AQA GCSE Physics 8463: Forces – 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 4.5 Forces (4.5.1 to 4.5.7) of the AQA GCSE Physics (8463) specification, Version 1.1 (30 September 2019), for teaching from September 2016 and exams from 2018 onwards. The topic is assessed on Paper 2, set at Foundation and Higher Tier. Questions 11 and 12, and parts 2(b), 6(b), 6(c), 8(b), 8(c), 10(c) and 10(d), test (HT only) content and are labelled Higher tier only. Questions 4 and 9 draw on required practicals 6 and 7. A calculator is allowed throughout.

Learn the content first in the Forces study guide and revision notes. See also the AQA GCSE Physics hub and printable checklist.

Questions

1. Sort these quantities into scalars and vectors: mass, velocity, displacement, speed, force, distance. [2]

2.

(a) The Earth pulls down on a book resting on a table. Describe the force that forms a Newton’s third law pair with this pull. [2] (b) (Higher tier only) Two forces act on a sledge: 12 N due east and 5.0 N due north. Use a scale drawing to find the size and direction of the resultant force. [3]

3. An astronaut’s equipment has a mass of 80 kg. Gravitational field strength on the Moon is 1.6 N/kg.

(a) Calculate the weight of the equipment on the Moon. [2] (b) State the mass of the equipment on the Moon. [1]

4. A student investigates how the extension of a spring depends on the force applied.

Force (N)	0	1.0	2.0	3.0	4.0	5.0	6.0
Extension (cm)	0	1.6	3.2	4.8	6.4	8.0	10.5

(a) Between which two forces was the limit of proportionality passed? Give a reason. [1] (b) Calculate the spring constant. [2] (c) Calculate the elastic potential energy stored when the force is 5.0 N. [2] (d) Describe how the student could check whether the spring had been inelastically deformed, and what result would show it. [2]

5. A uniform plank is balanced on a pivot at its centre. A child of weight 300 N sits 1.2 m to the left of the pivot.

(a) Calculate the moment of the child’s weight about the pivot. [1] (b) An adult of weight 450 N sits on the right. Calculate how far from the pivot the adult must sit for the plank to balance. [3] (c) A small gear drives a larger gear. Explain the effect on the moment at the larger gear. [2]

6. Use g = 9.8 N/kg.

(a) A crate of weight 900 N rests on a base measuring 0.30 m by 0.50 m. Calculate the pressure it exerts on the floor. [2] (b) (Higher tier only) A diver descends in seawater of density 1030 kg/m³. Calculate the increase in pressure between depths of 4.0 m and 12.0 m. [3] (c) (Higher tier only) Explain why a submerged object experiences upthrust. [2] (d) Explain why atmospheric pressure at the top of a mountain is lower than at sea level. [2]

7. A skydiver jumps from a plane and later reaches terminal velocity. Explain the changes in her velocity from the moment she jumps until she reaches terminal velocity, in terms of the forces acting. [5]

8. A cyclist starts from rest. She accelerates uniformly to 6.0 m/s in 4.0 s, travels at 6.0 m/s for 10 s, then decelerates uniformly to rest in 3.0 s.

(a) Calculate her acceleration during the first 4.0 s. [2] (b) (Higher tier only) Use the area under the velocity–time graph to calculate the total distance travelled. [2] (c) (Higher tier only) Calculate her average speed for the whole journey. [2]

9. A car of mass 1400 kg accelerates uniformly from 12 m/s to 20 m/s over a distance of 64 m.

(a) Calculate the acceleration. [2] (b) Calculate the resultant force. The resistive forces on the car are 900 N. Calculate the driving force. [2] (c) In required practical 7, a hanging mass pulls a trolley. Describe how to vary the force at constant total mass. [2]

10. A 1200 kg car travels at 25 m/s. The driver’s reaction time is 0.60 s; the brakes then decelerate the car at 6.25 m/s².

(a) Calculate the thinking distance. [2] (b) Calculate the braking distance and the stopping distance. [3] (c) (Higher tier only) Estimate the braking force. [2] (d) (Higher tier only) Show that the work done by the braking force equals the car’s initial kinetic energy. [2] (e) Explain why the brakes get hot, and give one danger of a very large deceleration. [2]

11. (Higher tier only) A trolley of mass 0.80 kg moves at 1.5 m/s. It collides with a stationary trolley of mass 0.40 kg, and they stick together.

(a) Calculate the momentum of the moving trolley before the collision. [1] (b) Calculate the velocity of the trolleys after the collision. [3] (c) State the condition needed for momentum to be conserved. [1]

12. (Higher tier only) A cyclist’s head (with helmet) has a mass of 5.0 kg and hits the ground at 6.0 m/s.

(a) Calculate the change in momentum of the head as it stops. [1] (b) Without the helmet’s foam, the head would stop in 0.012 s. With it, the head stops in 0.030 s. Calculate the force on the head in each case. [3] (c) Explain how the helmet reduces the force. [2]

Answers

1. Vectors: velocity, displacement, force [1]; scalars: mass, speed, distance [1] Examiner insight: One mark per complete group; a misplaced quantity costs both marks.

2. (a) The book pulls up on the Earth with a gravitational force [1]; of equal size to the Earth’s pull on the book [1] (b) Correct scale drawing with a stated scale, arrows tip to tail [1]; resultant 13 N [1]; direction about 23° north of east [1] Examiner insight: The table’s upward push is not the pair: a third-law pair acts on two different objects.

3. (a) W = mg = 80 × 1.6 [1]; = 128 N [1] (b) 80 kg [1] Examiner insight: A weight given in kg, or with no unit, loses the accuracy mark.

4. (a) Between 5.0 N and 6.0 N: the extension rose by 2.5 cm instead of 1.6 cm [1] (b) k = F/e = 5.0 / 0.080 [1]; = 62.5 N/m [1] (c) Eₑ = 0.5 × 62.5 × 0.080² [1]; = 0.20 J [1] (d) Remove the load and measure the length again [1]; if the spring does not return to its original length, it has been inelastically deformed [1] Examiner insight: Using 8.0 instead of 0.080 m loses the accuracy mark, though the method mark can still be earned.

5. (a) 300 × 1.2 = 360 N m [1] (b) Clockwise moment = anticlockwise moment [1]; 450 × d = 360 [1]; d = 0.80 m [1] (c) The larger gear has a larger radius, so the moment on it is larger [1]; but it turns more slowly [1] Examiner insight: Quoting the principle of moments earns the first mark in (b) even if the arithmetic then slips.

6. (a) A = 0.30 × 0.50 = 0.15 m² [1]; p = 900 / 0.15 = 6000 Pa [1] (b) Δh = 12.0 − 4.0 = 8.0 m [1]; Δp = 8.0 × 1030 × 9.8 [1]; = 80 752 Pa ≈ 8.1 × 10⁴ Pa [1] (c) Pressure increases with depth, so there is greater pressure on the bottom surface than the top [1]; this gives a resultant upward force [1] (d) Higher up there is less air above the surface, so less weight of air [1]; fewer air molecules collide with the surface, so pressure is lower [1] Examiner insight: In (b), the pressure at 12.0 m alone is not the answer; full marks need the difference.

7. At first only her weight acts significantly, so she accelerates downwards [1]; as speed increases, air resistance increases [1]; the resultant force falls, so acceleration falls [1]; eventually air resistance equals weight [1]; resultant force is zero, so velocity is constant (terminal) [1] Examiner insight: “The forces cancel, so she stops” is a common contradiction; zero resultant force means constant velocity, not rest.

8. (a) a = Δv / t = 6.0 / 4.0 [1]; = 1.5 m/s² [1] (b) Area = ½ × 4.0 × 6.0 + 10 × 6.0 + ½ × 3.0 × 6.0 [1]; = 12 + 60 + 9 = 81 m [1] (c) Average speed = 81 / 17 [1]; = 4.76 ≈ 4.8 m/s [1] Examiner insight: Allow error carried forward in (c): a wrong distance from (b) divided by 17 s still earns both marks.

9. (a) 20² − 12² = 2 × a × 64 [1]; a = 256 / 128 = 2.0 m/s² [1] (b) F = 1400 × 2.0 = 2800 N [1]; driving force = 2800 + 900 = 3700 N [1] (c) Move masses from the trolley to the hanger (or back) [1]; so the total mass being accelerated stays the same while the pulling force changes [1] Examiner insight: Writing (20 − 12)² for v² − u² gives 64 instead of 256 and loses the accuracy mark.

10. (a) Thinking distance = speed × reaction time = 25 × 0.60 [1]; = 15 m [1] (b) 0² − 25² = 2 × (−6.25) × s [1]; braking distance = 625 / 12.5 = 50 m [1]; stopping distance = 15 + 50 = 65 m [1] (c) F = ma = 1200 × 6.25 [1]; = 7500 N [1] (d) Eₖ = 0.5 × 1200 × 25² = 375 000 J [1]; work done = F × s = 7500 × 50 = 375 000 J, so they are equal [1] (e) Friction between the brakes and wheel does work, transferring kinetic energy to thermal energy of the brakes [1]; a large deceleration can overheat the brakes or cause loss of control [1] Examiner insight: A “show that” in (d) needs both values fully calculated; a bare 375 000 J earns nothing.

11. (a) p = 0.80 × 1.5 = 1.2 kg m/s [1] (b) Total momentum after = total momentum before = 1.2 kg m/s [1]; (0.80 + 0.40) × v = 1.2 [1]; v = 1.0 m/s [1] (c) The system must be closed (no external forces act) [1] Examiner insight: Using 0.40 kg instead of the combined 1.20 kg loses the last two marks.

12. (a) Δp = 5.0 × 6.0 = 30 kg m/s [1] (b) F = mΔv / Δt [1]; without foam: 30 / 0.012 = 2500 N [1]; with foam: 30 / 0.030 = 1000 N [1] (c) The foam increases the time taken for the head to stop [1]; so the rate of change of momentum, and therefore the force, is smaller [1] Examiner insight: “The foam absorbs the force” earns nothing; link longer time to smaller rate of change of momentum.

Where marks are usually lost

• Using centimetres in F = ke and Eₑ = ½ke².
• Naming the normal contact force as the third-law partner of weight.
• Forgetting to square both velocities in v² − u² = 2as.
• Missing the thinking distance, or adding the reaction time to the braking distance.
• (Higher tier only) Using the pressure at one depth when a pressure difference is asked for.
• (Higher tier only) Dropping the combined mass in a “stick together” momentum question.
• Leaving out units, or giving pressure in N instead of Pa.

Next steps

• Forces revision notes
• Forces study guide
• AQA GCSE Physics course hub
• Printable AQA GCSE Physics checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

AQA GCSE Physics (8463) specification, Version 1.1, 30 September 2019, for teaching from September 2016 and exams from 2018 onwards (AQA), section 4.5 Forces.