Question

A ball with momentum 0.5 kg m/s coming towards a batsman is hit by him such that it goes on the same path in opposite direction with momentum 0.3 kg m/s. If the time of contact of the ball with the bat is 0.02 s, find the force on the ball by the bat.
(A) 10 N
(B) 40 N
(C) 75 N
(D) 30 N

Answer 1

Answer:

10 N

Explanation:

Force = Change in momentum / Total time

= (0.5 kg m/s - 0.3 kg m/s) / (0.02 s)

= (0.2 kg m/s) / (0.02 s)

= 10 kg m/s²

= 10 N

Answer 2

Answer:

The right option is 40 N.

Explanation:

*$P_{1}$ = Initial momentum = - 0.5 kg m/s as it is moving in opposite direction.

*$P_{1}$ = Final momentum = 0.3 kg m/s.

*t = time = 0.02 s.

Force can be calculated by using the following formula:

F = $\frac{P_2 -P_1}{t}$

Plugging the above values we get:

=  $\frac{0.3 +0.5}{0.02}$

= 40 N.

The required force = 40 N.