← Back to Center of Mass , Momentum and Collision

Calculation of Center of Mass

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

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 3: moderate

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 no external force acts on the two-particle system, the acceleration of the centre of mass is zero. Since the system started from rest, the speed of the centre of mass remains zero.

💬 Discuss this Question

Question 4: 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.

💬 Discuss this Question

Question 5: 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.

💬 Discuss this Question