OCR A Level Physics A: Dynamics, free-body diagrams, drag and terminal velocity (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 OCR A Level Physics A (H556), sections 3.2.1 Dynamics and 3.2.2 Motion with non-uniform acceleration (specification Version 3.0, March 2026). Use g = 9.81 N kg⁻¹ unless told otherwise. Read the study guide and revision notes first. See the OCR A Level Physics hub, the printable checklist and the free 10-minute diagnostics.

Questions

1. Define the newton using F = ma. A net force acts on a 0.35 kg trolley and gives it an acceleration of 4.0 m s⁻². Calculate the force. [3]

2. An astronaut and suit have a combined mass of 82 kg. On Mars the gravitational field strength is 3.7 N kg⁻¹.

(a) Calculate the weight on Mars. [2] (b) State the mass on Mars. [1]

3. A block rests on a rough ramp and does not move.

(a) Name the three forces on the block and give the direction of each. [3] (b) State the net force on the block. [1]

4. A crane lifts a load of mass 350 kg using a cable. Find the tension in the cable when the load:

(a) accelerates upward at 0.80 m s⁻² [2] (b) moves upward at constant speed [1] (c) slows to rest, with a deceleration of 0.80 m s⁻² while moving upward [2]

5. A sledge of mass 24 kg is pulled along level snow by a rope at 30° above the horizontal. The tension is 80 N and friction on the sledge is 35 N.

(a) Calculate the horizontal component of the tension. [1] (b) Calculate the acceleration of the sledge. [2] (c) Calculate the normal contact force from the snow. [2]

6. A box of mass 7.5 kg is released from rest on a ramp at 32° to the horizontal. Friction is 12 N.

(a) Calculate the acceleration down the ramp. [3] (b) Calculate the speed after sliding 4.0 m. [2] (c) Calculate the acceleration if friction were zero. [1]

7. A ball is hit horizontally at 22 m s⁻¹ from a height of 2.4 m. Ignore drag.

(a) Calculate the time to reach the ground. [2] (b) Calculate the horizontal distance travelled. [1] (c) Calculate the speed on reaching the ground. [3]

8. (a) State three factors that affect the drag on a car moving through air, and say how each changes the drag. [3] (b) A steel ball bearing falls at terminal velocity in oil. Name the forces on it and state the relationship between them. [2]

9. A ball of mass 0.070 kg falls from rest through air.

(a) Calculate its weight. [1] (b) State its acceleration at the instant of release. [1] (c) Calculate its acceleration when the drag is 0.30 N. [2] (d) State the drag at terminal velocity. [1] (e) Describe the velocity–time graph for the whole fall. [2]

10. A skydiver of mass 68 kg falls at terminal velocity. She opens her parachute and the drag instantly becomes 3200 N.

(a) Calculate her weight. [1] (b) State the drag on her just before the parachute opens. [1] (c) Calculate her acceleration just after it opens, including direction. [3] (d) Explain why she eventually moves at a constant, smaller speed. [1]

11. Students drop stacks of identical paper cases, each case being the same shape, and time the final 1.00 m of the fall, when the speed is constant. Here, n is the number of cases.

n	1	2	3	4
time for 1.00 m / s	0.84	0.59	0.48	0.42

(a) Calculate the speed for n = 3 and n = 4. [2] (b) Show that the data support v² being proportional to n. [3] (c) Explain what this suggests about how drag depends on speed. [2] (d) Suggest one improvement to the method. [1]

Answers

1. One newton is the net force that produces an acceleration of 1 m s⁻² in a mass of 1 kg [1]. F = ma = 0.35 × 4.0 [1] = 1.4 N [1]. [3] Examiner insight: A definition that names both the 1 kg and the 1 m s⁻² is easier to credit than “the unit of force”.

2. (a) W = mg = 82 × 3.7 [1] = 3.0 × 10² N (303 N) [1]. (b) 82 kg, because mass does not depend on location [1]. [3] Examiner insight: Keep mass in kg and weight in N. Swapping them gives away an easy mark.

3. (a) Weight, vertically downward [1]. Normal contact force, perpendicular to the ramp surface and away from it [1]. Friction, parallel to the ramp and up the slope [1]. (b) Zero, because the block is at rest [1]. [4] Examiner insight: Direction is part of the answer: “friction” alone does not say which way it acts.

4. (a) T − mg = ma [1], so T = 350 × (9.81 + 0.80) = 3.7 kN (3713 N) [1]. (b) a = 0, so T = mg = 350 × 9.81 = 3.4 kN (3434 N) [1]. (c) The acceleration is downward: T = m(g − a) [1] = 350 × (9.81 − 0.80) = 3.2 kN (3154 N) [1]. [5] Examiner insight: Writing the equation with a sign convention before substituting makes the sign of a easy to check in part (c).

5. (a) 80 cos 30° = 69 N (69.3 N) [1]. (b) Net force = 69.3 − 35 = 34.3 N [1]; a = 34.3/24 = 1.4 m s⁻² [1]. (c) Vertical component of tension = 80 sin 30° = 40.0 N upward [1]. N = 24 × 9.81 − 40.0 = 195 N [1]. [5] Examiner insight: Show the weight and the vertical component in part (c). A bare 235 N, which is the weight alone, does not show your reasoning.

6. (a) Down-slope component = 7.5 × 9.81 × sin 32° = 39.0 N [1]. Net force = 39.0 − 12 = 27.0 N [1]. a = 27.0/7.5 = 3.6 m s⁻² [1]. (b) v² = 2as = 2 × 3.6 × 4.0 [1], so v = 5.4 m s⁻¹ [1]. (c) a = g sin 32° = 5.2 m s⁻² [1]. [6] Examiner insight: If you reuse your part (a) value in part (b), quote it with its units so the link is clear.

7. (a) s = ½gt², so t = √(2 × 2.4/9.81) [1] = 0.70 s [1]. (b) x = 22 × 0.70 = 15 m (15.4 m) [1]. (c) Vertical speed = 9.81 × 0.70 = 6.9 m s⁻¹ [1]. v = √(22² + 6.9²) [1] = 23 m s⁻¹ [1]. [6] Examiner insight: Say that the horizontal velocity stays constant because no horizontal force acts. A calculation without that reason leaves the physics unstated.

8. (a) Any three: higher speed gives greater drag [1]; larger cross-sectional area gives greater drag [1]; a less streamlined shape gives greater drag [1]; denser air gives greater drag. (b) Weight downward, and upthrust and drag upward [1]. Weight = drag + upthrust, because the net force is zero [1]. [5] Examiner insight: Each factor needs a direction of change. “Speed affects drag” says nothing useful.

9. (a) W = 0.070 × 9.81 = 0.69 N [1]. (b) 9.81 m s⁻², because the drag is zero when the speed is zero [1]. (c) Net force = 0.6867 − 0.30 = 0.387 N [1]; a = 0.387/0.070 = 5.5 m s⁻² [1]. (d) 0.69 N, equal to the weight [1]. (e) It starts at the origin with a gradient of 9.81 m s⁻² [1], then curves with a decreasing gradient and becomes horizontal at the terminal velocity [1]. [7] Examiner insight: For a graph description, state the starting gradient and the final gradient, not just “it curves”.

10. (a) W = 68 × 9.81 = 667 N [1]. (b) 667 N, upward, because she is at terminal velocity [1]. (c) Net force = 3200 − 667 = 2533 N upward [1]; a = 2533/68 [1] = 37 m s⁻² upward [1]. (d) As she slows, drag decreases until it again equals weight, so the net force is zero [1]. [6] Examiner insight: A deceleration should be labelled as an upward acceleration, because the direction carries the meaning.

11. (a) v = 1.00/0.48 = 2.1 m s⁻¹ [1]; v = 1.00/0.42 = 2.4 m s⁻¹ [1]. [2] (b) v² for n = 1 to 4: 1.42, 2.87, 4.34, 5.67 m² s⁻² [1]. v²/n = 1.42, 1.44, 1.45, 1.42 [1]. The ratio is constant within the scatter, so v² ∝ n [1]. [3] (c) At terminal velocity, drag equals weight, and weight is proportional to n [1]. So drag is proportional to v² [1]. [2] (d) Use more repeats for each n and average (or use light gates in place of a stopwatch) [1]. [1] Examiner insight: “Show that” questions need the calculated values written out. Stating only that the ratio is constant leaves your evidence unseen.

Where marks are usually lost

• Confusing mass with weight, or quoting weight in kg.
• Missing the direction of a vector answer, such as the sign of an acceleration.
• Putting the normal contact force vertical on a slope.
• Forgetting upthrust in a liquid.
• Using g as the acceleration while drag acts.
• Giving a final answer with too many or too few significant figures.
• Describing a graph as “curved” without the start and end gradients.
• Not saying why terminal velocity has been reached.

Next steps

• Revision notes
• Study guide
• OCR A Level Physics hub
• Printable checklist
• All free 10-minute diagnostics: /diagnostics/
• Book a free trial class: /trial/

Official syllabus

OCR A Level Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017, published by OCR. Sections 3.2.1 and 3.2.2.