Linear momentum and its conservation: past paper questions
Real questions on linear momentum and its conservation
The newest 13 of 13 written questions. Open one to see it on the real paper.
- 9702/23 · May/June 2025 · Q4a[2]
State the principle of conservation of momentum.
See it on the paper → - 9702/23 · May/June 2025 · Q4b(ii)[2]
the percentage of the total initial kinetic energy of the two objects that is transferred to other forms of energy during the collision. percentage = %
See it on the paper → - 9702/22 · February/March 2024 · Q4b(i)[2]
State the principle of conservation of momentum.
See it on the paper → - 9702/22 · October/November 2023 · Q4a(i)[2]
Calculate the component of the final momentum of ball Y in the direction perpendicular to line PQ. component of momentum = N s
See it on the paper → - 9702/22 · October/November 2023 · Q4a(ii)[1]
By considering the component of the final momentum of each ball in the direction perpendicular to line PQ, calculate m Z . m Z = kg
See it on the paper → - 9702/22 · October/November 2023 · Q4a(iii)[2]
During the collision, the average force exerted on Y by Z is F Y and the average force exerted on Z by Y is F Z . Compare the magnitudes and directions of F Y and F Z . Numerical values are not required. magnitudes: directions:
See it on the paper → - 9702/22 · October/November 2023 · Q4b[1]
Two blocks, A and B, move directly towards each other along a horizontal frictionless surface, as shown in the view from above in Fig. 4.3. 4 m s –1 6 m s –1 A B Fig. 4.3 The blocks collide perfectly elastically. Before the collision, block A has a speed of 4 m s –1 and block B has a speed of 6 m s –1 . After the colli...
See it on the paper → - 9702/22 · February/March 2023 · Q4b[1]
Determine the magnitude of the momentum of block Y after the collision. momentum = kg m s –1
See it on the paper → - 9702/22 · February/March 2023 · Q4c[2]
Block X exerts an average force of 7.7 N on block Y during the collision. Calculate the time that the blocks are in contact with each other. time = s
See it on the paper → - 9702/21 · October/November 2021 · Q2b(iii)[2]
Calculate the magnitude of the velocity of ball Y after the collision. velocity = m s –1
See it on the paper → - 9702/21 · October/November 2021 · Q2c[3]
On Fig. 2.3, sketch the variation with time t of the force F X exerted on ball X by ball Y during the collision in (b). 400 F X / N 200 0 0 1 2 3 4 5 t / ms –200 – 400 Fig. 2.3
See it on the paper → - 9702/22 · February/March 2019 · Q3a[2]
By considering the components of momentum in the direction from A to B, show that ball X has a mass of 4.0 kg.
See it on the paper → - 9702/22 · February/March 2019 · Q3b[2]
Calculate the common speed V of the two balls after the collision. V = m s –1
See it on the paper →
What examiners said
This question was reasonably well answered. Some candidates omitted the mass of the ball from their calculation and therefore determined the resolved component of velocity rather than momentum. Another common mistake was to use cos 27° instead of sin 27°, which gave the wrong component of momentum.
— Examiner report, October/November 2023, 9702/22 Q4(a)(i) (Cambridge)
Most candidates realised that the algebraic sum of the two momenta perpendicular to line PQ would be zero. Many candidates made an arithmetic error when calculating the mass of the ball.
— Examiner report, October/November 2023, 9702/22 Q4(a)(ii) (Cambridge)
It was expected that the candidates would use Newton’s third law of motion to deduce that the two forces would be equal and opposite. A common misconception was that the magnitude of FY was greater than the magnitude of FZ. It was insufficient to describe the directions as just ‘different’ rather than ‘opposite’. Candidates should be encouraged to use precise language in descriptive questions of t...
— Examiner report, October/November 2023, 9702/22 Q4(a)(iii) (Cambridge)
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