Question

A man running on a horizontal road has half the kinetic energy of a boy of half of his mass. When the man speeds up by m/s then his kinetic energy becomes equal to kinetic energy of the boy, the original speed of the man is
$(a){\text{ }}\sqrt 2 m/s \\\ (b){\text{ }}\left( {\sqrt 2 - 1} \right)m/s \\\ (c){\text{ 2m/s}} \\\ {\text{(d) }}\left( {\sqrt 2 + 1} \right)m/s \\\$

Answer 1

Hint: In this question let the speed and the mass of the man be x m/s and M Kg respectively. Use the direct formula for the kinetic energy that is $K.E = \dfrac{1}{2}m{v^2}$, let the speed of the boy be y m/s. Use the constraints of the question to formulate the relationship between x and y.

Complete step-by-step solution -
Let the speed of the man be x m/s.
And the mass of the man is M kg.
Now it is given that the kinetic energy of a man is half the kinetic energy of the boy when running on a horizontal road who has mass half of the man.
So the mass of the boy = $\dfrac{M}{2}$Kg.
Let the speed of the boy is y m/s.
Now as we know that the kinetic energy is given as
$\Rightarrow K.E = \dfrac{1}{2}m{v^2}$
Where, K.E = kinetic energy
m = mass of the body
v = velocity of the body by which it travels.
So the kinetic energy of the boy ${\left( {K.E} \right)_b}$= $\dfrac{1}{2}\left( {\dfrac{M}{2}} \right){y^2}$ Joule
And the kinetic energy of the man ${\left( {K.E} \right)_m}$ = $\dfrac{1}{2}M{x^2}$ Joule
Now it is given that the kinetic energy of a man is half the kinetic energy of the boy.
Therefore,
${\left( {K.E} \right)_m} = \dfrac{1}{2}{\left( {K.E} \right)_b}$
Now substitute the values we have,
$\Rightarrow \dfrac{1}{2}M{x^2} = \dfrac{1}{2}\left[ {\dfrac{1}{2}\left( {\dfrac{M}{2}} \right){y^2}} \right]$
Now simplify this we have,
$\Rightarrow {x^2} = \dfrac{1}{4}{y^2}$
Now take square root on both sides we have,
$\Rightarrow x = \sqrt {\dfrac{1}{4}{y^2}} = \dfrac{1}{2}y$................. (1)
Now when the man speeds up by 1 m/s then his kinetic energy becomes equal to the kinetic energy of the boy.
So the new speed of the man = (x + 1) m/s.
Now according to condition we have,
New Kinetic energy of the man = kinetic energy of the boy
$\Rightarrow \dfrac{1}{2}M{\left( {x + 1} \right)^2} = \dfrac{1}{2}\left( {\dfrac{M}{2}} \right){y^2}$
Now simplify this we have,
$\Rightarrow {\left( {x + 1} \right)^2} = \dfrac{1}{2}{y^2}$
Now take square root on both sides we have,
$\Rightarrow x + 1 = \sqrt {\dfrac{1}{2}{y^2}} = \dfrac{1}{{\sqrt 2 }}y$................. (2)
Now from equation (1) we have, y = 2x so substitute this value in equation (2) we have,
$\Rightarrow x + 1 = \dfrac{1}{{\sqrt 2 }}2x$
$\Rightarrow x + 1 = \sqrt 2 x$, $\left[ {\because \dfrac{2}{{\sqrt 2 }} = \sqrt 2 } \right]$
$\Rightarrow x = \dfrac{1}{{\sqrt 2 - 1}}$
Now rationalize this by $\left( {\sqrt 2 + 1} \right)$ we have,
$\Rightarrow x = \dfrac{1}{{\sqrt 2 - 1}} \times \dfrac{{\sqrt 2 + 1}}{{\sqrt 2 + 1}}$
Now the denominator will work as according to property $\left( {a - b} \right)\left( {a + b} \right) = \left( {{a^2} - {b^2}} \right)$ so we have,
$\Rightarrow x = \dfrac{{\sqrt 2 + 1}}{{{{\left( {\sqrt 2 } \right)}^2} - {1^2}}} = \dfrac{{\sqrt 2 + 1}}{{2 - 1}} = \sqrt 2 + 1$
So this is the required original speed of the man.
Hence option (D) is the correct answer.

Note: Kinetic energy can not only be depicted by the relation that is $K.E = \dfrac{1}{2}m{v^2}$, but however it can also be expressed in terms of momentum that is $K.E = \dfrac{1}{2}\dfrac{{{P^2}}}{m}$. The main reason behind gaining of the kinetic energy is that when an object works or speeds up then some net force is applied by the object onto the ground, this helps in production of energy.