Center of Mass , Momentum and Collision

Practice Questions:

Question 21: moderate

A boat of length \( 12\text{ m} \) and mass \( 840\text{ kg} \) is floating without motion in still water. A man of mass \( 60\text{ kg} \) standing at one end of it walks to the other end of it and stops. The magnitude of displacement of the boat relative to the ground is:

1. \( 50\text{ cm} \)

2. \( 80\text{ cm} \)

3. \( 120\text{ cm} \)

4. \( 150\text{ cm} \)

View Answer

Since no external horizontal force acts on the boat-man system, the center of mass does not move. The displacement of the boat is \( x = \frac{m L}{m + M} = \frac{60 \times 12}{60 + 840} = 0.8\text{ m} = 80\text{ cm} \).

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Question 22: moderate

Two bodies with masses \( m_1 \) and \( m_2 \) (\( m_1 > m_2 \)) are joined by a string passing over a fixed pulley. Assuming masses of the pulley and thread are negligible. Then the acceleration of the centre of mass of the system is:

1. \( \left(\frac{m_1 - m_2}{m_1 + m_2}\right) g \)

2. \( \left(\frac{m_1 - m_2}{m_1 + m_2}\right)^2 g \)

3. \( \frac{m_1 g}{(m_1 + m_2)} \)

4. \( \frac{m_2 g}{(m_1 + m_2)} \)

View Answer

The acceleration of each block is \( a = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) g \). The acceleration of the center of mass is \( a_{\text{cm}} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2} = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) a = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)^2 g \).

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Question 23: moderate

A ball of mass \( m \) approaches a wall of mass \( M \) (\( M \gg m \)) with speed \( 4\text{ m/s} \) along the normal to the wall. The speed of the wall is \( 1\text{ m/s} \) towards the ball. The speed of the ball after an elastic collision with the wall is:

1. \( 5\text{ m/s} \) away from the wall

2. \( 9\text{ m/s} \) away from the wall

3. \( 3\text{ m/s} \) away from the wall

4. \( 6\text{ m/s} \) away from the wall

View Answer

Using the coefficient of restitution \( e = 1 \), the relative velocity of separation equals the relative velocity of approach. The approach velocity is \( 4 - (-1) = 5\text{ m/s} \). After collision, the relative separation velocity is \( v' - 1 = 5 \implies v' = 6\text{ m/s} \) away from the wall.

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Question 24: easy

A bullet of mass \( m \) leaves the barrel of a gun of mass \( M \) with a velocity \( v \). The gun is known to recoil with a velocity \( V \). If \( k \) and \( K \) respectively denote the kinetic energies of the bullet and the gun respectively; then

1. \( K = \left(\frac{m}{M}\right)^2 k \)

2. \( K = \sqrt{\frac{m}{M}} k \)

3. \( K = \left(\frac{m}{M}\right) k \)

4. \( K = \left(\frac{M}{m}\right) k \)

View Answer

By conservation of momentum, the bullet and gun have equal momentum magnitude, \( p \). Since kinetic energy is \( K_{\text{E}} = \frac{p^2}{2\text{mass}} \), we have \( k = \frac{p^2}{2m} \) and \( K = \frac{p^2}{2M} \). Thus, \( K = \left(\frac{m}{M}\right) k \).

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Question 25: moderate

A block of mass \( m \) moving with a velocity \( v \) collides with another block of mass \( M \) at rest. The two blocks stick together due to the collision. The loss of K.E. expressed as a fraction of total initial kinetic energy is:

1. \( \frac{M}{m+M} \)

2. \( \frac{m}{m+M} \)

3. \( \frac{M^2}{m+M} \)

4. \( \frac{M-m}{m+M} \)

View Answer

By conservation of momentum, the final velocity after a completely inelastic collision is \( v_f = \frac{mv}{m+M} \). The fractional loss of kinetic energy is \( \frac{K_i - K_f}{K_i} = 1 - \frac{\frac{1}{2}(m+M)v_f^2}{\frac{1}{2}mv^2} = \frac{M}{m+M} \).

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