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Motion of Center of Mass

Practice Questions:

Question 1: moderate

Two persons of masses $55\text{ kg}$ and $65\text{ kg}$ respectively, are at the opposite ends of a boat. The length of the boat is $3.0\text{ m}$ and weighs $100\text{ kg}$. The $55\text{ kg}$ man walks up to the $65\text{ kg}$ man and sits with him. If the boat is in still water the center of mass of the system shifts by:

(2012 Pre)

1. $3.0\text{ m}$

2. $2.3\text{ m}$

3. Zero

4. $0.75\text{ m}$

View Answer

Since no external horizontal force acts on the system (boat + persons), the position of the centre of mass of the system remains unchanged. Thus, the shift in the centre of mass is zero. Option (c) is correct.

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Question 2: moderate

A man of $50\text{ kg}$ mass is standing in a gravity free space at a height of $10\text{ m}$ above the floor. He throws a stone of $0.5\text{ kg}$ mass downwards with a speed $2\text{ m/s}$. When the stone reaches the floor, the distance of the man above the floor will be :

(2010 Pre)

1. $9.9\text{ m}$

2. $10.1\text{ m}$

3. $10\text{ m}$

4. $20\text{ m}$

View Answer

In gravity-free space, no external force acts, so the centre of mass remains at its initial height of $10\text{ m}$. Using COM conservation: $M_m h_m + M_s h_s = (M_m + M_s) Y_{cm} \implies 50(h) + 0.5(0) = (50 + 0.5)(10) \implies h = 10.1\text{ m}$. Option (b) is correct.

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Question 3: easy

Two particles which are initially at rest, move towards each other under the action of their internal attraction. If their speeds are $v$ and $2v$ at any instant, then the speed of centre of mass of the system will be:

(2010 Pre)

1. $v$

2. $2 v$

3. Zero

4. $1.5 v$

View Answer

Concept: Since external force on the system is zero, the acceleration of the center of mass is zero. Formula: $v_{cm} = \frac{\sum m_i v_i}{\sum m_i}$. Solution: Since the system starts from rest and only internal forces act, velocity of center of mass remains zero.

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Question 4: easy

Consider a system of two particles having masses $m_1$ and $m_2$. If the particle of mass $m_1$ is pushed towards the mass centre of particles through a distance ‘$d$’ by what distance would the particle of mass $m_2$ move so as to keep the mass centre of particles at the original position:

(2004)

1. $\frac{m_1}{m_2} d$

2. $d$

3. $\frac{m_1}{m_2}$

4. $\frac{m_1}{m_1 + m_2} d$

View Answer

Concept: Shift in center of mass must be zero. Formula: $m_1 \Delta x_1 = m_2 \Delta x_2$. Solution: Substituting $\Delta x_1 = d$ gives $\Delta x_2 = \frac{m_1}{m_2}d$.

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