Question

A man of mass $m$ walks from end $A$ to the other end $B$ of a boat of mass $M$ and length $l$. The coefficient of friction between man and boat is $\mu$ and neglect any resistive force between boat and water.

• A. If man runs at his maximum acceleration the acceleration of boat is $\frac{m}{M}\mu g$.
• B. Minimum time take by man to reach other end of the boat is $\sqrt{\frac{2Ml}{(M+m)\mu g}}$
• C. Magnitude of displacement of centre of mass of boat is $\frac{Ml}{m+M}$
• D. Velocity of $CM$ of man and boat is zero.

Answer 1

Answer: (A), (B), (D)

The correct options are
A If man runs at his maximum acceleration the acceleration of boat is $\frac{m}{M}\mu g$.
B Minimum time take by man to reach other end of the boat is $\sqrt{\frac{2Ml}{(M+m)\mu g}}$
D Velocity of $CM$ of man and boat is zero.

As man walks on the boat only force acting between them is kinetic friction between man and mass.

From $f_{\text{max}}=\mu mg$, we get

$a_{\text{boat}}=\frac{\mu mg}{M}$

$a_{\text{man}}=\frac{\mu mg}{m}=\mu g$

As there is no external force COM of the system doesnot move $(V_{com}=0)$while man and boat move in opposite directions.
$a_{\text{man w.r.t boat}}=\mu g-(-\frac{\mu mg}{M})=\frac{\mu (M+m)g}{M}$

As man moves from one end of boat to another displacement of man relative to boat is $l$.
Using
$s=u\times t+\frac{1}{2}\times a{t}^2$
Here $(u=0)$ as initially man is at rest relative to boat
$l=\frac{1}{2}\frac{\mu (M+m)g}{M}{t}^2$
$t=\sqrt{\frac{2Ml}{\mu [M+m]g}}$