Question

A 8.0-kg ball is thrown towards a wall with a speed of 9.0 m/s. The ball hits the wall and rebounds backward with a speed of 2.0 m/s. What is the momentum change experienced by the ball?

Answer 1

DP = 56 [kg*m/s]

Step-by-step explanation:

We know that momentum is defined by the product of mass by Velocity. In this way by means of the following equation, we can calculate the impulse.

`P=m*v\\`

where:

P = momentum [kg*m/s]

m = mass [kg]

v = velocity [m/s]

Now replacing:

`P=8*9\\P=72[kg*m/s]`

And after the ball hits the wall:

`P_(2)=8*2\\P_(2)=16[kg*m/s]`

The change in momentum is equal to:

`DP=72-16\\DP = 56 [kg*m/s]`