Question

The kinetic energy of a man is half the kinetic energy of a boy of half of his mass. If the man increases his speed by $1\;m{s^{ - 1}}$ then his kinetic energy becomes equal to that of the body. Find the ratio of the velocity of the boy and that of the man?
A) $\dfrac{2}{1}$
B) $\dfrac{1}{2}$
C) $\dfrac{3}{4}$
D) $\dfrac{4}{3}$

Answer 1

Hint: Given that the kinetic energy of the man is half of the kinetic energy of the boy and the mass of the body is half of the mass of the man. Hence, by using the kinetic energy formula and equating it on the both of the sides, the ratio of the velocity of the boy and man can be calculated.

Useful formula:
The kinetic energy of the body, $K.E = \dfrac{1}{2}m{v^2}$
Where, $K.E$ is the kinetic energy of the body, $m$ is the mass of the body and $v$ is the velocity of the body.

Assume that,
The kinetic energy of the man is $K.{E_m}$
The kinetic energy of the boy is $K.{E_b}$
The mass of the man is $M$
The mass of the boy is $m$
The velocity the of man is $V$
The velocity of the boy is $v$

Given data:
The kinetic energy of a man is half the kinetic energy of a boy, $K.{E_m} = \dfrac{1}{2}K.{E_b}$
The mass of the boy is half of the mass of the man, $m = \dfrac{1}{2}M$

Complete step by step solution:
The kinetic energy of the man, $K.{E_m} = \dfrac{1}{2}M{V^2}$
The kinetic energy of the boy, $K.{E_b} = \dfrac{1}{2}m{v^2}$

From the question, the kinetic energy of the man is equal to half of the kinetic energy of the boy.
Thus,
$K.{E_m} = \dfrac{1}{2}K.{E_b}\;..............................\left( 1 \right)$
Substitute the values of $K.{E_m}$ and $K.{E_b}$ in equation (1), we get
$
  \dfrac{1}{2}M{V^2} = \dfrac{1}{2}\left[ {\dfrac{1}{2}m{v^2}} \right] \\
  M{V^2} = \dfrac{1}{2}m{v^2}\;.........................................\left( 2 \right) \\
 $
From the question, the mass of the boy is equal to the half of the mass of the man.
Thus,
$m = \dfrac{1}{2}M$
Substitute the above value in equation (2), we get
$
  M{V^2} = \dfrac{1}{2}\left( {\dfrac{1}{2}M} \right){v^2} \\
  M{V^2} = \dfrac{1}{4}M{v^2} \\
  {V^2} = \dfrac{1}{4}{v^2} \\
  \dfrac{{{V^2}}}{{{v^2}}} = \dfrac{1}{4} \\
 $
Taking square root on both sides, we get
$
  \sqrt {\dfrac{{{V^2}}}{{{v^2}}}} = \sqrt {\dfrac{1}{4}} \\
  \dfrac{V}{v} = \dfrac{1}{2} \\
 $
Taking reciprocal on both sides, we get
$\dfrac{v}{V} = \dfrac{2}{1}$

Hence, the option (A) is correct.

Note: Using the relation, the kinetic energy of the man is equal to half of the kinetic energy of the boy, the above solution is given. It can also derive by using another relation given in the question, which is that the kinetic energy of the man is equal to the kinetic energy of the boy when the man increases his velocity by $1\;m{s^{ - 1}}$.