Two particles of masses `2 kg` and `4 kg` are approaching towards each other with accelerations `1m//s^(2)` and `2m//s^(2)`, respectively, on a smooth horizontal surface. Then find the acceleration of centre of mass of the system and direction of acceleration of CM.
Text Solution
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The acceleration of CM of the system is given by `a_(CM)=(m_(1)a_(1)+m_(2)a_(2))/(m_(2)+m_(2))impliesa_(CM)=(2xx1+4(-2))/(2+4)=-1m//s^(2)` Negative sign indicates that aceleratioin of CM will be in the direction of acceleration of `4 kg` mass.
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