Question

A running man has half the kinetic energy of that of a boy of half of his mass. The man speeds up by $1m/s$. So as to have the same kinetic energy as that of the boy. The original speed of the man is:
A) $\sqrt 2 m/s$
B) $(\sqrt 2 - 1)m/s$
C) $\dfrac{1}{{\sqrt 2 }}m/s$
D) $\dfrac{1}{{\sqrt 2 - 1}}m/s$

Answer 1

Hint: Here, in this question we have asked to find out the original speed of the man. In order to do so you have to know the concept of kinetic energy. Here, two cases are given, the kinetic energy of man and boy. Remember that the kinetic energy of a moving body is equal to half of the product of the mass of the body and the square of its speed.

Formula used:
The kinetic energy of a moving body is given by,
$K.E = \dfrac{1}{2}M{v^2}$
Where, $M$ is the mass of the moving body
$v$ is the velocity of the moving body

Complete step by step solution:
Kinetic energy of a moving body is defined as the half of the product of mass of the body and the square of its speed.
Let us assume that the mass of the man is $M$ and is moving with the velocity of $v$ whereas the boy has the mass of $m$ and moving with the velocity of $v'$.
Now for the first case it is given that the mass of the boy is half of the mass of the mass.
So, $m = \dfrac{M}{2}$
And it is also given that the kinetic energy of man is half of that of boy half of his mass, which is given as
${(K.E)_m} = \dfrac{1}{2}{(K.E)_b}$………(i)
Now, the formula for kinetic energy of moving body is given by
$K.E = \dfrac{1}{2}M{v^2}$
Now, on applying above formula in equation (i), we get
$ \Rightarrow \dfrac{1}{2}M{v^2} = \dfrac{1}{2}\left( {\dfrac{1}{2}m{{v'}^2}} \right)$
On putting $m = \dfrac{M}{2}$ in the above equation, we get
$ \Rightarrow \dfrac{1}{2}M{v^2} = \dfrac{1}{2}\left( {\dfrac{1}{2}\dfrac{M}{2}{{v'}^2}} \right)$……….(ii)
On further solving we get the relation between both the velocities of man and boy as,
${v^2} = \dfrac{1}{4}{v'^2}$………(iii)
Now, for the second case it is given that the man speeds up by $1m/s$, so now the velocity of the man becomes,
$v_m = v + 1$
It is also given that by speeding up the kinetic energy of man and boy becomes equal and the mass of the boy is again the half that of man, so the relation of the kinetic energy is given as,
$ \Rightarrow \dfrac{1}{2}M{(v + 1)^2} = \dfrac{1}{2}\dfrac{M}{2}{v'^2}$
On further solving we get the relation between both the velocities of man and boy as,
${(v + 1)^2} = \dfrac{1}{2}{v'^2}$……….(iv)
Now, by dividing the equation (iv) by (iii), we get
$\dfrac{{{{(v + 1)}^2}}}{{{v^2}}} = \dfrac{{\dfrac{1}{2}{{v'}^2}}}{{\dfrac{1}{4}{{v'}^2}}}$
On further solving we get,
$ \Rightarrow {\left( {\dfrac{{v + 1}}{v}} \right)^2} = 2$
On taking square root on both sides, we get
$ \Rightarrow \dfrac{{v + 1}}{v} = \sqrt 2 $
On further simplifying, we get
$ \Rightarrow 1 + \dfrac{1}{v} = \sqrt 2 $
On further solving, we get the value of the speed of man as
$v = \dfrac{1}{{(\sqrt 2 - 1)}}m/s$
Hence, the speed of man is $\dfrac{1}{{\sqrt 2 - 1}}m/s$ .
Therefore, the correct option is (D).

Note: Now, as we have found the velocity of the man, we can also find the velocity of the boy by using any of the relation that we have calculated above that
${v^2} = \dfrac{1}{4}{v'^2}$
And ${(v + 1)^2} = \dfrac{1}{2}{v'^2}$
We have calculate, $v = \dfrac{1}{{(\sqrt 2 - 1)}}m/s$
By putting this in any of the above two relation, we get the velocity of the boy as,
$v' = \dfrac{1}{{2(\sqrt 2 - 1)}}m/s$.