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hodyreva [135]
4 years ago
11

A kid drops 1.2kg heavy ball vertically down. The ball hits ground and vertically bounce back. Speeds of the ball just before an

d 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
Physics
1 answer:
iVinArrow [24]4 years ago
8 0

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

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