A-LevelPhysicsDynamics9702/21 · Oct/Nov 2018 · Q26 Marks

A wooden block moves along a horizontal frictionless surface, as shown in Fig. 2.1. [Figure 2.1] The block has mass 85g and moves to the left with a velocity of 2.0ms⁻¹. A steel ball of mass 4.0g is fired to the right. The steel ball, moving horizontally with a speed of 45ms⁻¹, collides with the block and remains embedded in it. After the collision the block and steel ball both have speed v.

(a)Calculate v.[2]
(b)(i)For the block and ball, state 1. the relative speed of approach before collision, 2. the relative speed of separation after collision.[1]
(b)(ii)Use your answers in (i) to state and explain whether the collision is elastic or inelastic.[1]
(c)Use Newton's third law to explain the relationship between the rate of change of momentum of the ball and the rate of change of momentum of the block during the collision.[2]

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[2] (p =) mv or 4.0 × 45 or 2.0 × 85 or 89v
[2] (4.0 × 45) - (2.0 × 85) = 89 v
[2] v = 0.11 ms⁻¹
[1] 1. speed of approach = 47 m s⁻¹
[1] and 2. speed of separation = 0
[1] speed of separation less than/not equal to speed of approach and so inelastic collision
[2] force is equal to rate of change of momentum
[2] force on ball (by block) equal and opposite to force on block (by ball) so rates of change of momentum are equal and opposite
[2] or force on ball (by block) equal and opposite to force on block (by ball)
[2] force is equal to rate of change of momentum so rates of change of momentum are equal and opposite

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