Solution:

Center of Mass:
The centre of mass of a system is defined as the point where the entire mass of the system can be assumed to be concentrated.

Centre of Mass Equation:
Let there be a system of two particles with masses m1 and m2.

Their positions are given by r1 and r2 respectively.

Then the position of the centre of mass is given by -

R = (m1r1 + m2r2) / (m1 + m2)

Acceleration of Centre of Mass:

The acceleration of the centre of mass of a system is given by -

a = F_net / M

where F_net is the net force acting on the system and M is the total mass of the system.

Now, let's solve the given problem.

Given:

Two identical particles, one at rest and the other with an acceleration a.

We need to find the acceleration of the centre of mass of the system.

Solution:

First, let's find the position of the centre of mass of the system.

Let the mass of each particle be m.

Position of the first particle (at rest) is r1 = 0.

Position of the second particle (with acceleration a) is r2 = at^2 / 2, where t is the time elapsed.

Total mass of the system is M = 2m.

Using the Centre of Mass Equation, we get -

R = (m * 0 + m * at^2 / 2) / (2m)

R = at^2 / 4

Now, let's find the net force acting on the system.

The first particle is at rest, so there is no force acting on it.

The second particle has a force F = ma acting on it.

Using Newton's third law, we know that the first particle also exerts an equal and opposite force on the second particle.

Therefore, the net force on the system is F_net = ma - ma = 0.

Using the acceleration equation, we get -

a = F_net / M

a = 0 / 2m

a = 0

Therefore, the acceleration of the centre of mass of the system is 0.

This means that the centre of mass of the system remains at rest.

Alternative Method:

We can also use the fact that the acceleration of the centre of mass of a system is equal to the weighted average of the accelerations of the individual particles.

Let a1 and a2 be the accelerations of the first and second particle respectively.

Since the first particle is at rest, its acceleration is 0.

The acceleration of the second particle is a2 = a.

Using the formula for weighted average, we get -

a = (m * 0 + m * a) / (2m)

a = a / 2

Therefore, the acceleration of the centre of mass of the system is a / 2, which is option (C).