UPSC 2024 Maths Optional Paper 2 Q5d — Step-by-Step Solution

10 marks · Section B

Motion of rigid bodies in two dimensions · Mechanics & Fluid Dynamics · asked 4× in 13 yrs · Read the full method →

Question

A rough uniform board of mass $m$ and length $2a$ rests on a smooth horizontal plane and a man of mass $M$ walks on it from one end to the other. Find the distance covered by the board during this time.

Technique

Conservation of horizontal centre of mass (no external horizontal force on a smooth floor).

Solution

Step 1 — Set up coordinates.

Let the board’s centre initially be at $x=0$ (so the board occupies $[-a,a]$ ). The man starts at the right end $x=a$ .

Initial CG of the system:

X_{\text{cg}}^{(i)}=\frac{m\cdot 0+M\cdot a}{m+M}=\frac{Ma}{m+M}.

Step 2 — After the walk.

The smooth floor exerts no horizontal force, so the system’s CG is conserved. Let $\Delta$ be the displacement of the board (positive in the $+x$ direction). The board’s centre is now at $\Delta$ ; the man has walked to the other end, which is at $\Delta-a$ .

Final CG:

X_{\text{cg}}^{(f)}=\frac{m\Delta+M(\Delta-a)}{m+M}=\Delta-\frac{Ma}{m+M}.

Step 3 — Solve.

Setting $X_{\text{cg}}^{(i)}=X_{\text{cg}}^{(f)}$ :

\frac{Ma}{m+M}=\Delta-\frac{Ma}{m+M}\;\Rightarrow\;\Delta=\frac{2Ma}{m+M}.

The board moves in the direction opposite to the man’s walk.

Answer

\boxed{\;\text{Distance covered by board}=\frac{2Ma}{m+M}.\;}