Question

A man weighing 80 kg is standing on a trolley weighing 320 kg. The trolley is resting on frictionless horizontal rails. If the man starts walking on the trolley along the rails at a speed 1 ms${}^{-1}$ then after 4 second his displacement relative to the ground will be :

• A. 3.0 m
• B. 3.2 m
• C. 4.8 m
• D. 5 m

Answer 1

Answer: (B)

3.2 m

The external force acting on the trolley $+$ man system is zero. Hence, the momentum of the trolley $+$ man system is conserved.
Initially, both the trolley and the man are at rest.
Let the initial com of man and trolley be at a and b respectively.

$x_{cm}=\frac{80a+320b}{80+320}=\frac{80a+320b}{400}$

After 4 secs, the com will be at $x_{cm}=\frac{80(a-x_1+4)+320(b-x_1)}{400}$ where $x_1$ is the distance moved by the trolley towards left

when the man moves towards right.

Since there is no external force on the trolley $+$man system, $x_{cm}$ will not change.

Hence, $\frac{80a+320b}{400}=\frac{80(a-x_1+4)+320(b-x_1)}{400}$

Solving, $x_1=\frac{4}{5}$
Hence, displacement of the man relative to ground is $4-x_1=4-0.8=3.2m$