Consider the adjacent figure. The cube shaped 0.46 carrage ABCDEFGH of side l has a mass M and it can slide over two frictionless rails PQ and RS. A shot of mass m is thrown from corner A such that it lands at corner F. The angle of projection at seen from the carriage is `45^(@)`. While the shot is in the air. the velocity of carriage as seen from the ground is

Options

A. `(msqrt(glsqrt(2)))/(2(M+m))`
B. `(msqrt(2gl))/(2(M+m))`
C. `(msqrt(gl))/((M+m))`
D. `(msqrt(2gl))/((M+m))`

Explanation

Say v is the velocity of carriage w.r.t. ground and u is the velocity of shot w.r.t. carriage as shown in the figure.
`u^(2)sin(2xx45^(@))/(g)=lsqrt(2)`
`rArr u=sqrt(glsqrt(2))` … .(i)
`Mv=m[u(cos 45^(@))xxcos (45^(@))-v]`

`rArr (M+m)v=(mu)/(2)`
`rArr `T_(2)-(T_(1))/(T_(1))=(Deltah)/(h)`