A girl of mass 50kg jumps out of a going boat of mass 300kg on the bank ,with a horizontal velocity of 3m/s
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Answered by
796
m1=50 kg
m2=300 kg
v1= 3 m/s
v2=?
we have to find.
conservation of momentum=m1+v1+m2v2+0
0 is because no initial velocity is applied and the answer will be
50 x 3 + 300v2=0
i.e v2 = -150/300=-0.5m/s
so it is moving backward direction the velocity (v2) will be 0.5m/s in the backward direction.
IF IT HELPS MARK IT AS BRAINIEST......
ANY PROBLEM MESSAGE.....
m2=300 kg
v1= 3 m/s
v2=?
we have to find.
conservation of momentum=m1+v1+m2v2+0
0 is because no initial velocity is applied and the answer will be
50 x 3 + 300v2=0
i.e v2 = -150/300=-0.5m/s
so it is moving backward direction the velocity (v2) will be 0.5m/s in the backward direction.
IF IT HELPS MARK IT AS BRAINIEST......
ANY PROBLEM MESSAGE.....
Answered by
317
hopefully you'll get a lot of hint
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