Question

A ball weighing 10$gm$ hits a hard surface vertically with a speed of 5$m/s$ and rebounds with the same speed. The ball remains in contact with the surface for 0.01 sec. The average force exerted by the surface on the ball is:

• A. 10$N$
• B. 20$N$
• C. 400$N$
• D. 200$N$

Answer 1

Answer: (A)

10$N$

Given,

Mass of ball is $10gm$

Speed of ball is $5m/s$

Time of ball remain on surface is $0.01sec$

Here, ball is moving in upward direction with speed $5m/s$ and in downward direction speed is $5m/s$

We know that

Force exerted by the surface is impulse divided by time and impulse s change in momentum

Now,

Change in Momentum is,

$△P=m(v_f-v_i)$

Substitute all value in above equation

$△P=\frac{10}{1000}(5-(-5\left)\right)△P=\frac{10\times 10}{1000}=\frac{1}{10}$

$t=0.01sec$

Force exerted by the surface is

$F=\frac{△P}{t}F=\frac{\frac{1}{10}}{0.01}F=\frac{100}{10}=10N$

Hence, correct option is $\left(A\right)$