Question

A man of mass $m$ runs without sliding from rest from one end of a boat of mass $M$ and length $l$ with an acceleration a relative to the boat. If the friction between water and boat is neglected, find the
a) acceleration of $CM$ of the system $(M+M)$
b) acceleration of the man and boat
c) position of man at the time when he reaches to other end of the boat
d) frictional force
e) work done by friction
f) total work done by friction
g) velocities of man and boat when the man reaches other end of the boat.
h) work done by man.

Answer 1

a) Consider 'boat and man' a system. No external force acts on the system in horizontal direction and initially the system is at rest i.e. the centre of mass is at rest. The acceleration of centre of mass will be zero. There will be not be any displacement of centre of mass.
b) Considering the FBD of man and boat. If we consider man and boat will be responsible for the motion of man and boat. But as a whole the friction will be an internal force and will not move centre of mass of system.
Let the acceleration and displacement of man be ${\overrightarrow{a}}_m$ and ${\overrightarrow{x}}_m$ and that for boat be ${\overline{a}}_M$ and ${\overrightarrow{x}}_M$
${\overrightarrow{a}}_m={\overrightarrow{a}}_{m.M}+{\overrightarrow{a}}_M=-a\hat{i}+a_M\hat{i}$
$=(a_m-a)\hat{i}$
But ${\overrightarrow{a}}_{cm}=\frac{m{\overrightarrow{a}}_m+M{\overrightarrow{a}}_M}{(m+M)}$
or $0=\frac{m(a_M-a)\hat{i}+M{a}_M\hat{i}}{(m+M)}$
$⟹{\overrightarrow{a}}_M=\frac{ma}{(m+M)}\left(\hat{i}\right)$ (towards right)
Hence $a_m=\frac{ma}{(m+M)}-a$
$⟹{\overrightarrow{a}}_m=\frac{Ma}{(m+M)}(-\hat{i})$ (towards left)
c) We know the displacement of $CM$
${\overrightarrow{x}}_{cm}=\frac{m{\overrightarrow{x}}_m+M{\overrightarrow{x}}_M}{(m+M)}$
Here ${\overrightarrow{x}}_m={\overrightarrow{x}}_{m.M}+{\overrightarrow{x}}_M=(-l\hat{i}+x_M\hat{i})$
$⟹{\overrightarrow{x}}_m=(x_m-l)\hat{i}$
Hence $x_{cm}=O=\frac{m(x_M-l)\hat{i}+M{x}_M\hat{i}}{(m+M)}$
$⟹\overrightarrow{M}=-\frac{ml}{(m+M)}(\widehat{i)}$ [towards right]
and ${\overrightarrow{x}}_m=\frac{Ml}{(m+M)}\left(\hat{i}\right)$ [towards left]
Position of man from observer at the time when the man reaches to other end of the boat.
$\overrightarrow{x}=(x_0+l)\hat{i}+{\overrightarrow{x}}_m$
$=(x_0+l)\hat{i}-\frac{Ml}{(m+M)}\hat{i}=(x_0+\frac{ml}{(m+M)})\hat{i}$
d) From FBD of man, friction acting on the man,
$f=m{a}_m=m\left(\frac{Ma}{(m+M)}\right)=\frac{Mma}{(m+M)}$
e) Work done by friction force on boat
$(W_f{)}_{boat}=\overrightarrow{f}{\overrightarrow{x}}_m=\left[\frac{Mma}{(m+M)}\right]\left[\frac{ml}{(m+M)}\right]$
$⟹(W_f{)}_{boat}=\frac{M{m}^{2}al}{(m+M{)}^2}$
f) As friction between man and boat in static nature and total work done by static friction force on the system should be zero.
g) For finding the velocity of man when he reaches other end of the boat, we can use kinematics
Using ${\nu }_m^2=u_m^2+2a_m{x}_m=0+2\left(\frac{Ma}{(m+M)}\right)\left(\frac{Ml}{(m+M)}\right)$
${\nu }_m^2=\left(\frac{M}{m+M}\right)2al$
Hence velocity of man ${\nu }_m=\left(\frac{M}{m+M}\right)\sqrt{2al}$
Similarly velocity of boat ${\nu }_M=\left(\frac{M}{m+M}\right)\sqrt{2al}$
h) If we consider 'man + boat' on system, friction will do no work on system as it in static nature. But we can see the kinetic energy of the system will increase. The question arises here who does work to increase the kinetic energy of system. The answer is 'man'. The man does this work through his internal energy to increase the kinetic energy of the system.
Using work energy theorem on the system, we get
$W_{man}+W_{friction}=\Delta K=K_f-K_i$
We know $W_{friction}=0$
$W_{man}+0=(\frac{1}{2}m{\nu }_m^2+\frac{1}{2}M{V}_M^2)-0$
$W_{man}=\frac{1}{2}m\left[{\left(\frac{M}{m+M}\right)}^{2}2al\right]+\frac{1}{2}M\left[{\left(\frac{M}{m+M}\right)}^{2}2al\right]-0$
$⟹W_{man}=\frac{mMal}{(m+M)}$