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A bullet of mass 4.2 xx10^(-2)kg moving ...

A bullet of mass `4.2 xx10^(-2)`kg moving at a speed of 300 m/s, gets stuck into a block with a mass 9 times that of the bullet. If the block is free to move without any kind of friction, the heat generated in the process will be

A

45cal

B

405 cal

C

450 cal

D

1701 cal

Text Solution

Verified by Experts

The correct Answer is:
B
Let , mass of the bullet is m, mass of the block is M, speed of the bullet before collision is v and speed of the combined bullet and block system after collision is `v_1`.
According to the law of conservation of momentum,
`mv=(m+M)v_1 or, v_1(m)/(m+M)v`
Heat energy generated = loss of kinetic energy
`=1/2 mv^2=1/2 (m+M)v_1^2`
`=1/2 mv^2=1/2(m+M)xx(m^2v^2)/((m+M)^2)=1/2 (mMv^2)/((m+M))`
`=1/2 xx(mxx9mxx300^2)/((m+9m))=1/2xx9/(10)xx4.2xx10^(-2)xx300^2`
`=1701 J =(1701)/(4.2)cal =405 cal`
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