Question

A man of mass $62\,kg$ is standing on a stationary boat of mass $238\,kg$ . The man is carrying a sphere of mass $0.5\,kg$ in his hands. If the man throws the sphere horizontally with a velocity of $12\,m{s^{ - 1}}$ , find the velocity with which the boat will move (in magnitude)
A. $0.02\,m{s^{ - 1}}$
B. $0.5\,m{s^{ - 1}}$
C. $0.04\,m{s^{ - 1}}$
D. $0.06\,m{s^{ - 1}}$

Answer 1

Hint:This question utilizes the concept of conservation of momentum. The total mass of the system is given to us. Also, the mass of the object thrown is given and also the velocity it attains is given. Also, since the man throws the sphere horizontally, it involves only the x axis.

Formulae used:
$p = mv$
where $p$ is the linear momentum, $m$ is the mass of the body and $v$ is the velocity of the body

Complete step by step answer:
According to the given question
Mass of the man ${m_m} = 62\,kg$
Mass of the stationary boat ${m_b} = 238\,kg$
Mass of the sphere ${m_s} = 0.5\,kg$
Initial velocity of the sphere $u = 12\,m{s^{ - 1}}$

Here, we consider the boat – man – sphere as an isolated system which is devoid of any external force.From the Law of conservation of momentum, we deduce that the total momentum of an isolated system remains constant.Thus, to maintain this law, the boat – man system will need to move with a velocity $v'$ opposite to that of the sphere.

From the law of conservation of momentum, we get
$momentum{\kern 1pt} {\kern 1pt} of{\kern 1pt} {\kern 1pt} man \cdot boat{\kern 1pt} {\kern 1pt} system + momentum{\kern 1pt} {\kern 1pt} of{\kern 1pt} {\kern 1pt} sphere = 0$ (since the man-boat-sphere system is stationary)
Substituting the respective symbols, we get
$ \Rightarrow ({m_m} + {m_b})v' + {m_s}u = 0$
Now, putting in the values, we have
$\Rightarrow (62kg + 238kg)v' + 0.5kg \times 12m{s^{ - 1}} = 0 \\
\Rightarrow 300kg \times v' = - (0.5kg \times 12m{s^{ - 1}}) \\
\Rightarrow 300kg \times v' = - 6kgm{s^{ - 1}} \\
\Rightarrow v' = \dfrac{{ - 6kgm{s^{ - 1}}}}{{300kg}} \\
\therefore v' = - 0.02\,m{s^{ - 1}} \\ $
Thus, the correct answer to the question is option A.

Note:The negative sign shows that the velocity of the man-boat system is opposite to the direction of velocity of the sphere. Here, we consider that there is no friction between the body of the boat and the water, hence the boat moves. In normal circumstances, this wouldn’t be the case as the friction would stop it.