocity of the robot?
| Answer | Mark | Notes |
|---|---|---|
| Velocity = gradient = (60 − 12) ÷ (8.0 − 0) = 6.0 m s−1. | ✓ 1 | Answer B |
Answer: B
Common error: Dividing the final reading by the final time ignores the starting displacement. A gradient uses the changes in both quantities.
A train slows uniformly from 30 m s−1 to 12 m s−1 over a distance of 420 m.
What is the magnitude of its deceleration?
| Answer | Mark | Notes |
|---|---|---|
| v2 = u2 + 2as: 122 = 302 + 2a(420), so a = (144 − 900) ÷ 840 = −0.90 m s−2. Magnitude 0.90 m s−2. | ✓ 1 | Answer B |
Answer: B
Exam tip: Choose the equation that leaves out the quantity you are not given. Here the time is not given, so use v2 = u2 + 2as.
A marble rolls off the edge of a horizontal shelf 0.76 m above the floor with a speed of 1.5 m s−1. Air resistance is negligible. g = 9.81 m s−2.
How far from the foot of the shelf does the marble land?
| Answer | Mark | Notes |
|---|---|---|
| Vertical: 0.76 = ½ × 9.81 × t2, so t = 0.394 s. Horizontal: 1.5 × 0.394 = 0.59 m. | ✓ 1 | Answer C |
Answer: C
Common error: The time in the air depends only on the height. Find it from the vertical motion, then use it horizontally.
In an experiment to measure the acceleration of free fall, a student releases a ball from rest and measures the time t to fall through different heights h. A graph of h on the y-axis against t2 on the x-axis is a straight line through the origin with a gradient of 4.85 m s−2.
What is the student's value for the acceleration of free fall?
| Answer | Mark | Notes |
|---|---|---|
| From rest, h = ½gt2, so the gradient of h against t2 is g/2. g = 2 × 4.85 = 9.70 m s−2. | ✓ 1 | Answer C |
Answer: C
Exam tip: Compare the equation with y = mx + c before reading anything from a graph: here y = h, x = t2 and m = g/2.
A light aircraft is pointed due north and flies at 65 m s−1 relative to the air. A wind blows from the west at 18 m s−1.
Calculate the speed of the aircraft relative to the ground.
Determine the direction of travel of the aircraft relative to the ground.
The pilot wants the aircraft to travel due north over the ground. Explain how the pilot must change the direction in which the aircraft is pointed.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| velocities at right angles, so Pythagoras: √(652 + 182) | ✓ 1 | |
| = 67 m s−1 (67.4) | ✓ 1 | |
| Part (b) | ||
| tan−1(18 ÷ 65) | ✓ 1 | |
| = 15° east of north (15.5°) | ✓ 1 | |
| Part (c) | ||
| point the aircraft west of north / into the wind | ✓ 1 | |
| so that a component of its air velocity, westwards, cancels the 18 m s−1 eastward wind | ✓ 1 | |
Common error: A direction is incomplete without a reference: write “15° east of north”, not just “15°”.
A cable car leaves a station from rest. It accelerates uniformly to 6.0 m s−1 in 12 s, travels at 6.0 m s−1 for 150 s, and then decelerates uniformly to rest in 15 s. It travels along a straight cable.
Calculate the acceleration of the cable car during the first 12 s.
Describe the shape of the velocity–time graph for the whole journey.
Calculate the total distance travelled by the cable car.
Calculate the average speed of the cable car for the whole journey.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 6.0 ÷ 12 = 0.50 m s−2 | ✓ 1 | |
| Part (b) | ||
| straight line from the origin rising to 6.0 m s−1 at 12 s, then horizontal | ✓ 1 | |
| then a straight line falling to zero at 177 s, less steep than the first section | ✓ 1 | |
| Part (c) | ||
| distance = area under the velocity–time graph | ✓ 1 | |
| ½ × 12 × 6.0 + 150 × 6.0 + ½ × 15 × 6.0 = 36 + 900 + 45 | ✓ 1 | |
| = 981 m | ✓ 1 | |
| Part (d) | ||
| total time = 12 + 150 + 15 = 177 s | ✓ 1 | |
| 981 ÷ 177 = 5.5 m s−1 (5.54) — allow error carried forward from the distance | ✓ 1 | |
Model sketch
Exam tip: The area under a velocity–time graph can be split into triangles and rectangles. Write each area separately so that every step can be followed.
A car joins a motorway along a straight slip road of length 240 m. It accelerates uniformly from 14 m s−1 at the start of the slip road to 31 m s−1 at the end.
Calculate the acceleration of the car.
Show that the average speed of the car on the slip road is about 23 m s−1.
Calculate the time the car takes to travel along the slip road.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| v2 = u2 + 2as: 312 = 142 + 2a × 240 | ✓ 1 | |
| a = 1.6 m s−2 (1.59) | ✓ 1 | |
| Part (b) | ||
| (14 + 31) ÷ 2 = 22.5 m s−1 | ✓ 1 | |
| Part (c) | ||
| t = 240 ÷ 22.5 or t = (31 − 14) ÷ 1.59 | ✓ 1 | |
| = 11 s (10.7 s) | ✓ 1 | |
Exam tip: In a “show that” question, give your result to one more significant figure than the value quoted — 22.5 m s−1, not 23 m s−1.
Water leaves a horizontal outlet pipe in a garden wall at 3.2 m s−1. The outlet is 1.25 m above the surface of a pond. Treat each small mass of water as a projectile. Air resistance is negligible. g = 9.81 m s−2.
Calculate the time for the water to fall to the pond.
Calculate the horizontal distance from the wall at which the water reaches the pond.
Determine the velocity of the water as it reaches the pond.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 1.25 = ½ × 9.81 × t2 | ✓ 1 | |
| t = 0.50 s (0.505 s) | ✓ 1 | |
| Part (b) | ||
| 3.2 × 0.505 = 1.6 m (1.62 m) | ✓ 1 | |
| Part (c) | ||
| vertical component = 9.81 × 0.505 = 4.95 m s−1 | ✓ 1 | |
| speed = √(3.22 + 4.952) = 5.9 m s−1 | ✓ 1 | |
| at tan−1(4.95 ÷ 3.2) = 57° below the horizontal | ✓ 1 | |
Exam tip: “Determine the velocity” needs a magnitude and a direction. Combine the two components, then give the angle from a named reference.
A student determines the acceleration of free fall g. A steel ball held by an electromagnet is released from rest; switching off the magnet starts an electronic timer, and the ball stops the timer when it hits a trapdoor a height h below. She measures the fall time t for six values of h between 0.30 m and 1.30 m and plots a graph of h (y-axis) against t2 (x-axis).
Explain why the graph is expected to be a straight line through the origin.
The gradient of her line is 4.88 m s−2. Calculate her value of g.
Calculate the percentage difference between her value and 9.81 m s−2.
Explain why an electronic timer is used instead of a hand-operated stopwatch.
The student measured every height from the top of the ball instead of the bottom. State the effect of this on her value of g.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| from rest, h = ½gt2 | ✓ 1 | |
| so h is proportional to t2, which compares with y = mx with gradient g/2 | ✓ 1 | |
| Part (b) | ||
| 2 × 4.88 = 9.76 m s−2 | ✓ 1 | |
| Part (c) | ||
| (9.81 − 9.76) ÷ 9.81 × 100 = 0.5% (0.51%) | ✓ 1 | |
| Part (d) | ||
| human reaction time (about 0.2 s) is a large fraction of the fall times, which are only about 0.2–0.5 s | ✓ 1 | |
| Part (e) | ||
| no effect — every height is too large by the same amount (the diameter of the ball), which shifts the line up (an intercept on the h-axis) but does not change its gradient | ✓ 1 | |
Exam tip: When a graph is used, a constant systematic error usually moves the line without changing its gradient. Say which one — gradient or intercept — the error affects.
The graph shows how the displacement of a motorbike varies with time as it accelerates from rest. The straight line is the tangent to the curve at t = 8.0 s.
What is the velocity of the motorbike at t = 8.0 s?
| Answer | Mark | Notes |
|---|---|---|
| The instantaneous velocity is the gradient of the tangent: (48 − 0) ÷ (10.0 − 4.0) = 8.0 m s−1. | ✓ 1 | Answer C |
Answer: C
Exam tip: For a curved displacement–time graph, the velocity at an instant is the gradient of the tangent at that instant, not displacement ÷ time.
The diagram is a multiflash record of a ball rolling off the edge of a table. The images were taken at intervals of 0.050 s, and each grid square is 3.0 cm × 3.0 cm. Air resistance is negligible.
What is the horizontal component of the velocity of the ball?
| Answer | Mark | Notes |
|---|---|---|
| Each image is 2 squares = 6.0 cm further along than the one before, in 0.050 s. Horizontal velocity = 0.060 ÷ 0.050 = 1.2 m s−1. The equal horizontal spacing shows that this component is constant. | ✓ 1 | Answer B |
Answer: B
Exam tip: In a multiflash record, equal horizontal gaps mean constant horizontal velocity; growing vertical gaps mean the vertical velocity is increasing.
Two forces P and Q act on an object at O. They are drawn to scale on the grid, where one square represents 5.0 N.
What is the magnitude of the resultant force?
| Answer | Mark | Notes |
|---|---|---|
| Add the components: horizontal 4 + 2 = 6 squares, vertical 0 + 3 = 3 squares. Resultant = √(62 + 32) = 6.71 squares × 5.0 N = 34 N (33.5 N). | ✓ 1 | Answer D |
Answer: D
Exam tip: When vectors are drawn on a grid, add their horizontal and vertical components square by square, then combine the totals with Pythagoras.
A trolley is pushed up a ramp. The graph shows its velocity from the moment it is released, with up the ramp taken as positive.
What is the displacement of the trolley from the release point at t = 6.0 s?
| Answer | Mark | Notes |
|---|---|---|
| Area above the axis (0–4.0 s) = ½ × 4.0 × 3.0 = 6.0 m up the ramp. Area below the axis (4.0–6.0 s) = ½ × 2.0 × 1.5 = 1.5 m back down. Displacement = 6.0 − 1.5 = 4.5 m up the ramp. | ✓ 1 | Answer A |
Answer: A
Exam tip: Area below the time axis is negative displacement. Subtract it to find displacement; add it to find total distance.
The diagram shows an arrangement for measuring the acceleration of free fall. Each timer records the time for which the card blocks the beam of its light gate.
What value for the acceleration of free fall do these readings give?
| Answer | Mark | Notes |
|---|---|---|
| Speeds at the gates: 0.050 ÷ 0.0250 = 2.00 m s−1 and 0.050 ÷ 0.0145 = 3.45 m s−1. v2 = u2 + 2as: a = (3.452 − 2.002) ÷ (2 × 0.400) = 9.9 m s−2 (9.86). | ✓ 1 | Answer C |
Answer: C
Exam tip: Card length ÷ blocking time gives the speed at each gate. With two speeds and a distance, but no time between the gates, use v2 = u2 + 2as.
Two tugs pull a ship using ropes A and B. The diagram is a plan view, seen from above, showing the forces exerted by the ropes on the ship.
Explain why the resultant force of the two ropes is not 14 kN.
Calculate the total component of the two forces along the line of the ship.
Calculate the total component of the two forces at right angles to the line of the ship.
Determine the magnitude and direction of the resultant force on the ship.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| force is a vector and the two forces act in different directions, so their magnitudes cannot simply be added | ✓ 1 | |
| Part (b) | ||
| 8.0 cos 20° + 6.0 cos 35° = 7.52 + 4.91 | ✓ 1 | |
| = 12.4 kN | ✓ 1 | |
| Part (c) | ||
| 6.0 sin 35° − 8.0 sin 20° = 3.44 − 2.74 = 0.71 kN towards the side of rope B | ✓ 1 | |
| Part (d) | ||
| √(12.42 + 0.712) = 12 kN (12.5 kN; allow 12.4 kN from rounded components) | ✓ 1 | |
| tan−1(0.71 ÷ 12.4) = 3.2° from the line of the ship, towards rope B | ✓ 1 | |
Exam tip: With two forces on opposite sides of a line, give each perpendicular component a sign before adding. Here they partly cancel.
A drone takes off vertically from the ground. The graph shows how its vertical velocity varies with time, with upwards taken as positive.
Calculate the acceleration of the drone between t = 0 and t = 2.0 s.
Describe the motion of the drone between t = 10 s and t = 12 s.
Determine the maximum height of the drone above the ground.
Determine the height of the drone above the ground at t = 19 s.
Calculate the average velocity of the drone for the first 19 s.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 4.0 ÷ 2.0 = 2.0 m s−2 | ✓ 1 | |
| Part (b) | ||
| velocity is zero, so the drone is at rest / hovering at a fixed height | ✓ 1 | |
| Part (c) | ||
| area under the graph from 0 to 10 s = ½ × 2 × 4.0 + 6 × 4.0 + ½ × 2 × 4.0 | ✓ 1 | |
| = 32 m | ✓ 1 | |
| Part (d) | ||
| area below the axis from 12 s to 19 s = ½ × 1 × 3.0 + 5 × 3.0 + ½ × 1 × 3.0 = 18 m | ✓ 1 | |
| height = 32 − 18 = 14 m | ✓ 1 | |
| Part (e) | ||
| 14 ÷ 19 = 0.74 m s−1 upwards | ✓ 1 | |
Common error: Adding the area below the time axis to the area above it. Below the axis the drone is moving down, so that area is subtracted to find its height.
A firework shell is fired from a launch tube on level ground with the velocity shown in the diagram. It explodes at E, the highest point of its path. Air resistance is negligible. g = 9.81 m s−2.
Calculate the horizontal and vertical components of the launch velocity.
Calculate the time taken for the shell to reach E.
Calculate the height H of E above the ground.
Calculate the horizontal distance x of E from the launch tube.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| horizontal: 30 cos 70° = 10 m s−1 (10.3) | ✓ 1 | |
| vertical: 30 sin 70° = 28 m s−1 (28.2) | ✓ 1 | |
| Part (b) | ||
| at E the vertical velocity is zero: t = 28.2 ÷ 9.81 = 2.9 s (2.87 s) | ✓ 1 | |
| Part (c) | ||
| 0 = 28.22 − 2 × 9.81 × H | ✓ 1 | |
| H = 41 m (40.5 m) | ✓ 1 | |
| Part (d) | ||
| 10.3 × 2.87 = 29 m (29.5 m) | ✓ 1 | |
Exam tip: At the highest point the vertical velocity is zero but the horizontal velocity is not. Use the vertical component alone to find the time and the height.
A student measures the acceleration of free fall g. She releases a small ball from rest at a height h above a light gate. The light gate and a timer give the speed v of the ball as it passes through the beam. She repeats this for six heights and plots the graph shown.
Show that a graph of v2 against h should be a straight line through the origin, and state what its gradient represents.
Identify the anomalous result and state how she should deal with it.
Determine the gradient of the line.
Calculate her value for g.
Explain why a small ball gives a more reliable value of v at height h than a large ball.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| from rest, u = 0, so v2 = 2gh, which is of the form y = mx | ✓ 1 | |
| gradient = 2g | ✓ 1 | |
| Part (b) | ||
| the point at h = 0.65 m; repeat that measurement / leave it out when drawing the line | ✓ 1 | |
| Part (c) | ||
| large triangle, e.g. 19.5 ÷ 1.00 | ✓ 1 | |
| = 19.5 m s−2 (allow 19.3–19.7) | ✓ 1 | |
| Part (d) | ||
| 19.5 ÷ 2 = 9.8 m s−2 (9.75) (allow error carried forward) | ✓ 1 | |
| Part (e) | ||
| the gate measures the average speed while the ball crosses the beam; a small ball crosses in a shorter time, so this is closer to the instantaneous speed at height h | ✓ 1 | |
Exam tip: Take a gradient from two points on the best-fit line that are far apart, not from individual data points, and include the units of the gradient.
A swimmer can swim at 1.0 m s−1 relative to the water. She wants to cross the river shown, from A to B, which is directly opposite A. She heads upstream at an angle θ to AB, so that she moves along the line AB.
Calculate the angle θ.
Calculate her speed relative to the river bank.
Calculate the time she takes to cross from A to B.
On another day she heads straight across, along AB, at 1.0 m s−1 relative to the water. The current is the same. Calculate how far downstream of B she reaches the other bank.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the upstream component of her velocity must cancel the current: 1.0 sin θ = 0.60 | ✓ 1 | |
| θ = 37° (36.9°) | ✓ 1 | |
| Part (b) | ||
| √(1.02 − 0.602) = 0.80 m s−1 | ✓ 1 | |
| Part (c) | ||
| 36 ÷ 0.80 = 45 s | ✓ 1 | |
| Part (d) | ||
| time to cross = 36 ÷ 1.0 = 36 s | ✓ 1 | |
| distance downstream = 0.60 × 36 = 22 m (21.6 m) | ✓ 1 | |
Exam tip: Draw the vector triangle: her velocity relative to the water is the hypotenuse, so her speed across is found by subtracting squares, not adding them.
Two identical stones are thrown from level ground with the same speed, X at 30° and Y at 60° above the horizontal. Air resistance is negligible.
What is (time for Y to reach its highest point) ÷ (time for X to reach its highest point)?
| Answer | Mark | Notes |
|---|---|---|
| Time to the highest point = u sin θ ÷ g, so the ratio is sin 60° ÷ sin 30° = √3 = 1.7. | ✓ 1 | Answer D |
Answer: D
Exam tip: In a ratio question write the general expression first; the quantities that are the same for both cancel and the arithmetic becomes short.
A sprinter starts from rest. Her acceleration decreases uniformly from 8.0 m s−2 at t = 0 to zero at t = 2.0 s. After t = 2.0 s she runs at constant velocity.
State what is represented by the area under an acceleration–time graph.
Determine her velocity at t = 2.0 s.
Explain why the equations of uniformly accelerated motion cannot be used for the first 2.0 s.
Calculate the distance she runs between t = 2.0 s and t = 10.0 s.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| change in velocity | ✓ 1 | |
| Part (b) | ||
| area of triangle = ½ × 2.0 × 8.0 | ✓ 1 | |
| = 8.0 m s−1 (starting from rest) | ✓ 1 | |
| Part (c) | ||
| the acceleration is not constant / it changes with time | ✓ 1 | |
| Part (d) | ||
| 8.0 × 8.0 = 64 m (allow error carried forward from (b)) | ✓ 1 | |
Common error: Using v = u + at with a = 8.0 m s−2 gives 16 m s−1, which assumes the acceleration stays at its starting value.
A ball is thrown vertically upwards at 6.5 m s−1 from a balcony 8.0 m above the ground. It misses the balcony on the way down and lands on the ground. Air resistance is negligible. g = 9.81 m s−2.
Calculate the maximum height of the ball above the ground.
Calculate the speed of the ball as it reaches the ground.
Calculate the time from release until the ball reaches the ground.
State the velocity of the ball as it passes the balcony on the way down.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| rise above balcony = 6.52 ÷ (2 × 9.81) = 2.15 m | ✓ 1 | |
| height above ground = 8.0 + 2.15 = 10.2 m | ✓ 1 | |
| Part (b) | ||
| taking upwards as positive, s = −8.0 m: v2 = 6.52 + 2(−9.81)(−8.0) | ✓ 1 | |
| v = 14 m s−1 (14.1) | ✓ 1 | |
| Part (c) | ||
| v = u + at with v = −14.1 m s−1: −14.1 = 6.5 − 9.81t | ✓ 1 | |
| t = 2.1 s (2.10 s) | ✓ 1 | |
| Part (d) | ||
| 6.5 m s−1 downwards | ✓ 1 | |
Common error: Taking s = +8.0 m when the ground is below the point of release. Write the positive direction first and give every vector quantity its sign.
A long jumper leaves the take-off board at 9.4 m s−1 at 21° above the horizontal. Model the jumper as a point that lands at the same level as it took off. Air resistance is negligible. g = 9.81 m s−2.
Calculate the horizontal and vertical components of the take-off velocity.
Calculate the time the jumper spends in the air.
Calculate the horizontal distance of the jump.
In a real jump, the jumper's centre of mass is lower on landing than at take-off. State the effect of this on the horizontal distance.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| horizontal: 9.4 cos 21° = 8.8 m s−1 (8.78) | ✓ 1 | |
| vertical: 9.4 sin 21° = 3.4 m s−1 (3.37) | ✓ 1 | |
| Part (b) | ||
| time to the top = 3.37 ÷ 9.81 = 0.343 s | ✓ 1 | |
| time in the air = 2 × 0.343 = 0.69 s (0.687 s) | ✓ 1 | |
| Part (c) | ||
| 8.78 × 0.687 = 6.0 m (6.03 m) | ✓ 1 | |
| Part (d) | ||
| the jump is longer, because the jumper is in the air for longer | ✓ 1 | |
Exam tip: For a launch and landing at the same level the flight is symmetric: the time in the air is twice the time to reach the highest point.
A volleyball player hits the ball horizontally at 16 m s−1 from a height of 2.60 m above the floor. The net is 2.43 m high and is 9.0 m horizontally from the point where the ball is hit. Air resistance and spin are negligible. g = 9.81 m s−2.
Calculate the time for the ball to travel 9.0 m horizontally.
Determine whether the ball passes over the net.
Determine the minimum horizontal speed for the ball to pass over the net.
Suggest why a player usually hits the ball slightly upwards rather than horizontally.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 9.0 ÷ 16 = 0.56 s (0.5625 s) | ✓ 1 | |
| Part (b) | ||
| vertical drop = ½ × 9.81 × 0.56252 | ✓ 1 | |
| = 1.55 m, so the ball is 2.60 − 1.55 = 1.05 m above the floor | ✓ 1 | |
| 1.05 m < 2.43 m, so the ball hits the net / does not pass over | ✓ 1 | |
| Part (c) | ||
| the ball may drop only 0.17 m: t = √(2 × 0.17 ÷ 9.81) = 0.186 s | ✓ 1 | |
| speed = 9.0 ÷ 0.186 = 48.34 m s−1, so the minimum speed is 48.4 m s−1 (rounded up: 48 m s−1 would not clear the net) | ✓ 1 | |
| Part (d) | ||
| a speed of over 48 m s−1 is unrealistically fast; an upward component lets the ball rise over the net at a realistic speed | ✓ 1 | |
Common error: Comparing the drop (1.55 m) with the net height (2.43 m). Find the height of the ball above the floor first, then compare it with the net.
The graph shows the path of a projectile launched from level ground. Air resistance is negligible.
At what angle above the horizontal was the projectile launched?
| Answer | Mark | Notes |
|---|---|---|
| Up to the highest point, horizontal distance = uxt and height = ½uyt (average vertical velocity). So uy ÷ ux = 2 × 3.0 ÷ 8.0 = 0.75 and the angle is tan−1 0.75 = 37° (36.9°). | ✓ 1 | Answer B |
Answer: B
Common error: Taking the angle of the straight line from the launch point to the highest point. The path curves, so the launch direction is steeper than that line.
A glider on a sloping air track is given a push up the slope at t = 0 and then moves freely. The graph shows its displacement up the slope from its starting point. The straight line is the tangent to the curve at t = 2.0 s.
Use the graph to determine the velocity of the glider at t = 2.0 s.
State the velocity of the glider at t = 4.0 s. Give a reason for your answer.
The acceleration of the glider is uniform. Calculate its acceleration.
Determine the total distance travelled and the displacement of the glider between t = 0 and t = 8.0 s.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| gradient of the tangent = (15 − 3.0) ÷ (4.0 − 0) | ✓ 1 | |
| = 3.0 m s−1 (allow 2.8–3.2) | ✓ 1 | |
| Part (b) | ||
| zero | ✓ 1 | |
| the gradient of the graph is zero there / the curve is horizontal at its maximum | ✓ 1 | |
| Part (c) | ||
| a = (0 − 3.0) ÷ (4.0 − 2.0) | ✓ 1 | |
| = −1.5 m s−2 (down the slope) (allow error carried forward from (a)) | ✓ 1 | |
| Part (d) | ||
| distance = 12 + 12 = 24 m | ✓ 1 | |
| displacement = 0 (it is back at its starting point) | ✓ 1 | |
Exam tip: To find a gradient from a tangent, use two points on the tangent that are far apart and easy to read, not points on the curve.
A football is kicked from the ground towards a garden wall, with the velocity shown in the diagram. Treat the ball as a particle and ignore the thickness of the wall. Air resistance is negligible. g = 9.81 m s−2.
Calculate the horizontal and vertical components of the initial velocity.
Calculate the time for the ball to travel 20.0 m horizontally.
Determine whether the ball passes over the wall.
Deduce whether the ball is rising or falling as it passes the wall.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| horizontal: 18 cos 40° = 13.8 m s−1 | ✓ 1 | |
| vertical: 18 sin 40° = 11.6 m s−1 | ✓ 1 | |
| Part (b) | ||
| 20.0 ÷ 13.8 = 1.45 s | ✓ 1 | |
| Part (c) | ||
| s = ut + ½at2 = 11.6 × 1.45 − ½ × 9.81 × 1.452 | ✓ 1 | |
| = 6.5 m (6.46 m) | ✓ 1 | |
| 6.5 m > 4.0 m, so the ball passes over the wall (by about 2.5 m) | ✓ 1 | |
| Part (d) | ||
| vertical velocity = 11.6 − 9.81 × 1.45 = −2.7 m s−1 | ✓ 1 | |
| negative, so the ball is falling / has passed its highest point | ✓ 1 | |
Common error: Using the full 18 m s−1 in the vertical equation. Each direction uses only its own component of the velocity.
A car waits at traffic lights. As the lights change, at t = 0, a van passes the car at a constant velocity, and at the same instant the car starts to accelerate uniformly from rest in the same direction. The graph shows the velocities of the car and the van.
State the time at which the car and the van have the same velocity.
Explain, using the graph, why the car has not caught up with the van at this time.
Calculate the time at which the car draws level with the van.
Calculate the distance the car has travelled when it draws level with the van.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 8.0 s | ✓ 1 | |
| Part (b) | ||
| distance = area under each line; at 8.0 s the area under the van's line (96 m) is greater than the area under the car's line (48 m) | ✓ 1 | |
| the van is ahead by the area between the lines, 48 m | ✓ 1 | |
| Part (c) | ||
| car's acceleration = 12 ÷ 8.0 = 1.5 m s−2; level when 12t = ½ × 1.5 × t2 | ✓ 1 | |
| t = 16 s | ✓ 1 | |
| Part (d) | ||
| 12 × 16 = 192 m (allow error carried forward) | ✓ 1 | |
Common error: Thinking the car catches the van where the lines cross. Equal velocities only means the gap has stopped growing; the car draws level when the areas are equal.
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