Question

A 1.2 kg object moving with a speed of 8.0 m/s collides perpendicularly with a wall and bounces off with a speed of 6.0 m/s in the opposite direction. If the object is in contact with the wall for 2.0 milliseconds, what is the magnitude of the average force on the object by the wall

Answer 1

Answer:

Average force will be equal to 8400 N

Explanation:

We have give mass of the object m = 1.2 kg

Initial velocity of the object u = - 8 m/sec

And after collision to the wall final velocity of the object v = 6 m/sec

Change in momentum is equal to $P=m(v-u)=1.2\times (6-(-8))=1.2\times 14=16.8kgm/sec$

Time is given t = 2 milliseconds = 0.002 sec

We know that force is equal to rate of change of momentum

So $F=\frac{dP}{dt}=\frac{16.8}{0.002}=8400N$

So average force will be equal to 8400 N