Question

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

• A. $\frac{3L}{4}$
• B. $\frac{L}{4}$
• C. $\frac{4L}{5}$
• D. $\frac{L}{3}$

Answer 1

The correct option is A $\frac{3L}{4}$

Initially,

centre of mass of system$=\frac{\frac{m}{3}\left(\frac{1}{2}\right)+m\left(L\right)}{\frac{m}{3}+m}$

$COM=\frac{7L}{8}$

Finally,

$COM=\frac{m\left(d\right)+\frac{m}{3}(L/2+d)}{m+\frac{m}{3}}=\frac{4d+\frac{L}{2}}{4}$

COM position won't change as where is no external force acting on it

$\therefore \frac{(4d+\frac{L}{2})}{4}=\frac{7L}{8}$

$\Rightarrow d=\frac{3L}{4}$.