Forces and Momentum: Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

Related: Forces and Momentum revision notes, the study guide and the IB DP Physics syllabus guide.

Section A

1. State Newton’s third law. [1]

2. Define impulse and state its SI unit. [2]

3. State the condition under which centripetal force acting on a body is provided by tension alone. [1]

Section B

4. A 0.50 kg ball travelling at 8.0 m/s is struck by a bat and rebounds at 6.0 m/s in the opposite direction. The bat is in contact with the ball for 0.020 s.

(a) Calculate the change in momentum of the ball, taking the initial direction as positive. [2] (b) Calculate the average force exerted by the bat on the ball. [2]

5. A 1500 kg car travelling at 20 m/s collides head-on with a stationary 1000 kg car. After the collision the two cars move together.

(a) Calculate the common velocity of the two cars immediately after the collision. [2] (b) Determine, showing your working, whether the collision is elastic or inelastic. [3]

6. A 0.20 kg mass is attached to a string and swung in a horizontal circle of radius 0.50 m at a constant speed, completing one revolution every 1.0 s.

(a) Calculate the linear speed of the mass. [2] (b) Calculate the centripetal acceleration of the mass. [2] (c) State which named force provides the centripetal force in this scenario. [1]

Section C

7. A skydiver of mass 80 kg falls with an initial downward acceleration less than 9.81 m/s2 due to air resistance.

(a) Using Newton’s second law, explain why the presence of air resistance means the skydiver’s acceleration is less than g. [2] (b) Explain, in terms of the resultant force, what happens to the skydiver’s acceleration as speed increases and air resistance grows, up to the point terminal velocity is reached. [2] (c) Sketch (in words) the free-body diagram at terminal velocity, and state the relationship between the forces shown. [2] (d) After the parachute opens, the skydiver’s speed decreases rapidly for a short period. Explain, using F = Δp/Δt, why a rapidly decreasing speed corresponds to a large resultant force. [2]

Worked answers

1. Every action force has an equal and opposite reaction force acting on a different body. [1]

2. Impulse is the product of force and the time for which it acts (J = FΔt); it equals the resulting change in momentum. [1] SI unit: kg m/s (equivalently, N s). [1]

3. Tension alone provides the centripetal force when the circular motion is horizontal (or in the absence of any other force with a component towards the centre, such as when gravity acts perpendicular to the plane of motion, as in a conical pendulum’s horizontal case is an approximation – for a simple horizontal circle on a frictionless surface with only string tension acting, tension is the sole centripetal force). [1]

4. (a) Δp = m(v_final - v_initial) = 0.50 x (-6.0 - 8.0) = 0.50 x (-14.0) = -7.0 kg m/s. [2] (b) F = Δp/Δt = -7.0/0.020 = -350 N (magnitude 350 N, direction opposite to the ball’s initial motion). [2]

5. (a) By conservation of momentum: (1500)(20) + (1000)(0) = (1500+1000)v, so 30000 = 2500v, v = 12 m/s. [2] (b) KE before = 1/2(1500)(20^2) = 300000 J. KE after = 1/2(2500)(12^2) = 180000 J. Since KE fell from 300000 J to 180000 J (not conserved) while momentum was conserved, the collision is INELASTIC. [3] (1 mark for correct KE before, 1 for correct KE after, 1 for correct conclusion with reasoning)

6. (a) v = 2πr/T = 2π(0.50)/1.0 = 3.14 m/s. [2] (b) a = v^2/r = (3.14)^2/0.50 = 19.7 m/s2. [2] (c) Tension in the string. [1]

7. (a) The resultant force on the skydiver is weight minus air resistance (both acting, in opposite directions); since air resistance is non-zero and opposes motion, the resultant force is less than weight alone, so by F = ma the acceleration is less than g (which corresponds to weight being the only force). [2] (b) As speed increases, air resistance increases, so the resultant force (weight minus air resistance) decreases, meaning acceleration decreases; this continues until air resistance equals weight, at which point the resultant force is zero and acceleration reaches zero – this is terminal velocity. [2] (c) Two arrows of equal length acting on the skydiver: weight acting downward and air resistance acting upward, with the two forces equal in magnitude (resultant force = zero), consistent with Newton’s first law describing constant velocity. [2] (d) A rapidly decreasing speed means a large change in momentum in a short time interval (large Δp, small Δt), and since F = Δp/Δt, a large Δp over a small Δt gives a large resultant force – this is exactly why a parachute is designed to increase air resistance sharply, producing the large deceleration force needed to slow the skydiver quickly. [2]

Why question 7 goes further than the revision notes

Question 7 deliberately extends beyond the two worked examples in the revision notes (a collision classification and an impulse calculation) into a qualitative, multi-step free-fall scenario, because Paper 2 long-answer questions on A.2 routinely combine Newton’s laws with impulse-momentum reasoning in exactly this kind of extended scenario rather than testing each formula in isolation. Part (d) in particular tests whether a student can apply F = Δp/Δt to a situation where no numbers are given at all – a common HL-paper style that rewards conceptual fluency with the relationship over formula substitution.

Official syllabus

International Baccalaureate Organization, Diploma Programme Subject Brief – Sciences: Physics, first assessment 2025, published January 2022 – the same source cited by the Forces and Momentum revision notes and the IB DP Physics syllabus guide.