A 10kg bat hits a 50g ball with 100N force in 1milisecond then what is impulse?
Answers
Answer:
M=0.4 kg
h
1
=5m
h
2
=20m
g=10 m/s
2
velocity of ball after dropping 5m:u
velocity of ball after being hit by bat: v
u
2
=2gh
1
∴u=
2×10×5
u=10 m/s downwards
v
2
=2gh
2
∴v=
2×10×20
=20 m/s upwards
Change in momentum=impulse=m(v+u)
=0.4(10+20)
=0.4×30
=12 kg m/s
Force =100N=12/t
t=time of contact =0.12 sec.
Answer:
M=0.4 kg
h
1
=5m
h
2
=20m
g=10 m/s
2
velocity of ball after dropping 5m:u
velocity of ball after being hit by bat: v
u
2
=2gh
1
∴u=
2×10×5
u=10 m/s downwards
v
2
=2gh
2
∴v=
2×10×20
=20 m/s upwards
Change in momentum=impulse=m(v+u)
=0.4(10+20)
=0.4×30
=12 kg m/s
Force =100N=12/t
t=time of contact =0.12 sec.
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SIMILAR QUESTIONS
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A 150 g ball, moving horizontally at 20 m/s was hit straight back to bowler at 12 m/s. If contact with bat lasted for 0.04 sec, than average force exerted by the bat on the ball is:
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A cricket ball of mass 150 gm is moving with a velocity of 12 m/sec. and is hit by a bat so that the ball is tuned back with a velocity of 20 m/sec. The force of bat acts for 0.01 s on the ball then the average force exerted by the bat on the ball.
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