Q. A man of mass $M$ stands at one end of a plank of length which is at rest on a frictionless horizontal surface. The man walks to he other end of the plank. If mass of the plank is $M/3$, the distance that the man moves relative to ground is
A) $L$
B) $L/4$
C) $3L/4$
D) $L/3$
Explanation
(c) If speed of man relative to plank be $v$, then it can be shown easily that speed of man relative to ground
$v_{mg} = v\frac{M}{\left( M + \frac{M}{3}\ \right)} = \frac{3}{4}v$
$\therefore$ Distance covered by man relative to ground
$= L\frac{v_{mg}}{v} = \frac{L}{v}\frac{3}{4}v = \frac{3L}{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.