Center of Mass , Momentum and Collision

Practice Questions:

Question 21: easy

Two particles of mass $5\text{ kg}$ and $10\text{ kg}$ respectively are attached to the two ends of a rigid rod of length $1\text{ m}$ with negligible mass. The centre of mass of the system from the $5\text{ kg}$ particle is nearly at a distance of :

(2020)

1. $50\text{ cm}$

2. $67\text{ cm}$

3. $80\text{ cm}$

4. $33\text{ cm}$

View Answer

Use the centre of mass formula $x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$. Substituting $m_1 = 5\text{ kg}$, $x_1 = 0$, $m_2 = 10\text{ kg}$, $x_2 = 100\text{ cm}$, we get $x_{cm} = \frac{10 \times 100}{15} = 66.67\text{ cm} \approx 67\text{ cm}$. Option (b) is correct.

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Question 22: easy

Which of the following statements are correct?

A. Centre of mass of a body always coincides with the centre of gravity of the body

B. Centre of gravity of a body is the point at which the total gravitational torque on the body is zero

C. A couple on a body produce both translational and rotational motion in a body

D. Mechanical advantage greater than one means that small effort can be used to lift a large load

(2017-Delhi)

1. A and B

2. B and C

3. C and D

4. B and D

View Answer

Centre of gravity is the point where total gravitational torque is zero (Statement B is correct). Mechanical advantage greater than one implies a small effort lifts a large load (Statement D is correct). Thus, statements B and D are correct, making option (d) the right choice.

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Question 23: moderate

Two spherical bodies of mass $M$ and $5M$ and radii $R$ and $2R$ are released in free space with initial separation between their centres equal to $12R$. If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is :

(2015)

1. $4.5R$

2. $7.5R$

3. $1.5R$

4. $2.5R$

View Answer

The centre of mass remains stationary. Initial distance of COM from mass $M$ is $\frac{5M \times 12R}{M + 5M} = 10R$. At collision, distance between centers is $3R$, and the distance of $M$ from COM is $\frac{5}{6} \times 3R = 2.5R$. Thus, distance covered by $M$ is $10R - 2.5R = 7.5R$. Option (b) is correct.

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Question 24: moderate

Three masses are placed on the $x$-axis : $300\text{ g}$ at origin, $500\text{ g}$ at $x = 40\text{ cm}$ and $400\text{ g}$ at $x = 70\text{ cm}$. The distance of the center of mass from the origin is :

(2012 Mains)

1. $40\text{ cm}$

2. $45\text{ cm}$

3. $50\text{ cm}$

4. $30\text{ cm}$

View Answer

Use the centre of mass formula $x_{cm} = \frac{\sum m_i x_i}{\sum m_i}$. Substituting the given values: $x_{cm} = \frac{300(0) + 500(40) + 400(70)}{300 + 500 + 400} = \frac{48000}{1200} = 40\text{ cm}$. Option (a) is correct.

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Question 25: 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 26: 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 27: 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 28: easy

Two bodies of mass $1text{ kg}$ and $3text{ kg}$ have position vectors $\hat{i} + 2\hat{j} + \hat{k}$ and $-3\hat{i} – 2\hat{j} + \hat{k}$, respectively. The center of mass of this system has a position vector:

(2009)

1. $2\hat{i} - \hat{j} + \hat{k}$

2. $-2\hat{i} - \hat{j} + \hat{k}$

3. $-\hat{i} + \hat{j} + \hat{k}$

4. $-2\hat{i} + 2\hat{k}$

View Answer

Concept: Center of mass position vector formula. Formula: $\vec{r}_{cm} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2}$. Solution: Substituting the given masses and position vectors yields $-2\hat{i} - \hat{j} + \hat{k}$.

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Question 29: 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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Question 30: easy

The centre of mass of system of particles does not depend on:

(1997)

1. Position of the particles

2. Relative distances between the particles

3. Masses of the particles

4. Forces acting on the particle

View Answer

Concept: Definition and properties of center of mass. Formula: $\vec{R}_{cm} = \frac{\sum m_i \vec{r}_i}{\sum m_i}$.

Solution: Center of mass depends only on masses and positions, independent of internal or external forces acting on particles.

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