Question

A cricket ball of 200g. moving with a velocity of 10m/s. is brought to rest by a player in 0.5sec Calculate the impulse of the ball and average force applied by the player​

Answer 1

Answer:

Explanation:

Impulse is [ just] the change in momentum ,I = mv[2] - mv[1]

m=200 , v[2] =0 , v[1]= 10

I = 200[0] - 200[10] =- 2000 Ns

Force = RATE of change in momentum = [mv[2] - mv[1]] / [t[2]- t[1]]

F = m( v[2]-v[1])/t = m a , a =( v[2]-v[1] )/t

m=200 , a = [0–10]/0.05 =- 200 m/s^2

F = m a = 200 [200] = 40,000 N =40kN