Question

A man weighing $60\text{kg}$ is standing on a flat boat having ends $A$ and $B$, such that he is at a distance of $10\text{m}$ from one end $\left(B\right)$. The boat weighs $20\text{kg}$ and there is no friction between it and the water. If the man walks $4\text{m}$ relative to the boat towards the end $\left(B\right)$ and stops, then how far is the man from the end $B$ ?

• A. $1\text{m}$
• B. $7\text{m}$
• C. $3\text{m}$
• D. $9\text{m}$

Answer 1

The correct option is D $9\text{m}$

$\text{Mass of man}\left(m\right)=60\text{kg}$
$\text{Mass of boat}\left(M\right)=20\text{kg}$
Initially, the system is at rest
$\therefore {\overrightarrow{v}}_{\text{CM}}=0$
As there is no external force
$\therefore {\overrightarrow{v}}_{\text{CM}}=\text{constant}=0$
Hence displacement of centre of mass,
$\Delta x_{\text{CM}}=0$
$\Rightarrow \frac{m_1\Delta x_1+m_2\Delta x_2}{m_1+m_2}=0$
$\therefore m\Delta x_1+M\Delta x_2=0....\left(i\right)$
where $\Delta x_1=$ displacement of man w.r.t ground
& $\Delta x_2=$displacement of boat w.r.t ground

Let the boat move by distance $d_2$ in a direction opposite to man (direction of motion of man is taken $+ve$)
Hence, $\Delta x_2=-d_2$
Since man walks $+4\text{m}$ relative to boat, displacement of man relative to ground
$\Delta x_1=4+\Delta x_2=4-d_2$
Substituting in Eq $\left(i\right)$:
$m(4-d_2)+M(-d_2)=0$
$60(4-d_2)+20(-d_2)=0$
$\therefore d_2=3\text{m}$
Displacement of man towards end $B$ w.r.t ground will be:
$\Delta x_1=4-d_2=4-3=1\text{m}$
$\therefore$Man is now at a distance of $(10-1)=9\text{m}$ from end $B$.