1. **Problem statement:** A gun of mass $M$ fires a shell of mass $m$ which travels along the barrel with speed $V$. We need to find the recoil speed of the barrel in two cases: (i) when the barrel is horizontal, (ii) when the barrel is inclined at an angle $30^\circ$ to the horizontal. 2. **Relevant principle:** We use the law of conservation of linear momentum. Initially, the system (gun + shell) is at rest, so total initial momentum is zero. 3. **Case (i): Barrel horizontal** - Let the recoil speed of the gun be $v$. - Momentum of shell after firing: $mV$ (to the right, say). - Momentum of gun after firing: $Mv$ (to the left, opposite direction). By conservation of momentum: $$ Mv = mV $$ Since they move in opposite directions, we write: $$ Mv = -mV $$ Taking magnitudes and solving for $v$: $$ v = \frac{mV}{M} $$ However, the shell moves inside the barrel, so the total mass moving forward is $M + m$ (gun + shell), so the recoil speed is: $$ v = \frac{mV}{M + m} $$ 4. **Case (ii): Barrel inclined at $30^\circ$** - The shell velocity $V$ is along the barrel, so its horizontal component is: $$ V_x = V \cos 30^\circ = V \times \frac{\sqrt{3}}{2} $$ - The recoil speed $v$ of the gun is horizontal (opposite direction). By conservation of horizontal momentum: $$ Mv = m V_x = m V \frac{\sqrt{3}}{2} $$ Solving for $v$: $$ v = \frac{m V \frac{\sqrt{3}}{2}}{M + m} = \frac{\sqrt{3} m V}{2 (M + m)} $$ **Final answers:** (i) $v = \frac{mV}{M + m}$ (ii) $v = \frac{\sqrt{3} m V}{2 (M + m)}$