Numerical Problems Based on Conservation of Linear Momentum for Class 11 Physics

Mass of the bullet, $m_1 = 30\text{ g} = 0.03\text{ kg}$

Velocity of the bullet, $v_1 = 300\text{ ms}^{-1}$

Recoil velocity of the rifle, $v_2 = -0.60\text{ ms}^{-1}$ (negative due to opposite direction)

Let the mass of the rifle be $m_2$. Since the system is initially at rest, the initial momentum is zero.

According to the law of conservation of momentum:

Initial mass of the shell, $M = 40\text{ kg}$

Initial velocity, $V = 72\text{ kmh}^{-1} = 72 \times \frac{5}{18} = 20\text{ ms}^{-1}$

Mass of first piece, $m_1 = 15\text{ kg}$; Velocity of first piece, $v_1 = 0$ (it stops)

Mass of second piece, $m_2 = M – m_1 = 40 – 15 = 25\text{ kg}$

Let the velocity of the second piece be $v_2$. By conservation of momentum:

Note: The calculated recoil velocity is $2.4\text{ ms}^{-1}$, correcting a likely typo in the original textbook’s answer key ($2.4\text{ cms}^{-1}$).

Mass of the gun, $M = 4\text{ kg}$

Mass of the bullet, $m = 80\text{ g} = 0.08\text{ kg}$

Velocity of the bullet, $v = 120\text{ ms}^{-1}$

1. Recoil Velocity ($V$):
Using conservation of momentum (Initial Momentum = Final Momentum):

The recoil velocity of the gun is $2.4\text{ ms}^{-1}$.

2. Combined Momentum:
Since both were initially at rest, the initial momentum is zero. By the law of conservation of momentum, the combined momentum of the isolated system remains constant, so the final momentum is also zero.

Mass of the car, $m_1 = 1000\text{ kg}$; Initial speed of the car, $u_1 = 30\text{ ms}^{-1}$

Mass of the lorry, $m_2 = 9000\text{ kg}$; Initial speed of the lorry, $u_2 = 0$

This is a perfectly inelastic collision where both move with a common velocity, $v$.

By conservation of momentum:

Mass of the bullet, $m = 7\text{ g} = 0.007\text{ kg}$

Mass of the block, $M = 7\text{ kg}$

Common velocity after impact, $v = 70\text{ cms}^{-1} = 0.7\text{ ms}^{-1}$

Initial velocity of the block, $U = 0$. Let the initial velocity of the bullet be $u$.

By conservation of momentum:

Mass of first truck, $m_1 = 2 \times 10^4\text{ kg}$; Velocity, $u_1 = 0.5\text{ ms}^{-1}$

Mass of second truck, $m_2 = \frac{1}{2}m_1 = 1 \times 10^4\text{ kg}$

Velocity of second truck, $u_2 = -0.4\text{ ms}^{-1}$ (negative sign denotes opposite direction)

Let their common velocity after coupling be $v$.

By conservation of momentum: