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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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.
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