Question

Two particles A and B, initially at rest, move towards each other under a mutual force of attraction. At the instant when the speed of A is v and the speed B is 2v, the speed of the centre of mass is:
A. zero
B. v
C. 1.5v
D. 3v

Answer 1

Hint: The acceleration of the centre of mass of a system of particles is equal to ${{a}_{com}}=\dfrac{{{F}_{net}}}{M}$. ${{F}_{net}}$ is the net force acting on the system and M is the total mass of the system. Calculate the net force acting on the given system. Remember that particles under mutual gravity attract each other with forces of equal magnitudes.

Complete step by step answer:
Centre of mass of a body is that point on the body, where all the mass of the body appears to be concentrated. When we apply a force on any point on the body, the translation motion of the body is the same when we apply the same force on the centre of mass of the body. The meaning of this is that we can consider the body as a point sized mass positioned at its centre of mass. Dealing with a point sized body is easy.
Similarly, we have a centre of mass for a system of particles. The position centre of mass of a system of n particles is given as $\overrightarrow{{{r}_{com}}}=\dfrac{{{m}_{1}}\overrightarrow{{{r}_{1}}}+{{m}_{2}}\overrightarrow{{{r}_{2}}}+{{m}_{3}}\overrightarrow{{{r}_{3}}}+......+{{m}_{n}}\overrightarrow{{{r}_{n}}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}+.......+{{m}_{n}}}$.
When we differentiate the position with respect to time we get the velocity of the centre of mass.
i.e. $\overrightarrow{{{v}_{com}}}=\dfrac{{{m}_{1}}\overrightarrow{{{v}_{1}}}+{{m}_{2}}\overrightarrow{{{v}_{2}}}+{{m}_{3}}\overrightarrow{{{v}_{3}}}+......+{{m}_{n}}\overrightarrow{{{v}_{n}}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}+.......+{{m}_{n}}}$ .
And when differentiate velocity with time we get,
$\overrightarrow{{{a}_{com}}}=\dfrac{{{m}_{1}}\overrightarrow{{{a}_{1}}}+{{m}_{2}}\overrightarrow{{{a}_{2}}}+{{m}_{3}}\overrightarrow{{{a}_{3}}}+......+{{m}_{n}}\overrightarrow{{{a}_{n}}}}{{{m}_{1}}+{{m}_{2}}+{{m}_{3}}+.......+{{m}_{n}}}$
When more than one force is applied on a point sized particle, we can easily calculate the net force on the body. However, when we have a system of particles, we consider the forces acting on the centre of mass.
Therefore, acceleration of the centre of mass of the system is equal to ${{a}_{com}}=\dfrac{{{F}_{net}}}{M}$,
where M is the total of mass of the system.
The given system consists of two particles. They are attracted towards each other by the force of gravity. There are two forces acting on the system, both equal in magnitude and opposite in direction. Therefore, the net force acting on the system is zero.
As discussed earlier, the net force on the system affects the motion of the centre of mass. However, in this case the net force is zero. Therefore, the acceleration of the centre of mass is zero.
If the acceleration is zero then the centre of mass of the two particles will have constant velocity i.e. the velocity before being affected by the forces.
It is given that the two particles were at rest. Therefore, their centre of mass will also be at rest. And since the centre mass is not accelerated, it will continue to stay at rest.
Hence, the correct option is A.
Note:
Here the centre of mass of the two particles is at rest or we can also say the centre mass has zero acceleration. This does not mean that the particles of the system are at rest or zero acceleration. The two particles will accelerate due to the gravitational force acting on them, which is equal to $F=\dfrac{G{{M}_{1}}{{M}_{2}}}{{{r}^{2}}}$. However, the particles accelerate in such a way that their centre of mass is at rest.