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Question: A man of mass \(80Kg\) is riding on a small cart of mass \(40Kg\) which is rolling along a level flo...

A man of mass 80Kg80Kg is riding on a small cart of mass 40Kg40Kg which is rolling along a level floor at a speed of 2ms−12m{{s}^{-1}}. He is running on the cart so that his velocity relative to the cart is 3ms−13m{{s}^{-1}} in the direction opposite to motion of cart. What is the speed of the centre of mass of the system?
A. 1.5ms−11.5m{{s}^{-1}}
B. 1ms−11m{{s}^{-1}}
C. 3ms−13m{{s}^{-1}}
D. 0ms−10m{{s}^{-1}}

Explanation

Solution

The centre of mass of an object is the point where its entire mass seems to exist and hence governs all the typical movements of the body. Motion of the centre of mass of the body leads to its smooth balanced motion. Firstly, we will find the velocity of the man with respect to the ground. Then we shall use the general equations of motion of the centre of mass of the system.

Complete step-by-step solution:
The relative velocity of man on the cart is given as:
VMg=VMm−Vmg{{V}_{Mg}}={{V}_{Mm}}-{{V}_{mg}}
Where,
Vmg={{V}_{mg}}= velocity of cart with respect to ground
VMg={{V}_{Mg}}= velocity of man with respect to ground
VMm={{V}_{Mm}}= velocity of man with respect to cart
Here, we have
Vmg=2ms−1{{V}_{mg}}=2m{{s}^{-1}} and VMm=3ms−1{{V}_{Mm}}=3m{{s}^{-1}}

& \Rightarrow {{V}_{Mg}}=\left( 3-2 \right)m{{s}^{-1}} \\\ & \Rightarrow {{V}_{Mg}}=1m{{s}^{-1}} \\\ \end{aligned}$$ Velocity of centre of mass of a system of $'n'$ bodies is given by: ${{V}_{cm}}=\dfrac{{{m}_{1}}{{V}_{1}}+{{m}_{2}}{{V}_{2}}+--+{{m}_{n}}{{V}_{n}}}{{{m}_{1}}+{{m}_{2}}+--+{{m}_{n}}}$ The velocity of centre of mass of the man-cart system is given as: ${{V}_{cm}}=\dfrac{{{V}_{M}}{{m}_{M}}+{{V}_{m}}{{m}_{m}}}{{{m}_{M}}+{{m}_{m}}}$ Where, ${{V}_{cm}}=$ velocity of centre of mass of system ${{V}_{M}}=$ velocity of man $=1m{{s}^{-1}}$ ${{m}_{M}}=$ mass of man $=80Kg$ ${{V}_{m}}=$ velocity of cart $=2m{{s}^{-1}}$ ${{m}_{m}}=$ mass of cart $=40Kg$ Substituting all the values, we get $\Rightarrow {{V}_{cm}}=\dfrac{\left( 1 \right)\left( 80 \right)+\left( -2 \right)\left( 40 \right)}{80+40}$. The negative sign has been used to imply the fact that the velocity of the man is opposite to the velocity of the cart. $\Rightarrow {{V}_{cm}}=0m{{s}^{-1}}$ Therefore, the speed of centre of mass of the system is $0m{{s}^{-1}}$. **The correct option is (D) $0m{{s}^{-1}}$.** **Note:** The velocity of the centre of mass of the system does not change if the system is closed. A closed system means that no external force is acting on the system. The system moves like all the mass is concentrated at a single point. This point is the centre of mass.