Two particles of masses 2 kg and 4 kg are approaching each other with acceleration `1 ms^(-2)` and `2 ms^(-2)`, respectively, on a smooth horizontal surface. Find the acceleration of center of mass of the system.

Two particles of masses 2 kg and 4 kg are approaching each other with

1.	Two particles of masses 2 kg and 4 kg are approaching each other with acceleration `1 ms^(-2)` and `2 ms^(-2)`, respectively, on a smooth horizontal surface. Find the acceleration of center of mass of the system.
Answer» The acceleration of center of mass of the system `a_(CM) = \|m_(1)a_(2) + m_(2)a_(2)\|/(m_(1) + m_(2))` `rArr a_(CM) = \|m_(1)a_(2) + m_(2)a_(2)\|/(m_(1) + m_(2))` `=(2xx 1-4xx2)/(2+4) = -1 ms^(2)` [-ve sign because direction of 4 kg is opposite to that of 2kg] Since `m_(2)a_(2) gt m_(1)a_(1)` so the direction of acceleration of center of mass will be directed towards `a_(1)`

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Two particles of masses 2 kg and 4 kg are approaching each other with acceleration `1 ms^(-2)` and `2 ms^(-2)`, respectively, on a smooth horizontal surface. Find the acceleration of center of mass of the system.

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