Question

Consider a system of two identical particles. One of the particles is at rest and the other has an acceleration a. The centre of mass has an acceleration
A) Zero
B) $${\displaystyle \frac{a}{2}}$$
C) a
D) 2a

Answer 1

Hint: Before proceeding with the solution of the given problem, let us discuss the centre of mass in brief. The centre of mass is a position defined relative to an object or system of objects. It is the average position of all the parts of the system, weighted according to their masses. For simple rigid objects with uniform density, the centre of mass is located at the centroid.

Formula Used:
$$a_{CM}={\displaystyle \frac{m_1{a}_1+m_2{a}_2+.......+m_n{a}_n}{m_1+m_2+.......+m_n}}$$

Complete step by step solution:
Let the acceleration of the centre of mass of the two particles be $$a_{CM}$$.
From the concept of the acceleration of the mass, we can say that the acceleration of the centre of mass is equal to the ratio of the sum of the product of individual masses and their accelerations to the sum of the masses.
Mathematically, we can express this as $$a_{CM}={\displaystyle \frac{m_1{a}_1+m_2{a}_2}{m_1+m_2}}$$
Now, we have been told that the particles are identical; this means that the mass of both the particles is same, that is $$m_1=m_2=m$$
Substituting this in the expression for the acceleration of the centre of mass, we get
$$a_{CM}={\displaystyle \frac{m(a_1+a_2)}{2m}}$$
Further simplifying this equation, we get
$$a_{CM}={\displaystyle \frac{(a_1+a_2)}{2}}$$
We have been told that one of the particles is at rest. This means that the acceleration of this particle will be zero, that is $$a_1=0$$. The acceleration of the other particle has been given as a.
Substituting the values of the acceleration, we get
$$a_{CM}={\displaystyle \frac{(0+a)}{2}}={\displaystyle \frac{a}{2}}$$

Hence the acceleration of the centre of mass of the system of particles will be $${\displaystyle \frac{a}{2}}$$.

Note: We have to keep an eye out for terms that might help us. For example, the word identical tells us that the masses of the particles are the same. For this, a keen reading of the question is needed. The centre of mass is a useful reference point for calculations in mechanics that involve masses distributed in space, such as the linear and angular momentum of planetary bodies and rigid body dynamics. In orbital mechanics, the equations of motion of planets are formulated as point masses located at the centres of mass.