A bullet of mass `m` moving with velocity `v` strikes a block of mass `M` at rest and gets embedded into it. The kinetic energy of the composite block will be

Options

A. `(1)/(2) mv^(2) xx (m)/((m+M))`
B. `(1)/(2) mv^(2) xx (M)/( (m+M))`
C. `(1)/(2) mv^(2) xx ((M+m))/(m)`
D. `(1)/(2) Mv^(2) xx (m)/((m+M))`

Answer

A

Explanation

By conservation of momentum, `mv + M xx 0 = (m + m)V`
Velocity of composite block `V = ((m)/(m+M))v`
K.E. of composite block `= (1)/(2)(M+m)V^(2)`
`= (1)/(2)(M +m)((m)/(M+m))^(2) v^(2) = (1)/(2) mv^(2)((m)/(m+M))`