A child is standing at one end of a long trol | vedclass

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(D) The centre of mass of a system moves according to the net external force acting on it,given by the equation $F_{ext} = M_{total} \cdot a_{cm}$.Sin

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$v$

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(D) The centre of mass of a system moves according to the net external force acting on it,given by the equation $F_{ext} = M_{total} \cdot a_{cm}$.Since the floor is smooth (frictionless) and there are no other external horizontal forces acting on the system (trolley + child),the net external force $F_{ext} = 0$.Consequently,the acceleration of the centre of mass $a_{cm} = 0$,which means the velocity of the centre of mass remains constant.Initially,the trolley and the child are both moving with speed $v$,so the velocity of the centre of mass is $v$.Since the child running on the trolley involves only internal forces,the velocity of the centre of mass remains $v$ throughout the motion.

$A$ child is standing at one end of a long trolley moving with a speed $v$ on a smooth horizontal floor. If the child starts running towards the other end of the trolley with a speed $u$,the centre of mass of the system (trolley + child) will move with a speed:

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