Question

A 0.89 kg ball is moving horizontally with a speed of 3.8 m/s when it strikes a vertical wall. The ball rebounds with a speed of 1.3 m/s. What is the magnitude of the change in linear momentum of the ball?

Answer 1

Answer:

The change in momentum, p = -4.539 kg-m/s

Explanation:

It is given that,

Mass of the ball, m = 0.89 kg

Initial speed of the ball, u = 3.8 m/s

Final speed of the ball, v = -1.3 m/s (as it rebounds)

Let p is the change in linear momentum of the ball. We know that the linear momentum is equal to the product of mass and velocity. It is given by :

$p=m(v-u)$

$p=0.89\ kg\times ((-1.3)-3.8)\ m/s$

p = -4.539 kg-m/s

So, the change in linear momentum of the ball is 4.539 kg-m/s. Hence, this is the required solution.