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
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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
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