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Motion of Center of Mass

Practice Questions:

Question 1:

Two bodies of masses 2 kg and 4 kg, initially at rest, start moving towards each other due to mutual gravitational attraction. At a certain instant their speeds are 2 m/s and 1 m/s respectively. The speed of their centre of mass at that instant is

1. 5 m/s

2. 6 m/s

3. 8 m/s

4. zero (0)

View Answer

As Centre of mass was initially at rest and no external force act on it, Center of mass will remain at rest.

Note: Gravitational force here is internal force acting between two objects.

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Question 2: easy

A boat of mass 40 kg is at rest. A dog of mass 4 kg starts moving in the boat with velocity of 10 m/s. What is the speed of boat ?

1. 1 m/s

2. 2 m/s

3. 4 m/s

4. 8 m/s

View Answer

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

As Center of mass will remain at rest \[ (m_{1}v_{1}+ m_{2}v_{2}) = 0 \]

4 × 10 + 40 × v = 0

v = -1 m/s

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Question 3: easy

A system consists of two identical particles. One particle is at rest and the other particle has an acceleration a. The centre of mass of the system has an acceleration

1. 2a

2. a

3. a/2

4. a/4

View Answer

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

\[ a_{cm} = \frac{(m\times a+ m\times 0)}{(m+ m)}= \frac{a}{2} \]

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Question 4: 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.

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Question 5: easy

Two particles \(A\) and \(B\) initially at rest, move towards each other under mutual force of attraction. At an instance when the speed of \(A\) is \(v\) and speed of \(B\) is \(3v\), the speed of centre of mass is

1. \(v\)

2. \(4v\)

3. \(2v\)

4. Zero

View Answer

Since there is no external force acting on the system, the acceleration of the centre of mass is zero. Since the system was initially at rest, the velocity of the centre of mass remains zero.

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

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

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