First slide

Motion of centre of mass

Question

At t = 0, the positions and velocities of two particles are as shown in the figure. They are kept on a smooth surface and being mutually attracted by gravitational force. Find the position of centre of mass at t = 2s.

Moderate

A

X = 5 m
B

X = 7 m
C

X = 3 m
D

X = 2 m

Solution

As no extemal force are acting on the system. The velocity of centre of mass will be constant.
initial position of CM is $X_{cm} = \frac{1 \times 0 + 1 \times 10}{1 + 1} = 5 m$
$v_{cm} = \frac{1 \times 5 - 1 \times 3}{1 + 1} = 1 m / s$
Displacement of centre of mass $∆ x_{cm} = 1 \times 2 = 2 m$
Hence position of centre of mass $∆ x_{cm} = 5 + 2 = 7 m$

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