A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system?
The speed of the centre of mass of the (trolley + child) system remains V (unchanged).
Since the floor is smooth, there is no external horizontal force acting on the (trolley + child) system. The only external forces acting are gravity (downward) and normal reaction (upward), both of which are vertical and cancel each other. These forces do not affect the horizontal motion of the centre of mass.
When the child runs about on the trolley, the forces exerted between the child and the trolley are internal forces. According to the law of conservation of linear momentum, internal forces cannot change the total momentum of the system. Since the total mass of the system remains constant and no external horizontal force acts, the velocity of the centre of mass must remain constant. Hence, the centre of mass continues to move with the same speed V.
Explanation
This question tests the understanding of centre of mass motion and conservation of linear momentum. The key insight is that on a smooth (frictionless) horizontal surface, there is no external horizontal force on the system. The child's movement creates only internal forces between the child and trolley, which cannot change the system's total momentum. Therefore, the centre of mass velocity remains constant at V.
Solution Steps
Step 1: Identify that the floor is smooth, meaning no friction - hence no external horizontal force acts on the system.
Step 2: Recognize that the forces between child and trolley are internal forces.
Step 3: Apply the principle that internal forces cannot change the total momentum of a system.
Step 4: Conclude that since total momentum and total mass are conserved, the velocity of centre of mass remains V.