Question

A cricket ball of mass 0.2 kg moving with a velocity of 40 ms-1 is brought to rest by a player in 0.02 seconds. What is the impulse of a ball and the average force applied by the player
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Answer 1

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impulse=change in momentum=0.2×20ms=4kg-m/s

force average= impulse/time=4/0.1=40n

Answer 2

Given:

A cricket ball of mass 0.2 kg moving with a velocity of 40 ms-1 is brought to rest by a player in 0.02 seconds

TO find:

the impulse of a ball and the average force applied by the player​

Solution:

We have given the parameters as:

mass = 0.2 kg = 200 g

v1 = 40 m/sec

v2 = 0 m/sec

Time = 0.02 sec

Impulse is given by the change in momentum of the body

Impulse = m[v2-v1]

Impulse = 200[0 - 40]

Impulse = -8000 Ns

Force is given by the product of the mass and the acceleration of the ball

a = v2 - v1/t

a = 0 - 40/0.02

a = -2000 m/sec²

F = ma

F = 200×(-2000)

F = -4,00,000 N

F = -400 kN

The impulse of a ball and the average force applied by the player​ are -8000Ns and -400 kN respectively.