Question

7.
A Cricket ball of mass 150 gram moving with a velocity of 12 m/s strikes against the bat. It
rebounds with a velocity of 20 m/s. The ball remains in touch with the bat for 0.01 second the
average force applied by the bat on the ball is
[ ]
A) 840 N
B) 480 N
C) 804 N
D) 408 N
tan Penetrate a​

Answer 1

Explanation:

$\bf Given:-$

• Mass of the cricket ball = 150g = 0.15kg

• Initial velocity = 12m/s

• Final velocity = 20m/s

• Time = 0.01 sec

$\bf To\:Find :-$

• The force applied by the bat on the ball.

$\bf Solution:-$

$\sf \: Force \: = \dfrac{Impulse}{Time}$

As we know that

Impulse = change in linear momentum

Impulse = mass × acceleration

(Acceleration = change in velocity)

Impulse = mass × change in velocity

Therefore:-

$\sf \: Force \: = \dfrac{Mass \times change \: in \: velocity}{Time}$

Change in velocity = Final velocity - initial velocity

= -20 - 12 ( Here we take final velocity as negative because the ball rebounds back)

Change in velocity = -32m/s

Therefore:-

$\sf \: Force \: = \dfrac{0.15 \times - 32}{0.01}$

$\sf \: Force \: = {15 \times - 32}$

$\sf \: Force \: = - 480$

Hence;

$\boxed{ \therefore\sf \: Force \:applied \: by \: the \: bat = 480N}$