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
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Impulse is given by change in momentum ie 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 . t is the time given.
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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
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