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Displacement of COM:Application

Four particle...

Question

Four particles A,B,C and D with masses $m_{A} = m, m_{B} = 2 m, m_{C} = 3 m$ and $m_{D} = 4 m$ are at the corners of a square. They have accelerations of equal magnitude with directions as shown. The acceleration of the centre of mass of the particles in ${ms}^{- 2}$ is

\frac{a}{5} (^i -^j)

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a (^i +^j)

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Zero

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\frac{a}{5} (^i +^j)

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Solution

The correct option is A $\frac{a}{5} (^i -^j)$
For a system of discrete masses, acceleration of centre of mass (CM) is given by
$a_{C M} = \frac{m_{A} a_{A} + m_{B} a_{B} + m_{C} a_{C} + m_{D} a_{D}}{m_{A} + m_{B} + m_{C} + m_{D}}$

where, $m_{A} = m, m_{B} = 2 m, m_{C} = 3 m$ and $m_{D} = 4 m$ ,
$| a_{A} | = | a_{B} | = | a_{C} | = | a_{D} | = a$ (according to the question)
$_{C M} = \frac{- m a^i + 2 m a^j + 3 m a^i - 4 m a^j}{m + 2 m + 3 m + 4 m}$
$= \frac{2 a^i - 2 a^j}{10} = \frac{a}{5} . (^i -^j) {ms}^{- 2}$

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