Question

A plank of mass $M$ and length $L$ is at rest on a frictionless floor. The top surface of the plank has friction. At one end of it a man of mass $m$ is standing as shown in Figure. If the man walks towards the other end, find the distance, which the plank moves (a) till the man reaches the centre of the plank, (b) till the man reaches the other end of the plank.

• A. $\frac{2mL}{(m+M)}$ and $\frac{mL}{2(m+M)}$
• B. $\frac{mL}{2(m+M)}$ and $\frac{2mL}{(m+M)}$
• C. $\frac{2mL}{(m+M)}$ and $\frac{mL}{2(m+M)}$
• D. $\frac{mL}{2(m+M)}$ and $\frac{mL}{(m+M)}$

Answer 1

The correct option is D $\frac{mL}{2(m+M)}$ and $\frac{mL}{(m+M)}$

Since no external force is acting on the system, therefore net momentum$\left(P\right)$ of the system will remain conserved and COM of the system will remain stationary. Hence from frame of reference of COM,
$P_{man}=P_{plank}$
For certain time, the above equation can be written as
$m{x}_m=M{x}_M$
Let man moves forward by a distance $x$ w.r.t plank and as aresult plank moves backward by a distance $L_{plank}$.
$m(x-L_{plank})=M{L}_{plank}$
Rearranging this we get,
$L_{plank}=\frac{mx}{m+M}$
Case 1
When man reaches centre of plank, $x=\frac{L}{2}$
$L_{plank}=\frac{mL}{2(m+M)}$
Case 2
When man reaches other end of plank, $x=L$
$L_{plank}=\frac{mL}{(m+M)}$