Question

A man weighing 80kg is standing at the centre of a flat boat and he is $20m$ from the shore. He walks $8m$ on the boat towards the shore and then halts. The boat weighs $200kg$. How far is he from the shore at the end of this time?
(A) $11.2m$
(B) $13.8m$
(C) $14.3m$
(D) $15.4m$

Answer 1

Hint: There is no external force on the system. So the displacement of the centre of mass must be zero. From this information, we can find out the backward displacement of the boat, from which the displacement of the man can be found out.

Complete step-by-step answer:
Let us consider the man and the boat together as a system. There is no external force acting on this system. Therefore the acceleration of the centre of mass of the system will be zero. Initially, according to the question, the man and the boat were at the state of rest. This means that the initial velocity of the centre of mass of the system should be zero. As the acceleration of the centre of mass is zero, the velocity of the centre of mass will always remain zero. This means that the displacement of the centre of mass is zero, that is,
${x_{com}} = 0$ ……………….(1)
Let us assume the direction from the centre of the boat to the shore to be positive.
As the man walks towards the shore, the boat will displace away from the shore. Let the displacement of the boat be $x$. According to our sign convention, this displacement is negative. Therefore the displacement of the boat is
${x_1} = - x$ …………………...(2)
According to the question, the man walks $8m$ towards the shore relative to the boat. So his displacement with respect to the ground is
${x_2} = \left( {8 - x} \right)$ …………………...(3)
Now, we know that the displacement of the centre of mass is given by
${x_{com}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2}}}{{{m_1} + {m_2}}}$
From (1), (2) and (3)
$0 = \dfrac{{{m_1}\left( { - x} \right) + {m_2}\left( {8 - x} \right)}}{{{m_1} + {m_2}}}$
$ \Rightarrow {m_1}x = {m_2}\left( {8 - x} \right)$
On rearranging the terms, we have
$\left( {{m_1} + {m_2}} \right)x = 8{m_2}$
$ \Rightarrow x = \dfrac{{8{m_2}}}{{\left( {{m_1} + {m_2}} \right)}}$
According to the question, the mass of the boat is ${m_1} = 200kg$, and the mass of the man is ${m_2} = 80kg$. Substituting these above we get
$x = \dfrac{{8 \times 80}}{{280}}$
$ \Rightarrow x = 2.29m$ …………..(4)
Substituting (4) in (3) we get the displacement of the man as
${x_2} = \left( {8 - 2.29} \right)m$
$ \Rightarrow {x_2} = 5.71m$ ………………..(5)
So the man has walked $5.71m$ towards the shore.
The initial distance of the shore from the man is given to be
$d = 20m$ …………………..(6)
So the final distance of the man becomes
$d' = d - {x_2}$
Putting (5) and (6)
$d' = 20 - 5.71$
$ \Rightarrow d' = 14.29m \approx 14.3m$
Hence, the man is $14.3m$ away from the shore at the end.

So the correct answer will be option C.

Note: Choosing the sign convention is very important to be assumed before starting the solution. Also, do not take the displacement of the man to be $8m$. It is with respect to the boat, while we are concerned with the displacement with respect to the ground.