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A bullet of mass m moving horizontally w...

A bullet of mass m moving horizontally with a velocity v strikes a block of wood of mass M and gets embedded in the block. The block is suspended from the ceiling by a massless string. The height to which block rises is

A

`v^2/(2g)(m/(M+m))^2`

B

`v^2/(2g)((M+m)/(m))^2`

C

`v^2/(2g)(m/M)^2`

D

`v^2/(2g)(M/m)^2`

Text Solution

Verified by Experts

The correct Answer is:
A
The situation is as shown in the figure.
Let V be velocity of the block- bullet system just after collision. Then by the law of conservation of linear momentum, we get
mv=(m+M)V
`V=(mv)/(m+M)`
Let the block rises to a height h.
According to law of conservation of mechanical energy, we get
`1/2(m+M)V^2=(m+M)gh`
`h=V^2/(2g)=v^2/(2g)(m/(M+m))^2`
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