Question

A man of mass $M$ stands at one end of a plank of length $L$ which lies at rest on a frictionless horizontal surface. The man walks to the other end of the plank. If the mass of the plank is ($M/3$), the distance that the man moves relative to 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 man is standing at the starting of plank( end A), at X=0 end of plank(end B), at X=L

Cm of man is at x=0

Cm of man is at x=L/2

Position of Cm of system $=\frac{M\times 0+\left(M/3\right)\times \left(L/2\right)}{M+\left(M/3\right)}=\frac{ML/6}{4M/3}=\frac{L}{8}$

Let after walking to end A from end B, he travels a distance z.

Cm of man is at x=z

Cm of plank will shift (L-z) towards left,

Cm of plank is at$=[\frac{L}{2}-(L-z\left)\right]=(z-\frac{L}{2})$

Since there is no external force, Cm of system won't change its position.

$\frac{L}{8}=\frac{(M\times z)+\left(M/3\right)\times (z-(L/2\left)\right)}{M+M/3}$

$\frac{L}{8}=\frac{Mz+Mz/3-ML/6}{4M/3}$

$\frac{L}{8}=\frac{4Mz/3-ML/6}{4M/3}$

$\frac{L}{8}=\frac{4Mz/3}{4M/3}-\frac{ML/6}{4M/3}$

$\frac{L}{8}=z-\frac{L}{8}$

$z=\frac{L}{8}+\frac{L}{8}=\frac{L}{4}$