A bullet of mass `4.2 xx 10^(-2)` kg, moving at a speed of `300 ms^(- 1)`, 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

Options

A. 45 cal
B. 405 cal
C. 450 cal
D. 1701 cal

Answer

B

Explanation

If V is the velocity of the combined system (i.e. block+ bullet) after collision and vis the velocity of bullet before collision.
By the law of conservation of momentum,
`mv+Mxx0=(M+m)V` or `V=(mv)/(m+M)` …(i)
Loss of kinetic energy= Heat generated in the process
`thereforeDeltaK=1/2mv^(2)-1/2(M+m)V^(2)` [From eqn. (i) ]
`=1/2mv^(2)-1/3(M+m)(m^(2)v^(2))/((M+m)^(2))`
`=1/2mv^(2)-1/2(m^(2)v^(2))/((M+m))=1/2mv^(2)(1-m/(M+m))`
`=1/2mv^(2)((M+m-m)/(M+m))`
`=1/2mv^(2)(M/(m+m))=(mMv^(2))/(2(M+m))` ...(ii)
`becausem=4.2xx10^(-2)kg,v=300ms^(-1)`
`M=9m=9xx4.2xx10^(-2)kg`
Substituting the values in eqn. (ii), we get
`DeltaK=1701J=1701/(4.2)` cal=405 cal