A bag of sand of mass M is suspended by a string. A bullet of mass m is fired at it with velocity v and gets embedded into it. The loss of kinetic energy in this process is

Options

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

Answer

C

Explanation

Linear momentum is conserved.
`(M + m ) v . = mv`
`therefore v . = ( mv )/( M + m )`
`KE = (1)/(2) (M + m ) v ^(2)`
`therefore KE = (1)/(2) (M + m ) ( m ^(2) v ^(2))/( ( M + m ) ^(2))= (1)/(2) ( m ^(2) v ^(2))/( (M + m ))`
Loss of KE = KE of bullet - KE of embedded bag
`= (1)/(2) mv ^(2) - (1)/(2) ( m ^(2) v ^(2))/( ( M + m ) ^(2)) = (1)/(2) ( m ^(2) v ^(2))/( ( M + m ))`
Loss of KE = KE of bullet - KE of embedded bag
`= (1)/(2) mv ^(2) - (1)/(2) ( m ^(2) v ^(2))/(( M + m )) = ( m M v ^(2))/( 2 ( M + m ))`