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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 accelerationA. zeroB. $\dfrac{1}{2} a$C. aD. 2a

Last updated date: 23rd Apr 2024
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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:
$\vec{a}= \dfrac{m_1 .\vec{a_1} +m_2. \vec{a_2}}{m_1+m_2}$

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).