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

Answer:

8400 N

Explanation:

We are given that

Mass,m=1.2 kg

Initial speed,u=-8 m/s

Final speed,v=6 m/s

Time,$\Delta t=2 ms=2\times 10^{-3} s$

$1 ms=10^{-3} s$

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,$F_{avg}=\frac{\Delta p}{\Delta t}$

Using the formula

$F_{avg}=\frac{16.8}{2\times 10^{-3}}=8400 N$

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