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) 0.000 kg . m/s
B) 0.0280 kg . m/s upwards
C) 0.0280 kg . m/s downwards
D) 0.308 kg . m/s downwards
E) 0.308 kg . m/s upwards

Question

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)  0.000 kg . m/s
B)  0.0280 kg . m/s  upwards
C)  0.0280 kg . m/s  downwards
D)  0.308 kg . m/s downwards
E)  0.308 kg . m/s upwards

Quizplus · Accepted Answer

The change in momentum is calculated using the formula Δp = m(v_final - v_initial). Before the bounce, the velocity is downwards at 1.20 m/s (taking down as negative), and after the bounce, it is upwards at 1.00 m/s (up is positive). Thus, Δp = 0.140 kg * (1.00 - (-1.20)) m/s = 0.140 kg * 2.20 m/s = 0.308 kg*m/s upwards.

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