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

Answer

D

Explanation

Initial kinetic energy of bullet `= (1)/(2) mv^(2) `
After inelastic collision system moves with velocity V By the conservation of momentum
` mv + 0 ( m+M) V rArr V= (mv)/( m+M) `
Kinetic energy of system `= (1)/(2) (m+M) V^(2) `
` (1)/(2) (m+M) ((mv)/(m+M) )^(2) `
Loss of kinetic energy = `(1)/(2) mv^(2) - (1)/(2) (m+M) ((mv)/(m+M))^(2) `
= `(1)/(2) mv^(2) ((M)/( m+M))`