OCR A Level Physics A: Kinematics, equations of motion and projectiles (H556) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 3.1 (Motion) of OCR A Level Physics A (H556), Version 3.0 (March 2026): kinematics, linear motion and projectile motion. Take g = 9.81 m s⁻² unless a question gives another value, and ignore air resistance. Learn the content first from the study guide and the revision notes. The course hub is OCR A Level Physics and the tick-list is the printable checklist.

Questions

1. A swimmer swims two lengths of a 50 m pool and finishes where she started, in 58 s. Calculate her average speed. [2] State her average velocity and explain your answer. [1]

2. A train leaves a station. Its speed rises uniformly from 0 to 18 m s⁻¹ in 30 s, stays constant for 120 s, then falls uniformly to rest in 45 s.

(a) Calculate the total distance travelled. [3] (b) Calculate the average speed for the whole journey. [2] (c) Calculate the magnitude of the deceleration in the final stage. [1]

3. A cyclist’s speed is recorded every 5.0 s: 0, 4.5, 7.0, 8.2 and 8.8 m s⁻¹ at 0, 5, 10, 15 and 20 s. Estimate the distance travelled in the 20 s. [3] State, with a reason, whether your estimate is too high or too low. [1]

4. A ball is thrown vertically upwards from the ground at 21 m s⁻¹.

(a) Calculate the maximum height reached. [2] (b) Calculate the time it spends in the air. [2] (c) Calculate its speed when it is 10 m above the ground. [2]

5. A stone dropped from a bridge takes 2.6 s to reach the water. Calculate the height of the bridge [2] and the speed of the stone on impact. [1]

6. In an experiment to find g, a steel ball is held by an electromagnet. Switching off the magnet starts a timer, and the ball stops the timer when it opens a trapdoor. The drop height h and time t are:

h / m	0.50	0.75	1.00	1.25	1.50
t / s	0.319	0.391	0.452	0.505	0.553

(a) Calculate t² for each height. [1] (b) Explain why a graph of h against t² should be a straight line and what its gradient represents. [1] (c) Use the first and last points to find the gradient and hence g. [2] (d) Suggest two ways to improve the reliability or accuracy of the experiment. [2]

7. A card 0.050 m wide is attached to a trolley accelerating down a slope. It blocks the beam of gate A for 0.0500 s and gate B for 0.0250 s. The gates are 0.90 m apart.

(a) Calculate the speed of the trolley at A and at B. [2] (b) Calculate the acceleration. [2] (c) Explain why the speeds in (a) are only approximate. [1]

8. A car travels at 25 m s⁻¹. The driver’s reaction time is 0.72 s and the car decelerates uniformly at 6.4 m s⁻² once the brakes are applied.

(a) Calculate the thinking distance, the braking distance and the stopping distance. [4] (b) The driver is distracted and the reaction time rises by 0.30 s. Calculate the new stopping distance. [2] (c) With the original reaction time, the speed is raised to 30 m s⁻¹. Calculate the stopping distance and the percentage increase on the 25 m s⁻¹ value. [2]

9. A ball rolls off the edge of a table 1.10 m high at 2.4 m s⁻¹.

(a) Calculate the time to reach the floor. [2] (b) Calculate the horizontal distance from the table edge to the landing point. [1] (c) Calculate the speed on impact and the angle below the horizontal. [3] (d) A second ball is dropped from rest at the table edge at the same moment. State and explain which lands first. [1]

10. A golf ball is hit from level ground at 38 m s⁻¹ at 28° above the horizontal.

(a) Calculate the horizontal and vertical components of the initial velocity. [2] (b) Calculate the time of flight. [2] (c) Calculate the range. [1] (d) Calculate the maximum height. [2] (e) State the speed at the highest point and explain your answer. [1]

11. A stone is thrown from the top of a 15 m cliff at 12 m s⁻¹, 40° above the horizontal, and lands in the sea at the foot of the cliff.

(a) Calculate the horizontal and vertical components of the initial velocity. [2] (b) Calculate the time of flight. [3] (c) Calculate the horizontal distance travelled. [1] (d) Calculate the speed on impact. [2] (e) Calculate the angle of the velocity below the horizontal on impact. [1]

Answers

1. Average speed = total distance / time = 100 / 58 [1] = 1.7 m s⁻¹ [1]. Average velocity = zero, because her displacement is zero [1]. Examiner insight: name displacement in the explanation; “she returned to the start” alone is weaker than “displacement is zero”.

2. (a) Phase 1: ½ × 30 × 18 = 270 m [1]. Phase 2: 120 × 18 = 2160 m [1]. Phase 3: ½ × 45 × 18 = 405 m, total = 2835 m [1]. (b) Total time = 30 + 120 + 45 = 195 s [1]. Average speed = 2835 / 195 = 14.5 m s⁻¹ [1]. (c) a = 18 / 45 = 0.40 m s⁻² [1]. Examiner insight: split the area into shapes and label each; a single unexplained total is hard to give credit to if one slip occurs.

3. Use trapezia of width 5.0 s [1]: 5 × (0 + 4.5)/2 + 5 × (4.5 + 7.0)/2 + 5 × (7.0 + 8.2)/2 + 5 × (8.2 + 8.8)/2 = 11.25 + 28.75 + 38.0 + 42.5 [1] = 120.5, so about 120 m [1]. Too low: the graph is concave (gradient falls), so each chord lies below the curve [1]. Examiner insight: the direction of the error needs a reason that links to the shape of the curve.

4. (a) v² = u² + 2as with v = 0: s = 21² / (2 × 9.81) [1] = 22.5 m [1]. (b) Time up = 21 / 9.81 = 2.14 s [1]; total = 4.28 s [1]. (c) v² = 21² − 2 × 9.81 × 10 = 244.8 [1], v = 15.6 m s⁻¹ [1]. Examiner insight: write the sign convention down; mixing up the sign of g is the usual way to lose credit here.

5. s = ½gt² = 0.5 × 9.81 × 2.6² [1] = 33 m (33.2 m) [1]. v = gt = 9.81 × 2.6 = 26 m s⁻¹ (25.5 m s⁻¹) [1]. Examiner insight: carry the unrounded height forward and round only the final answer.

6. (a) t²: 0.1018, 0.1529, 0.2043, 0.2550, 0.3058 s² [1]. (b) With u = 0, h = ½gt², which has the form y = mx with gradient g/2 [1]. (c) Gradient = (1.50 − 0.50) / (0.3058 − 0.1018) = 4.90 m s⁻² [1]; g = 2 × 4.90 = 9.80 m s⁻² [1]. (d) Any two: repeat each drop and average [1]; measure h with a set square or plumb line [1]; use a larger range of heights; allow for the ball’s release being delayed by residual magnetism. Examiner insight: “do it more carefully” earns nothing; name the technique and what it improves.

7. (a) v = d/Δt: at A, 0.050 / 0.0500 = 1.00 m s⁻¹ [1]; at B, 0.050 / 0.0250 = 2.00 m s⁻¹ [1]. (b) a = (v² − u²) / 2s = (4.00 − 1.00) / (2 × 0.90) [1] = 1.7 m s⁻² [1]. (c) The card gives the average speed over its width, not the speed at the instant it reaches the gate [1]. Examiner insight: name the quantity that is averaged; “it is not accurate” is not enough.

8. (a) Thinking = 25 × 0.72 = 18 m [1]. Braking = 25² / (2 × 6.4) [1] = 48.8 m [1]. Stopping = 18 + 48.8 = 66.8 m [1]. (b) Extra thinking distance = 25 × 0.30 = 7.5 m [1]; new stopping distance = 74.3 m [1]. (c) Thinking = 30 × 0.72 = 21.6 m; braking = 900 / 12.8 = 70.3 m; stopping = 91.9 m [1]. Increase = (91.9 − 66.8) / 66.8 = 38% [1]. Examiner insight: show the separate parts so that one arithmetic slip does not hide correct method.

9. (a) 1.10 = ½ × 9.81 × t², t = √(2.20 / 9.81) [1] = 0.474 s [1]. (b) x = 2.4 × 0.474 = 1.14 m [1]. (c) vᵧ = 9.81 × 0.474 = 4.65 m s⁻¹ [1]; speed = √(2.4² + 4.65²) = 5.2 m s⁻¹ [1]; angle = tan⁻¹(4.65 / 2.4) = 63° [1]. (d) They land together, because both start with zero vertical velocity and the horizontal motion does not affect the vertical motion [1]. Examiner insight: the reason matters more than the verdict; name independence of the two motions.

10. (a) vₓ = 38 cos 28° = 33.6 m s⁻¹ [1]; vᵧ = 38 sin 28° = 17.8 m s⁻¹ [1]. (b) T = 2vᵧ / g [1] = 2 × 17.84 / 9.81 = 3.64 s [1]. (c) Range = 33.55 × 3.637 = 122 m [1]. (d) H = vᵧ² / 2g [1] = 17.84² / 19.62 = 16.2 m [1]. (e) 33.6 m s⁻¹, because vᵧ = 0 there and vₓ stays constant [1]. Examiner insight: “the speed is zero at the top” is the common wrong answer; the ball is still moving horizontally.

11. (a) vₓ = 12 cos 40° = 9.19 m s⁻¹ [1]; vᵧ = 12 sin 40° = 7.71 m s⁻¹ [1]. (b) Taking up as positive, s = −15 m: −15 = 7.71t − 4.905t² [1]. Rearranged: 4.905t² − 7.71t − 15 = 0, so t = (7.71 + √(7.71² + 4 × 4.905 × 15)) / (2 × 4.905) [1]; t = 2.70 s (the negative root is rejected) [1]. (c) x = 9.19 × 2.70 = 24.9 m [1]. (d) vᵧ = 7.71 − 9.81 × 2.70 = −18.8 m s⁻¹ [1]; speed = √(9.19² + 18.8²) = 20.9 m s⁻¹ [1]. (e) Angle = tan⁻¹(18.8 / 9.19) = 64° below the horizontal [1]. Examiner insight: the negative root must be rejected explicitly, and the sign of s must match the sign convention.

Where marks are usually lost

• Giving a distance where a displacement or a vector is wanted.
• Mixing signs for g and u without a stated positive direction.
• Reading speeds off a graph without using the area of each shape.
• Using the final vertical velocity alone as the landing speed instead of combining it with the horizontal component.
• Forgetting that the horizontal velocity is non-zero at the top of the flight.
• Plotting h against t when t² is needed, or halving the gradient incorrectly.
• Treating thinking distance as proportional to speed squared.
• Rounding too early.
• Giving no unit, or an unlikely number of significant figures.

Next steps

• Re-read the revision notes and the study guide.
• Return to the OCR A Level Physics hub and tick off the printable checklist.
• Try all the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR, A Level GCE Physics A H556 Specification, Version 3.0 (March 2026), first assessment 2017, section 3.1 Motion (3.1.1 to 3.1.3) in Module 3: Forces and motion.