A 0.10-kg ball,traveling horizontally at 25 m/s,strikes a wall and rebounds at 19 m/s.What is the magnitude of the change in the momentum of the ball during the rebound?
A)1.2 kg ∙ m/s
B)1.8 kg ∙ m/s
C)4.4 kg ∙ m/s
D)5.4 kg ∙ m/s
E)72 kg ∙ m/s

Question

Quizplus · Accepted Answer

The magnitude of the change in momentum is given by:

Δp = p_f - p_i

where p_f is the final momentum of the ball after the rebound and p_i is the initial momentum of the ball before striking the wall.

Since momentum is a vector quantity, we need to consider its direction as well. The initial momentum of the ball is in the positive x-direction, while the final momentum of the ball is in the negative x-direction.

Thus, we have:

p_i = m*v_i = (0.10 kg)*(25 m/s) = 2.5 kg∙m/s (in the positive x-direction)

p_f = m*v_f = (0.10 kg)*(-19 m/s) = -1.9 kg∙m/s (in the negative x-direction)

Δp = p_f - p_i = (-1.9 kg∙m/s) - (2.5 kg∙m/s) = -4.4 kg∙m/s

The magnitude of Δp is therefore:

|Δp| = |-4.4 kg∙m/s| = 4.4 kg∙m/s

Therefore, the best choice is C.

A 010-Kg Ball,traveling Horizontally at 25 M/s,strikes a Wall and Rebounds