A glass marble whose mass is 100gm falls from a height of 40m and rebounds to a height of 10m. find the impulse and the average force between the marble and the floar. If the time during which they are in contact 0.2 sec
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m = 100 gm
h = 40 m
velocity before touching the floor = u m/s
u² = 2 g h = 2 * 9.8m/s² * 40 m= 784
u = - 28 m/s downwards.
height of rebound = 10 m
velocity after touching the ground = v m/s
v² = 2 g h = 2* 9.8 * 10 = 196
v = 14 m/s
Impulse is equal to the change in linear momentum during the impact/collision
= m v - m u = 100 /1000 kg * (14 - (-28)) = 4.2 kg-m/s or N-sec
AVerage force during the impact = 4.2 N-sec/0.2 sec = 21 Newtons.
h = 40 m
velocity before touching the floor = u m/s
u² = 2 g h = 2 * 9.8m/s² * 40 m= 784
u = - 28 m/s downwards.
height of rebound = 10 m
velocity after touching the ground = v m/s
v² = 2 g h = 2* 9.8 * 10 = 196
v = 14 m/s
Impulse is equal to the change in linear momentum during the impact/collision
= m v - m u = 100 /1000 kg * (14 - (-28)) = 4.2 kg-m/s or N-sec
AVerage force during the impact = 4.2 N-sec/0.2 sec = 21 Newtons.
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11
Answer:0.42 Newton/s .....4.2Newton
Explanation:
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