Alex throws a 0.10-kg rubber ball down onto the floor.The ball's speed just before impact is 7.5 m/s,and just after is 2.5 m/s.What is the change in the magnitude of the ball's momentum?

Question

Quizplus · Accepted Answer

The change in magnitude of momentum is equal to the final momentum minus the initial momentum, or Δp = p_f - p_i. 
The initial momentum is p_i = m*v_i = 0.10 kg * 7.5 m/s = 0.75 kg*m/s. 
The final momentum is p_f = m*v_f = 0.10 kg * 2.5 m/s = 0.25 kg*m/s. 
Therefore, Δp = 0.25 kg*m/s - 0.75 kg*m/s = -0.50 kg*m/s. The negative sign indicates that the momentum decreased, which is expected since the ball bounced and lost some of its energy. The magnitude of this change is |-0.50 kg*m/s| = 0.50 kg*m/s. 
Therefore, the best answer is B) 1 kg*m/s, since it is the closest option to 0.50 kg*m/s.

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