A 0.140-kg baseball is dropped and reaches a speed of 1.20 m/s just before it hits the ground and bounces.It rebounds with an upward velocity of 1.00 m/s.What is the change of the ball's momentum during the bounce?
A)0.0280 kg ∙ m/s upwards
B)0.0280 kg ∙ m/s downwards
C)0.308 kg ∙ m/s upwards
D)0.308 kg ∙ m/s downwards
E)0.000 kg ∙ m/s

Question

Quizplus · Accepted Answer

The change in momentum ($\Delta p$) is calculated using the formula $\Delta p = m(v_f - v_i)$, where $m$ is the mass of the baseball, $v_f$ is the final velocity, and $v_i$ is the initial velocity. Before the bounce, the velocity is downwards at 1.20 m/s, and after the bounce, it is upwards at 1.00 m/s. Therefore, the change in velocity is $1.00 m/s - (-1.20 m/s) = 2.20 m/s$. The mass of the baseball is 0.140 kg, so the change in momentum is $0.140 kg \times 2.20 m/s = 0.308 kg \cdot m/s$ upwards.

A 0140-Kg Baseball Is Dropped and Reaches a Speed of 1