Question

A railway flat car has an artillery gun installed on it. The combined system has a mass $M$ and moves with a velocity $V$. The barrel of the gun makes an angle $\alpha$ with the horizontal. A shell of mass $m$ leaves the barrel at a speed $v$ relative to the barrel. The speed of the flat car so that it may stop after the firing is :

• A. $\frac{m}{M+m}$
• B. $\left(\frac{Mv}{M+m}\right)cos\alpha$
• C. $\left(\frac{mv}{M+m}\right)cos\alpha$
• D. $(M+m)vcos\alpha$

Answer 1

The correct option is C $\left(\frac{mv}{M+m}\right)cos\alpha$

Initially, the car + gun system along with the shell moves in the x direction vith velocity $V$.

$\therefore$ Initial momentum $P_i=(M+m)V$

After firing, the car stops and the shell moves in the direction making an angle $\alpha$ with x axis with a velocity $v$.

$\therefore$Component of velocity of shell in x axis $v^{\prime }=vcos\alpha$

Final momentum $P_f=M\left(0\right)+mvcos\alpha =mvcos\alpha$

Using conservation of linear momentum in x direction : $P_i=P_f$

$\therefore$ $(M+m)V=mvcos\alpha$

$⟹V=\frac{mv}{M+m}$ $cos\alpha$