Question

Two particles of mass $1\text{kg}$ and $3\text{kg}$ move towards each other under their mutual force of attraction. No other forces act on them. When the velocity of approach of the two particles is $2\text{m/s}$, their centre of mass has a velocity of $0.5\text{m/s}$. When the velocity of approach becomes $3\text{m/s}$, the velocity of the centre of mass of the system of particles is

• A. $0.75\text{m/s}$
• B. $0.5\text{m/s}$
• C. $1.25\text{m/s}$
• D. $1.5\text{m/s}$

Answer 1

The correct option is B $0.5\text{m/s}$

Mutual forces of attraction are internal forces. The sum of internal forces for a system is always zero.


Hence, there is no net external force acting on the system.
$F_{\text{ext}}=0$
From Newton's second law,
$F_{\text{ext}}=m{a}_{\text{com}}$
$\Rightarrow a_{\text{com}}=0$
$\Rightarrow v_{\text{com}}=\text{constant}$
Therefore, velocity of the centre of mass will not change. Velocity of centre of mass will remain constant at $0.5\text{m/s}$ throughout the motion.