A Level Physics 3.2 Non-Uniform Motion MCQs – 30 Practice Questions with Answers (Cambridge 9702)
Free practice questions for Cambridge International AS & A Level Physics (9702), Section 3.2 Non-Uniform Motion. These questions cover resistive forces, terminal velocity, motion graphs with changing acceleration, and applying Newton’s laws when acceleration is not constant – every answer is explained.
Formula recap: Resultant force F = ma. As speed increases, resistive (drag) force increases, so resultant force and acceleration decrease. At terminal velocity, drag = weight, resultant force = 0 and acceleration = 0. Instantaneous acceleration = gradient of the tangent to a velocity-time graph; displacement = area under a velocity-time graph. Revise all the key equations on our A Level Physics Formula Sheet.
Resistive Forces & Terminal Velocity
Q1. A ball falls through air and its speed increases. What happens to the air resistance acting on the ball, and why?
- (A) It decreases, because the ball’s weight decreases as it falls
- (B) It increases, because the ball collides with more air molecules each second
- (C) It stays constant, because air resistance does not depend on speed
- (D) It increases, because the acceleration of free fall increases with height fallen
Show Answer
Answer: B – Drag grows with speed: at higher speed the ball meets (and pushes aside) more air molecules per unit time, transferring more momentum to the air each second, so the resistive force is larger.
Q2. A small steel sphere is released from rest at the surface of a tall tank of oil. It accelerates downwards at first, but its acceleration gradually decreases. Why?
- (A) The viscous drag on the sphere acts upwards and grows as the speed increases, so the resultant force decreases
- (B) The weight of the sphere decreases as it sinks deeper into the oil
- (C) The upthrust on the sphere increases as its speed increases
- (D) The mass of the sphere decreases because oil sticks to its surface
Show Answer
Answer: A – The resultant force is weight − upthrust − viscous drag. Weight and upthrust are constant, but drag increases with speed, so the resultant force – and therefore the acceleration – gets smaller and smaller.
Q3. A skydiver is falling at terminal velocity. Which statement about the skydiver is correct?
- (A) The air resistance is greater than the weight, so the speed is constant
- (B) The acceleration is 9.81 m s−2 because the speed is large
- (C) The resultant force is zero, so the acceleration is zero even though the speed is large
- (D) The air resistance is zero because the speed no longer changes
Show Answer
Answer: C – At terminal velocity the upward drag exactly balances the downward weight. Resultant force = 0, so by Newton’s second law the acceleration is zero and the speed stays constant.
Q4. A skydiver steps out of a stationary aircraft. What is her acceleration in the first moment after leaving the aircraft?
- (A) Zero, because she has not yet reached terminal velocity
- (B) Exactly 9.81 m s−2, and it stays at this value until the parachute opens
- (C) Greater than 9.81 m s−2, because she is moving towards the Earth
- (D) Slightly less than 9.81 m s−2, because a small air resistance already acts upwards
Show Answer
Answer: D – Her speed is still small, so drag is small but not zero. The resultant force is weight minus a small drag, giving an acceleration just below g. It then decreases further as her speed rises.
Q5. A skydiver falling at a terminal velocity of 55 m s−1 opens her parachute. What happens immediately after the parachute opens?
- (A) She accelerates downwards faster because the parachute adds mass
- (B) She continues at 55 m s−1 because she is at terminal velocity
- (C) Her speed stays the same but her acceleration becomes 9.81 m s−2 upwards
- (D) The air resistance suddenly becomes much larger than her weight, so she decelerates
Show Answer
Answer: D – The open parachute has a huge surface area, so at 55 m s−1 the drag is far bigger than her weight. The resultant force is upwards, producing a large deceleration until she slows to a new, lower terminal velocity.
Q6. After opening her parachute, a skydiver settles at a new terminal velocity much lower than before. Why is the new terminal velocity lower?
- (A) The larger area means the drag force can equal her weight at a lower speed
- (B) The parachute reduces her weight, so less speed is needed
- (C) The air becomes thinner as she descends, reducing the drag
- (D) Her mass decreases when the parachute opens
Show Answer
Answer: A – Terminal velocity is reached when drag = weight. With the parachute’s large area, that balance is achieved at a much lower speed, giving a safe landing speed.
Q7. Two spheres of the same diameter, one made of steel and one made of wood, are dropped from a tall building. Air resistance is significant. Which sphere has the higher terminal velocity, and why?
- (A) The wooden sphere, because it is lighter and is slowed down less
- (B) Both are the same, because terminal velocity does not depend on mass
- (C) The steel sphere, because its greater weight needs a greater drag – and so a greater speed – to balance it
- (D) The wooden sphere, because it reaches terminal velocity sooner
Show Answer
Answer: C – Both spheres experience the same drag at the same speed (same size and shape). The steel sphere’s weight is larger, so drag only catches up with the weight at a higher speed.
Q8. A falling object has its cross-sectional area doubled while its mass stays the same (for example, a skydiver spreading out into a star shape). What is the effect on its terminal velocity?
- (A) It doubles
- (B) It stays the same
- (C) It increases slightly
- (D) It decreases
Show Answer
Answer: D – A larger area produces more drag at any given speed, so the drag equals the (unchanged) weight at a lower speed. This is why a spread-eagle skydiver falls more slowly than one in a head-down dive.
Q9. An object is dropped in a vacuum chamber on the Earth. Which statement describes its motion correctly?
- (A) It keeps accelerating at g and never reaches a terminal velocity
- (B) It reaches terminal velocity when its speed is high enough
- (C) Its acceleration decreases gradually to zero
- (D) It falls at constant speed because there is no air
Show Answer
Answer: A – With no air there is no drag, so the resultant force is always the weight. The acceleration stays at g and the speed keeps increasing – terminal velocity is purely an effect of resistive forces.
Q10. A small raindrop quickly reaches its terminal velocity, which is only a few metres per second. Why is the terminal velocity of a raindrop so low?
- (A) Raindrops are streamlined, so they experience no air resistance
- (B) The drop’s weight is very small, so drag equals the weight at a low speed
- (C) The drop evaporates as it falls, so its speed cannot increase
- (D) Air resistance on a drop is constant and equal to its weight at all speeds
Show Answer
Answer: B – Terminal velocity occurs when drag = weight. A raindrop’s weight is tiny, so only a small drag is needed to balance it, and that drag is produced at a low falling speed.
Graphs of Non-Uniform Motion
Q11. The velocity-time graph of a falling object curves and becomes less steep as time passes. What does the decreasing gradient of the curve tell us?
- (A) The displacement is decreasing
- (B) The velocity is decreasing
- (C) The acceleration is decreasing
- (D) The resultant force is increasing
Show Answer
Answer: C – The gradient of a velocity-time graph is the acceleration. A curve that flattens has a decreasing gradient, so the acceleration is falling – exactly what happens as drag builds up towards terminal velocity.
Q12. A velocity-time graph for a sprinter is curved. A tangent to the curve at t = 5 s passes through the points (0 s, 6 m s−1) and (10 s, 26 m s−1). What is the sprinter’s instantaneous acceleration at t = 5 s?
- (A) 0.50 m s−2
- (B) 5.2 m s−2
- (C) 3.2 m s−2
- (D) 2.0 m s−2
Show Answer
Answer: D – Instantaneous acceleration = gradient of the tangent = (26 − 6) / (10 − 0) = 20 / 10 = 2.0 m s−2. On a curve, the chord between two points would only give the average acceleration.
Q13. A student estimates the displacement of an object from its curved velocity-time graph by counting squares. One large square on the graph represents 2.0 s × 10 m s−1, and the area under the curve is approximately 9 large squares. What is the displacement?
- (A) 180 m
- (B) 90 m
- (C) 18 m
- (D) 360 m
Show Answer
Answer: A – Displacement = area under a velocity-time graph. Each square = 2.0 × 10 = 20 m, so 9 squares × 20 m = 180 m. Counting squares works even when the acceleration is not constant.
Q14. The displacement-time graph of an object is a curve that gets steeper and steeper. What does this show about the motion?
- (A) The acceleration is negative
- (B) The speed is increasing
- (C) The speed is constant but large
- (D) The object is slowing down
Show Answer
Answer: B – The gradient of a displacement-time graph is the velocity. A steepening curve means the gradient – and therefore the speed – is increasing: the object is accelerating.
Q15. A car brakes and its displacement-time graph gradually levels off to a horizontal line. What does the levelling off indicate?
- (A) The car’s acceleration is increasing
- (B) The car is moving at a high constant speed
- (C) The car’s speed is decreasing towards zero
- (D) The car has started moving backwards
Show Answer
Answer: C – The gradient of the displacement-time graph is falling towards zero, so the speed is falling towards zero. Once the graph is horizontal, the displacement no longer changes – the car has stopped.
Q16. An object is dropped from rest and falls through air until it reaches terminal velocity. Which description matches its acceleration-time graph?
- (A) It starts at 9.81 m s−2 and decreases, curving down to zero
- (B) It starts at zero and increases to 9.81 m s−2
- (C) It stays constant at 9.81 m s−2 and then drops suddenly to zero
- (D) It starts at 9.81 m s−2 and increases as the speed increases
Show Answer
Answer: A – At the instant of release the speed is zero, so drag is zero and a = g. As speed builds, drag grows, the resultant force shrinks and the acceleration falls smoothly towards zero at terminal velocity.
Q17. A ball is thrown vertically upwards and caught again, with air resistance acting throughout. How does the magnitude of the gradient of its velocity-time graph on the way up compare with that on the way down?
- (A) They are equal, both being g
- (B) It is smaller on the way up, because the ball is slowing down
- (C) It is zero at the top, so the two cannot be compared
- (D) It is larger on the way up, because weight and drag act in the same direction
Show Answer
Answer: D – Going up, both weight and drag point downwards, so the deceleration is greater than g. Coming down, drag acts upwards against the weight, so the acceleration is less than g. The velocity-time graph is steeper on the way up.
Q18. Why can the acceleration at one instant on a curved velocity-time graph not be found from the change in velocity divided by the total time?
- (A) Because velocity is a vector and time is a scalar
- (B) Because that calculation gives the average acceleration; the instantaneous value needs the gradient of the tangent at that instant
- (C) Because the acceleration is zero at every instant on a curve
- (D) Because velocity-time graphs can only be used for constant acceleration
Show Answer
Answer: B – Δv/Δt over a whole interval averages the changing acceleration. The instantaneous acceleration at a point is the gradient of the tangent drawn to the curve at that point.
Q19. For an object moving with changing acceleration, what does the area under its acceleration-time graph represent?
- (A) The distance travelled
- (B) The resultant force
- (C) The change in velocity
- (D) The kinetic energy gained
Show Answer
Answer: C – Since a = Δv/Δt, the area under an acceleration-time graph (a × t) equals the total change in velocity, just as the area under a velocity-time graph gives the displacement.
Q20. A stone is dropped from a tall tower and reaches terminal velocity before hitting the ground. Which velocity-time graph shape is correct?
- (A) A straight line of constant positive gradient from the origin
- (B) A straight line that becomes vertical at terminal velocity
- (C) A curve that rises with increasing gradient and then levels off
- (D) A curve that rises steeply at first, with decreasing gradient, levelling off at the terminal velocity
Show Answer
Answer: D – Initially a = g, so the gradient is steep. As drag grows the gradient (acceleration) falls, and the curve flattens to a horizontal line at terminal velocity where the acceleration is zero.
Applying Newton’s Laws When Acceleration Changes
Q21. A car’s engine provides a constant driving force of 4000 N. Air resistance and friction increase as the car’s speed increases. The car starts from rest. When is its acceleration greatest?
- (A) At the start, when the resistive forces are smallest
- (B) At the top speed, when the engine works hardest
- (C) Halfway to top speed, when the forces are balanced
- (D) The acceleration is constant because the driving force is constant
Show Answer
Answer: A – Resultant force = driving force − resistive force. At the start the speed is zero, so resistance is near zero and the resultant force – and hence the acceleration – is at its maximum. It falls as speed builds.
Q22. The car in Q21 eventually reaches a steady top speed, even though the engine still provides 4000 N. Why?
- (A) The engine force has been used up by the motion
- (B) The resistive force has grown until it equals 4000 N, so the resultant force is zero
- (C) The mass of the car has increased at high speed
- (D) The car’s acceleration becomes negative at top speed
Show Answer
Answer: B – As speed rises, resistance rises. When resistance = driving force, the resultant force is zero, acceleration is zero and the car cruises at constant (maximum) speed – the same physics as terminal velocity.
Q23. A ball bearing dropped into a cylinder of oil reaches terminal velocity almost immediately, while the same ball bearing falling in air keeps accelerating for much longer. Why the difference?
- (A) Oil has no upthrust, so the ball cannot accelerate
- (B) The ball’s weight is greater in oil than in air
- (C) Oil is very viscous, so a large drag force is produced even at very low speeds
- (D) Air resistance does not depend on speed, but viscous drag does
Show Answer
Answer: C – In a viscous liquid the resistive force rises rapidly with speed, so it balances the (weight − upthrust) after only a tiny increase in speed. In air the drag is much weaker, so a far higher speed is needed.
Q24. The same object falls in two fluids: air and a thick oil. How do its terminal velocities compare?
- (A) The terminal velocity is the same in both fluids
- (B) The terminal velocity in oil is higher, because oil is denser
- (C) There is no terminal velocity in a liquid
- (D) The terminal velocity in oil is much lower than in air
Show Answer
Answer: D – Oil provides much larger resistive forces (viscous drag, plus a larger upthrust because oil is denser than air), so the forces balance at a much lower falling speed.
Q25. A projectile is fired at an angle in air. Air resistance is not negligible. Compared with the same launch in a vacuum, which statement is correct?
- (A) The range and the maximum height are both increased
- (B) The maximum height is lower and the range is shorter
- (C) The maximum height is the same but the range is shorter
- (D) The impact speed is equal to the launch speed
Show Answer
Answer: B – Drag acts opposite to the motion throughout the flight, removing energy. The projectile cannot climb as high, cannot travel as far, and lands with less speed than it was launched with.
Q26. With air resistance, the trajectory of a projectile is not a symmetrical parabola. Which description of the real trajectory is correct?
- (A) The descending part of the path is steeper than the ascending part
- (B) The ascending part of the path is steeper than the descending part
- (C) The path is symmetrical but shorter
- (D) The projectile lands further away than the highest point’s midpoint suggests, because drag pushes it forwards
Show Answer
Answer: A – Drag slows the horizontal motion throughout, and on the way down the vertical speed is limited by drag. The projectile comes down more steeply and lands closer than the vacuum parabola predicts – a classic qualitative 9702 point.
Q27. Newton’s second law can be written as: resultant force = rate of change of momentum. A variable force acts on an object. What does the area under the force-time graph give?
- (A) The work done by the force
- (B) The distance moved while the force acts
- (C) The average value of the force
- (D) The change in momentum of the object
Show Answer
Answer: D – Since F = Δp/Δt, the product F × Δt accumulates the momentum change. The area under an F-t graph is the impulse, which equals the total change in momentum – true even when F and the acceleration vary.
Q28. A student tries to analyse a falling skydiver using the SUVAT equations (v = u + at, s = ut + ½at2, and so on) for the whole fall. Why will this give wrong answers?
- (A) SUVAT equations only work for motion in a straight line upwards
- (B) SUVAT equations cannot be used when the mass changes
- (C) SUVAT equations are only valid when the acceleration is constant, but here the acceleration decreases as drag grows
- (D) SUVAT equations ignore gravity
Show Answer
Answer: C – The equations of motion are derived assuming constant acceleration. A skydiver’s acceleration falls from g towards zero, so SUVAT can at best be used for short intervals where a is approximately constant – otherwise graphs or momentum methods are needed.
Q29. An object of mass m falls vertically through air. At a certain speed v the drag force is D. Which expression gives its acceleration at that instant?
- (A) g − D/m
- (B) D/m
- (C) g + D/m
- (D) (mg − D) × m
Show Answer
Answer: A – Resultant force = mg − D (downwards positive). Dividing by m: a = (mg − D)/m = g − D/m. As v and D grow, a falls, reaching zero when D = mg.
Q30. A stone is dropped into a deep tank of water. Its acceleration is close to g at entry but falls rapidly to zero as it sinks at terminal velocity. What is the main reason the acceleration falls?
- (A) The weight of the stone decreases as it goes deeper
- (B) The viscous drag increases as the stone’s speed increases, reducing the resultant force
- (C) The upthrust increases in proportion to the stone’s speed
- (D) The mass of the stone increases as water is absorbed
Show Answer
Answer: B – Weight and upthrust are (almost) constant once the stone is submerged. It is the speed-dependent viscous drag that grows, shrinking the resultant force until it reaches zero at terminal velocity.
Keep practising: Work through more Cambridge 9702 topics on our Physics MCQs hub and keep the A Level Physics Formula Sheet open while you revise – every equation for AS and A Level in one place.
All questions are original practice questions written for QuaidianPoint – not Cambridge past-paper questions. QuaidianPoint is not affiliated with Cambridge Assessment International Education.