Question

A kid drops 1.2kg heavy ball vertically down. The ball hits ground and vertically bounce back. Speeds of the ball just before and after are 8 m/s and 6 m/s respectively. The ball was in contact with the ground for 2ms (milli seconds). What is the magnitude of the average force on the ball during this collision

Answer 1

8400 N

Step-by-step explanation:

We are given that

Mass,m=1.2 kg

Initial speed,u=-8 m/s

Final speed,v=6 m/s

Time,
`\Delta t=2 ms=2* 10^(-3) s`

`1 ms=10^(-3) s`

We have to find the magnitude of the average force on the ball during this collision.

Change in momentum,=
`\Delta p=m(v-u)=1.2(6+8)=16.8 kgm/s`

Average force,
`F_(avg)=(\Delta p)/(\Delta t)`

Using the formula

`F_(avg)=(16.8)/(2* 10^(-3))=8400 N`

Hence, the magnitude of the average force on the ball during this collision=8400 N