Question

A gun is mounted on a railroad car. The mass of the car with all of its components is 80m and mass of each shell to be fired is 5m. If the muzzle velocity of the shells is 100 $ms^{-1}$ horizontally w.r.t gun, then calculate the recoil speed of the car after second shot (in $ms^{-1}$). Consider car to be at rest initially.

• A. 6.25 $ms^{-1}$
• B. 12.5 $ms^{-1}$
• C. 25 $ms^{-1}$
• D. 50 $ms^{-1}$

Answer 1

Answer: (B)

12.5 $ms^{-1}$

Answer 2

The problem is solved by applying the law of conservation of linear momentum sequentially.

1. First Shot: The car starts from rest (momentum = 0). After firing the first shell, the momentum of the car and the shell must sum to zero. Let $v_1$ be the car's velocity. The shell's velocity relative to the ground is $v_1 + 100$. Conservation of momentum: $(80m-5m)v_1 + 5m(v_1+100) = 0$. Solving for $v_1$ gives $v_1 = -25/4 \, m/s$.
2. Second Shot: The car is now moving with $v_1 = -25/4 \, m/s$. Its mass is $70m$. Applying conservation of momentum for the second shot: $70m \times v_1 = 70m \times v_2 + 5m(v_2+100)$, where $v_2$ is the final velocity of the car. Solving for $v_2$ gives $v_2 = -25/2 \, m/s$.
3. The recoil speed after the second shot is the magnitude of $v_2$, which is $25/2 \, m/s$ or $12.5 \, m/s$.