Center of Mass , Momentum and Collision

Practice Questions:

Question 1: moderate

The centre of mass of three particles of masses 1 kg, 2 kg and 3 kg is at (3, 3, 3) with reference to a fixed coordinate system. Where should a fourth particle of mass 4 kg be placed so that the centre of mass of the system of all particles shifts to the point (1, 1, 1)

1. (–1, –1, –1)

2. (–2, –2, –2)

3. (2, 2, 2)

4. (–3, –3, –3)

View Answer

Center of mass of 1 kg, 2 kg and 3 kg is at (3, 3, 3) so, 6kg can be assumed to present at (3, 3, 3).

Assume 4 kg is placed at (x,y,z) Center of mass is at (1,1,1)

1= (4x+6×3)/10

x=-2

similarly y=-2 and z=-2

COM(-2,-2,-2)

💬 Discuss this Question

Question 2: moderate

Two persons of masses 55 kg and 65 kg are at the opposite ends of a boat. The length of the boat is 3.0 m and its mass is 100 kg. The 55 kg man walks upto the 65 kg man and sits with him. If the boat is in still water, the centre of mass of the system shifts by

1. 0.75 m

2. 2.3 m

3. 3.0 m

4. zero

View Answer

As the boat was initially at rest and no external force acts on it centre of mass will remain at rest.

💬 Discuss this Question

Question 3: moderate

In carbon monoxide molecules, the carbon and the oxygen atoms are separated by a distance of 1.2 × 10^–10 m. The distance of the centre of mass from the carbon atom is

1. 0.48 × 10^–10 m

2. 0.51 × 10^–10 m

3. 0.69 × 10^–10 m

4. 0.56 × 10^–10 m

View Answer

\[ x_{cm}= \frac{(m_{1}x_{1} + m_{2}x_{2})}{m_{1}+m_{2}} \]

\[ x_{cm}= \frac{12(0)+ 16 (1.2 × 10^{–10})}{12+16}= 0.69 \times 10^{–10} m \]

💬 Discuss this Question

Question 4: moderate

A stationary body of mass m explodes into 3 parts with mass ratio of \(1 : 3 : 3\). The two fragments with equal mass move at right angles to each other with velocity of \(15\text{ ms}^{-1}\). The velocity of the third fragment is (in \(\text{ms}^{-1}\)):

1. \(15\sqrt{2}\)

2. 5

3. \(20\sqrt{2}\)

4. \(45\sqrt{2}\)

View Answer

The ratio of masses is \(m' : 3m' : 3m'\). The combined momentum of the two perpendicular \(3m'\) masses is \(P = \sqrt{(3m' \times 15)^2 + (3m' \times 15)^2} = 45\sqrt{2} m'\). Conservation of momentum requires the third fragment \(m'\) to balance this: \(m' v_3 = 45\sqrt{2} m' ⇒ v_3 = 45\sqrt{2}\text{ ms}^{-1}\).

💬 Discuss this Question

Question 5: moderate

A body with kinetic energy K moving in +X direction splits up into two parts A and B of equal mass on its own. Part ‘A’ moves back in -X direction with a velocity equal in magnitude to the initial velocity of the body. The kinetic energy of part B will be:

1. K

2. 4K

3. \(\frac{K}{2}\)

4. \(\frac{9}{2}K\)

View Answer

Let the total mass be \(2m\), so \(K = mv_0^2\). Under momentum conservation, \(2mv_0 = m(-v_0) + mv_B ⇒ v_B = 3v_0\). The kinetic energy of part B is \(K_B = \frac{1}{2}m(3v_0)^2 = \frac{9}{2}mv_0^2 = \frac{9}{2}K\).

💬 Discuss this Question

Question 6: 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 \( x = \frac{m L}{m + M} = \frac{60 \times 12}{60 + 840} = 0.8\text{ m} = 80\text{ cm} \).

💬 Discuss this Question

Question 7: 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 \( a_{\text{cm}} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2} = \left(\frac{m_1 - m_2}{m_1 + m_2}\right) a = \left(\frac{m_1 - m_2}{m_1 + m_2}\right)^2 g \).

💬 Discuss this Question

Question 8: 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.

💬 Discuss this Question

Question 9: 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 \( \frac{K_i - K_f}{K_i} = 1 - \frac{\frac{1}{2}(m+M)v_f^2}{\frac{1}{2}mv^2} = \frac{M}{m+M} \).

💬 Discuss this Question