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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. $\dfrac{1}{2} a$
C. a
D. 2a

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Last updated date: 23rd Apr 2024
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Answer
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Hint: The centre of mass of a system of two particles (or many particles), has coordinates, velocity or acceleration that is the relative average of the two particles (or many particles involved).

Formula used:
The acceleration of the centre of mass is given by:
$\vec{a}= \dfrac{m_1 .\vec{a_1} +m_2. \vec{a_2}}{m_1+m_2}$

Complete step by step answer:
For a system of two particles with masses $m_1$ and $m_2$, and position vector $\vec{r_1}$ and $\vec{r_2}$ respectively, the position vector of the centre of mass will be given by:
$\vec{R}= \dfrac{m_1 \vec{r_1} +m_2 \vec{r_2}}{m_1+m_2}$

Based on this formula, the formula for acceleration is just:
$\vec{a}= \dfrac{m_1 \vec{a_1} +m_2 \vec{a_2}}{m_1+m_2}$
As, acceleration is just a second order differential of position with respect to time.

We are given that the two particles are identical, which means both have equal masses. One is accelerating and other has zero acceleration so, we may write:
$\vec{a_{CM}}= \dfrac{m\vec{a} +m \vec{0}}{m+m}$
which gives us just:
$\vec{a_{CM}}= \dfrac{\vec{a} }{2}$

Therefore, the centre of mass of the two particles travels with an acceleration of a/2 along the direction of the accelerating particle.

The correct answer is option (B).

Additional Information:
The center of mass coordinates are very useful in many body problems. In case of gravitation problems, it can reduce two body motion problems to just one body motion i.e., motion of center of mass. The relative motion is only what matters to us after all.

Note:
The answer comes without any formula or calculation if we consider the fact that the average acceleration for the two particles, (since they have the same mass) in our case is just (0+a)/2. Two in the denominator is used because we are averaging for 2 particles.