Question

Consider a system of two identical particles. One of the particles is at rest, and others have acceleration$\vec a$. The center of mass has an acceleration of
A. Zero
B. $\dfrac{1}{2}\vec a$
C. $\vec a$
D. $2\vec a$

Answer 1

If the net external force on a system of the particle is zero, the center of mass of the system will not accelerate. The net force on a system of particles equals the mass of the system times the acceleration of the system’s center of mass, and the system behaves as if all of its mass were located at the system’s center of mass. Where the center of mass for n number of particles is given as
${x_{cm}} = \dfrac{{{x_1}{m_1} + {x_2}{m_2} + {x_3}{m_3} + ..... + {x_n}{m_n}}}{{{m_1} + {m_2} + {m_3} + ..... + {m_n}}}$
Centre of the mass of an object where mass is distributed is the unique point where the weighted relative position of the distributed mass system sums to zero.

Complete step by step answer:
Given that one of the particles is at rest ${a_1} = 0$and another particle is moving with the acceleration ${a_2} = a$
Where the acceleration of the center of mass is given as:
${a_{cm}} = \dfrac{{{a_1}{m_1} + {a_2}{m_2}}}{{{m_1} + {m_2}}}$
Also, two particles are identical hence their mass will also be equal
${m_1} = {m_2} = m$
Now we can write acceleration of the center of the mass equation as

$${a_{cm}} = \dfrac{{{a_1}{m_1} + {a_2}{m_2}}}{{{m_1} + {m_2}}} \\\ = \dfrac{{am + \left( {0 \times m} \right)}}{{m + m}} \\\ = \dfrac{{am}}{{2m}} \\\ = \dfrac{a}{2} \\\$$

Hence the center of mass has an acceleration of$= \dfrac{{\vec a}}{2}$, where acceleration is a vector quantity.

So, the correct answer is “Option B”.

Note:
The center of mass is the location where all the mass of the system could be considered to be located. The center of mass of a solid body does not necessarily lie within the body.