Question

A cricket ball of mass 150g is moving with a velocity of 12m/s and hitted by a bat so that the ball is turned back with a velocity of 20m/s. if the duration of contact btw the ball and bat is 0.01s.then: 1. the impulse is and the average force exerted on the ball by the bat is method

Answer 1

Impulse is given by change in momentum ie  $mv_1-mv_2$where m is the mass,v1 is incoming velocity, v2 is outgoing velocity. Since the ball is turned back the the second velocity will have minus sign so this becomes additive.Average force is  $\frac{ mv_1 -mv_2}{t}$. t is the time given.

Answer 2

Explanation:

A 150 g cricket ball moving at a speed of 12 m s-1 is

hit by a bat and turned back at a speed of 20 m s-1.

The duration of impact is 0.01 s. What is the average

force exerted on the ball by the bat?

Given :

m = 150

g = 0.15 kg,

u = 12 m/s^-1

v= -20 m/s^-1

t = 0.01 s,

To find :

F = ?

Solution :

As,

➸ F x t = Impulse = change in momentum

➸ F x 0.01

➸ m (v - u)

➸ 0.150 (- 20 - 12)

➸ F = -0.15 x 32/0.01

....➸ F = - 480N