A bullet of mass 10 g moving with a velocity of 400 m/s gets embedded in a freely suspended wooden block of mass 900 g . What is the velocity acquired by the wood
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m1u1 + m2u2 = (m1 + m2)v
Since the first body combines with the second.
10x400= 910 v
on solving:
v= 400/91
= 4.4 m/s
Since the first body combines with the second.
10x400= 910 v
on solving:
v= 400/91
= 4.4 m/s
sky365plus:
That's correct!
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