A two particle system having masses, M1 and M2 are placed in a gravity-free space. If the first particle is shifted towards the centre of mass of the system by a distance d , by how much distance should the second particle be shifted so that the centre of mass remains at its initial position?

Question

A two particle system having masses, M1 and M2 are placed in a gravity-free space. If the first particle is shifted towards the centre of mass of the system by a distance d , by how much distance should the second particle be shifted so that the centre of mass remains at its initial position? (A) (M1/M2)d (B) (M2/M1)d (C) (M1/M1 + M2)d (D) d

Tardigrade Admin · Accepted Answer

Let us suppose that first and second particle are pushed by the distances

Δ x_{1} &Δ x_{2}

respectively.
Then, the change in the position of the centre of mass,

Δ x_{c m} = \frac{M _{1} Δ x _{1} + M _{2} Δ x _{2}}{M _{1} + M _{2}}

As given,

Δ x_{c m} = 0,

we have

\frac{M _{1} Δ x _{1} + M _{2} Δ x _{2}}{M _{1} + M _{2}} = 0

or

M_{1} Δ x_{1} + M_{2} Δ x_{2} = 0

Here,

Δ x_{1} = d

∴ M_{1} \times d + M_{2} Δ x_{2} = 0

Δ x_{2} = - \frac{M _{1}}{M _{2}} d

The negative sign shows that mass,

M_{2}

will move in opposite direction.

$\frac{M _{1}}{M _{2}} d$

$\frac{M _{2}}{M _{1}} d$

$\frac{M _{1}}{M _{1} + M _{2}} d$

$d$

M2​M1​​d

M1​M2​​d

M1​+M2​M1​​d

d

$\frac{M _{1}}{M _{2}} d$

$\frac{M _{2}}{M _{1}} d$

$\frac{M _{1}}{M _{1} + M _{2}} d$

$d$