Answer to Question #258882 in Classical Mechanics for Gabgan

Question

A 0.140-kg baseball is dropped and reaches a speed of 1.20 m/s just before it hits the ground. It rebounds with a speed of 1.00 m/s. What is the change of the ball's momentum?

Answer 1

Before the bounce, the ball is moving downward, so let's say its velocity is $-1.2m/s$ (negative number to indicate downward motion). So it's momentum is ;

$0.14kg ×(-1.2m/s) = -0.168 kg m/s$ (using a negative number to indicate downward momentum).

After the bounce, the ball's momentum is $0.14kg × 1 m/s = 0.14 kg m/s$ (now using positive numbers to indicate upward velocity/momentum

Change in ball's momentum is final momentum minus initial momentum:

$0.14 - (-0.168) =0.308 kg m/s$