First slide

Location of centre on mass for multiple point masses

Question

Two particles are separated by a distance d. When the positions of the two particles are interchanged, the shift in the position of their centre of mass is d/3. If the mass of lighter particle is m, the mass of heavier particle is

Moderate

A

$\frac{3 m}{2}$
B

$\frac{4 m}{3}$
C

$2 m$
D

$3 m$

Solution

$\begin{array}{l} let m_{1} be smaller mass, t hen \\ x_{1} = \frac{m_{2} d}{m_{1} + m_{2}} a n d x_{2} = \frac{m_{1} d}{m_{1} + m_{2}} \\ x_{1} - x_{2} = \frac{(m_{2} - m_{1}) d}{m_{1} + m_{2}} = \frac{d}{3} \\ \Rightarrow m_{2} = 2 m_{1} = 2 m \end{array}$

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