OxfordAQA IGCSE Physics: Forces and Their Effects — 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 — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Forces and Their Effects revision notes

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Questions

1. State what is meant by a resultant force, and what happens when it is zero. [2]

2. Classify as scalar or vector: distance, velocity, mass, force, speed, momentum. [3]

3. A cyclist accelerates uniformly from rest to 12 m s⁻¹ in 6.0 s, then travels at constant speed for 20 s, then decelerates uniformly to rest in 4.0 s.

(a) Calculate the acceleration in the first stage. [2] (b) Calculate the total distance travelled. [4] (c) Calculate the average speed for the whole journey. [2]

4. A parachutist of mass 80 kg reaches terminal velocity. (g = 10 N kg⁻¹)

(a) State the air resistance at terminal velocity, with a reason. [2] (b) Explain, in four steps, how terminal velocity is reached. [4] (c) Explain what happens when the parachute opens. [3]

5. A 0.045 kg ball travelling at 22 m s⁻¹ is struck and returns at 26 m s⁻¹ in 5.0 ms.

(a) Calculate the change in momentum. [3] (b) Calculate the average force. [2]

6. A uniform 1.2 m plank of weight 60 N is pivoted at its centre. A 40 N weight is placed 0.40 m from the pivot.

(a) Calculate the moment of the 40 N weight. [2] (b) Calculate where a 25 N weight must be placed to balance it. [3] (c) Explain why the plank’s own weight can be ignored. [2]

7. State Newton’s third law, and explain why weight and the normal contact force on a book resting on a table are not a third-law pair. [3]

8. A 1000 kg car accelerates from rest to 20 m s⁻¹ in 5.0 s. If the drag on the car at that moment is 300 N, calculate the driving force. [4]

9. State the equation for stopping distance, and explain what happens to thinking distance and braking distance separately when speed doubles. [3]

10. Explain why “an object at terminal velocity has no forces on it” is an incorrect description. [2]

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Answers

1. The single force that has the same effect as all the forces acting [1]. When it is zero the object remains at rest or continues at constant velocity [1].

2. Scalars: distance, mass, speed [1] [1]. Vectors: velocity, force, momentum [1].

3. (a) a = 12 ÷ 6.0 [1] = 2.0 m s⁻² [1]. (b) Stage 1: ½ × 6.0 × 12 = 36 m [1] Stage 2: 12 × 20 = 240 m [1] Stage 3: ½ × 4.0 × 12 = 24 m [1] Total = 300 m [1]. (c) Total time = 30 s [1]; average speed = 300 ÷ 30 = 10 m s⁻¹ [1].

4. (a) 800 N [1], because at terminal velocity the resultant force is zero, so air resistance equals weight [1]. (b) Weight acts downwards so the parachutist accelerates [1]. As speed increases, air resistance increases [1]. When air resistance equals weight the resultant force is zero [1]. Acceleration becomes zero, so the parachutist falls at constant velocity [1]. (c) Air resistance increases suddenly and exceeds the weight [1], giving a resultant upward force, so the parachutist decelerates [1]. As speed falls, air resistance falls until it again equals weight, giving a new, lower terminal velocity [1].

5. (a) Taking the initial direction as positive: Initial p = 0.045 × 22 = +0.99; final p = 0.045 × (−26) = −1.17 [1] Δp = −1.17 − 0.99 [1] = −2.16 kg m s⁻¹ (magnitude 2.16) [1]. (b) F = 2.16 ÷ (5.0 × 10⁻³) [1] = 432 N [1].

6. (a) 40 × 0.40 [1] = 16 N m [1]. (b) 25 × d = 16 [1] d = 0.64 m from the pivot, on the other side [1] [1]. (c) The plank is uniform, so its weight acts at its centre of mass, which is at the pivot [1]; its moment about the pivot is therefore zero [1].

7. Newton’s third law: forces come in pairs, equal in size and opposite in direction, acting on two different bodies [1]. Weight and normal contact force both act on the same object (the book), so they are a first-law balance, not a third-law pair [1]; the genuine third-law pair is the book pushing down on the table and the table pushing back up on the book [1].

8. a = (20 − 0) ÷ 5.0 = 4.0 m s⁻² [1]. Resultant force F = ma = 1000 × 4.0 = 4000 N [1]. Driving force = resultant + drag = 4000 + 300 = 4300 N [1], since the resultant is what remains after drag is subtracted [1].

9. Stopping distance = thinking distance + braking distance [1]. Doubling speed doubles thinking distance, since it is proportional to speed [1]; it quadruples braking distance, since that depends on speed squared [1].

10. At terminal velocity the resultant force is zero, not the individual forces themselves [1] — weight and drag are both still acting on the object, equal and opposite, so describing it as having “no forces” is factually wrong even though its motion is unchanging [1].

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Where marks are usually lost

• Ignoring the vector nature of momentum in a rebound.
• Saying the parachutist stops at terminal velocity.
• Forgetting to convert milliseconds to seconds.
• Not explaining why a uniform plank’s weight has no moment about a central pivot.
• Naming weight and normal contact force on the same object as a third-law pair.
• Forgetting to add drag back on to the resultant force to find the driving force.
• Saying braking distance doubles (rather than quadruples) when speed doubles.