Consider a two particle system with particles having masses m₁ and m_{2 .} If the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle be moved, so as to keep the centre of mass at the same position ?

Option 1)
$d$
Option 2)
$\frac{m_{2}}{m_{1}}d$

Option 3)
$\frac{m_{1}}{m_{1}+m_{2}}d$
Option 4)
$\frac{m_{1}}{m_{2}}d$

Answers (1)

A Aadil Khan

As we learnt in

Centre of Mass of a system of N discrete particles -

$x_{cm}=\frac{m_{1}x_{1}+m_{2}x_{2}.........}{m_{1}+m_{2}.......}$

$y_{cm}=\frac{m_{1}y_{1}+m_{2}y_{2}+m_{3}y_{3}.........}{m_{1}+m_{2}+m_{3}.......}$

$z_{cm}=\frac{m_{1}z_{1}+m_{2}z_{2}+m_{3}z_{3}.........}{m_{1}+m_{2}+m_{3}.......}$

- wherein

m₁, m₂ ........... are mass of each particle x₁, x₂ ..........y₁, y₂ ............ z₁, z₂ are respectively x, y, & z coordinates of particles.

$\Delta x_{cm}=\frac{m_{1}.\Delta x_{1}+m_{2}.\Delta x_{2}}{m_{1}+m_{2}}$

Let $m_{2}$ be moved by $x$ so as to keep the centre of mass at the same position

$\Delta x_{1}=-d$ and $\Delta x_{CM}=0$

$\therefore \; \; \; m_{1}(-d)+m_{2}(x_{2})=0$ ,

$or\; \; \; m_{1}d=m_{2}x_{2}; \; \; or\; \; \; x_{2}=\frac{m_{1}}{m_{2}}d$

Option 1)

$d$

Option 2)

$\frac{m_{2}}{m_{1}}d$

Option 3)

$\frac{m_{1}}{m_{1}+m_{2}}d$

Option 4)

$\frac{m_{1}}{m_{2}}d$

Option 1) $d$	Option 2) $\frac{m_{2}}{m_{1}}d$
Option 3) $\frac{m_{1}}{m_{1}+m_{2}}d$	Option 4) $\frac{m_{1}}{m_{2}}d$

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