Question

A wooden plank of mass $30\text{kg}$ is resting on a smooth horizontal floor. A man of mass $70\text{kg}$ starts moving from one end of the plank to the other end. The length of plank is $20\text{m}$. Find the displacement of the plank over the floor when the man reaches the other end of the plank.

• A. $10\text{m}$
• B. $12\text{m}$
• C. $8.5\text{m}$
• D. $14\text{m}$

Answer 1

The correct option is D $14\text{m}$

Displacement of $\text{COM}$ of system in horizontal direction is ZERO.
$\therefore \Delta x_{\text{CM}}=\mathrm{0.....}\left(i\right)$
due to absence of any external force on the system of (man +plank) in horizontal direction.


When man reaches the other end of plank, $L=20\text{m}$ is the displacement of man w.r.t plank in $+ve$ x-direction

Let $x\text{m}$ be the displacement of plank towards $-ve$ x-direction w.r.t ground to keep net displacement of $\text{COM}$ as zero.

Assuming the origin, i.e., $x=0$ at the position shown in figure.


$x_{\text{CM}}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$
$\therefore \Delta x_{\text{CM}}=\frac{m_1\Delta x_1+m_2\Delta x_2}{m_1+m_2}....\left(ii\right)$

$\Delta x_1=$ displacement of man w.r.t ground $=(+20-x)=(20-x)\text{m}$
$m_1=70\text{kg}$
and
$\Delta x_2=$ displacement of plank's COM w.r.t ground $=-x\text{m}$
$m_2=30\text{kg}$

From Eq. ($i)$ and ($ii)$:
$\Rightarrow 0=\frac{(70\times (20-x\left)\right)+(30\times (-x\left)\right)}{70+30}$
$1400-70x-30x=0$
$\therefore x=14\text{m}$

Hence plank will be displaced by $14\text{m}$ in opposite direction.