A bullet of mass `'m'` is fired from a gum of mass `M` with a (muzzle) velocity `u`. If the cart on which gun is fixed can move on the smooth horizontal floor as shown

Energy of explosion (or) change in kinetic energy of the system is

Options

A. `(m(M+msin^(2) theta)u^(2))/(2(M+m))`
B. `(M(M+m cos^(2) theta)u^(2))/((M+m))`
C. `(m(M+m)u^(2))/(M)`
D. None of these

Answer

A

Explanation

`Mv = m( u cos theta-v), v = ((m u cos theta)/(M+m))`
`(1)/(2)kx^(2) = (1)/(2)Mv^(2) , x^(2) = (M)/(K)v^(2)`
`x = sqrt((M)/(K)) xx v , x = sqrt((M)/(K)) xx ((m u cos theta)/(M+m))`
`Delta k = (1)/(2)Mv^(2)+(1)/(2)m[u^(2) sin^(2) theta+(u cos theta - v)^(2)]`
`= ((M+m)/(2))((m u cos theta)/(M+m))+(m u^(2))/(2)-m u cos theta xx (m u cos theta)/(M+m)`
`= (m^(2)u^(2) cos^(2) theta)/(2(M+m)) - (m^(2)u^(2) cos theta)/(M+m)+(m u^(2))/(2)`
`= -(m u^(2) cos^(2) theta)/(2(M+m))+(m u^(2))/(2) = (m u^(2))/(2)(1-(m cos^(2) theta)/(M+m))`
`Delta K = (m(M+m sin^(2) theta)u^(2))/(2(M+m))`