Question

A man of mass $M$ stands at one end of a plank of length $L$ which is at rest on a frictionless horizontal surface. The man walks to the other end of the plank. If mass of the plank is $\frac{M}{3}$, the distance that the man moves relative to the ground is

• A. $L$
• B. $\frac{L}{4}$
• C. $\frac{3L}{4}$
• D. $\frac{L}{3}$

Answer 1

The correct option is B $\frac{L}{4}$

Mass of man = $M$
Mass of plank = $M/3$
Let the man be standing at $x=0$
Then, end of the plank (end B) is at $x=L$
Therefore, COM of man is at $x=0$ and
COM of plank is at $x=L/2$

Position of COM of system $=\frac{M\times 0+M/3\times L/2}{M+M/3}=\frac{L}{8}$

Let after walking from end A to end B, the man travels a distance $z$ w.r.t ground.
So, COM of man is at $x=z$
Now, COM of plank will shift $(L-z)$ towards left,
$\therefore$ COM of plank is at $=[L/2-(L-z\left)\right]=(z-L/2)$

Since there is no external force, COM of system won't change its position.
$\therefore$ $\frac{L}{8}=\frac{M\times z+M/3\times (z-L/2)}{M+M/3}$
$\Rightarrow \frac{L}{8}=\frac{M\times z+Mz/3\times -ML/6}{M+M/3}$
$\Rightarrow \frac{L}{8}=\frac{4Mz/3-ML/6}{4M/3}$
$\Rightarrow \frac{L}{8}=z-\frac{L}{8}$
$\Rightarrow z=\frac{L}{4}$
So, distance that man moves w.r.t ground is $\frac{L}{4}$