A player in 0.05sec brings a cricket ball of mass 200g moving with a velocity of 15m/s to rest. What is the impulse of the ball and the average force applied by the player?
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Answered by
29
m = 0.2 kg, u = 15 m/s, v = 0
Change in momentum = m(v - u) = -0.2 x 15 = -3 Ns
As change in momentum = impulse = -3 Ns
Impulse(J) = average force(F) x time(t)
F = J/t = 3/0.05 = 60 N in _ve direction
Answered by
16
Impulse = Mass × Change in velocity
= 200 × 10^-3 kg × (0 - 15) m/s
= -3 kgm/s
Force = Impulse / Time
= -3 kgm/s / (0.05 s)
= -60 N
[Note: Negative sign indictes that Force and impulse are in the direction opposite to the motion of cricket ball]
= 200 × 10^-3 kg × (0 - 15) m/s
= -3 kgm/s
Force = Impulse / Time
= -3 kgm/s / (0.05 s)
= -60 N
[Note: Negative sign indictes that Force and impulse are in the direction opposite to the motion of cricket ball]
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