Question

Two spheres of masses 2M and M are initially at rest at a distance R apart. Due to the mutual force of attraction, they approach each other. When they are at a separation $\frac{R}{2}$ ,the acceleration of the centre of mass of spheres would be

• A. 0 $m{s}^{-2}$
• B. $1m{s}^{-2}$
• C. $3m{s}^{-2}$
• D. $12m{s}^{-2}$

Answer 1

The correct option is A 0 $m{s}^{-2}$

Given,

Mass of sphere $1=M_1=2M$

Mass of sphere $2=M_2=M$

Distance between both sphere initially $=R$

To find, acceleration of centre of mass (CM) at distance $R/2$,

Here,

$\Rightarrow$ Intial velocity of both sphere is $Dm{s}^{-1}$

Here, Momentum is

$\therefore M_1{u}_1+M_1{u}_2=(M_1+M_2)V$

$\therefore M_1\left(0\right)+M_2\left(0\right)=(M_1+M_2)V$ .... ($V\Rightarrow$ velocity of CM i.e. whole system)

$\therefore V=0$

$\Rightarrow$ There is mutual force of attraction.

$\therefore$ Force $f_1=\frac{G{M}_1{M}_2}{r^2}$ $f_2=\frac{G{M}_1{M}_2}{r^2}$

As both forces are same, & no external force,

$\therefore f_{net}=0$

Now for acceleration of CM

$\Rightarrow f=Ma$

$\therefore Ma=0$

$\therefore a=0m/s^2$

Hence, acceleration is $0m/s^2$

Hence, option (A) is correct.