A-LevelPhysicsDynamics9702/23 · May/June 2017 · Q29 Marks

A constant resultant force F acts on an object A. The variation with time t of the velocity v for the motion of A is shown in [Figure 2.1]. The mass of A is 840 g. The force F is removed at t = 4.0s. Object A continues at constant velocity before colliding with an object B, as illustrated in [Figure 2.2]. Object B is initially at rest. The mass of B is 730 g. The objects A and B join together and have a velocity of 4.7ms⁻¹.

(a)State Newton's second law of motion.[1]
(b)(i)the change in momentum of A,[2]
(b)(ii)the force F.[1]
(c)(i)By calculation, show that the changes in momentum of A and of B during the collision are equal and opposite.[2]
(c)(ii)Explain how the answers obtained in (i) support Newton's third law.[2]
(c)(iii)By reference to the speeds of A and B, explain whether the collision is elastic.[1]

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Mark scheme

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2(a) (resultant) force is proportional/equal to the rate of change of momentum
2(b)(i) change in momentum = m(v₂ - v₁)
2(b)(i) = 0.84 × (8.8 – 4.2)
2(b)(i) = 3.9 (3.86) kg m s⁻¹
2(b)(ii) F = (3.9/4.0) = 0.97 (0.965) N
2(c)(i) change in momentum for A: 0.84 × (4.7 – 8.8) = −3.4 (3.44)
2(c)(i) change in momentum for B: 0.73 × (4.7 – 0) = 3.4 (3.43)
2(c)(i) change in momentum for B is equal and opposite to A
2(c)(ii) change in momentum equal (for A and B)
2(c)(ii) force is change in momentum / time and time (of collision) is the same
2(c)(ii) hence force on A and B equal and opposite as for Newton's third law
2(c)(iii) inelastic as relative speed of approach not equal to relative speed of separation

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