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Calculation of Center of Mass

Question 1:
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)

(–2, –2, –2)

(2, 2, 2)

(–3, –3, –3)

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Solution:

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)

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Question 2:
Center of mass is closer to

Heavier Mass

Closer Mass

Mid Point

Any one of the above

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Solution:

Center of Mass is the weighted mean of masses so it is closer to heavier mass.

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Question 3:
Centre of mass is the point which divides the line joining the masses in

ratio of masses

reverse ratio of masses

1:1

1:2

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Solution:

Centre of mass is the point which divides the line joining the masses in reverse ratio of masses.

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Question 4:
Masses 2 kg and 5kg are located at origin and (7,0) respectively. Centre of mass of the arrangement is located at

(0,5)

(5,0)

(0,0)

(7,0)

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Solution:

Centre of mass is the point which divides the line joining the masses in reverse ratio of masses.

Here ratio of mass is 2:5 so distance from 2kg mass located at origin will be 5/7 of 7 cm so

\[ x_{cm}= \frac{2(0)+5(7)}{5+2}= 5 \]

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Question 5:
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

0.48 × 10^–10 m

0.51 × 10^–10 m

0.69 × 10^–10 m

0.56 × 10^–10 m

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Solution:

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

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

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