Question

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

Answer 1

Here two cases are given relating the kinetic energy of the man and the boy. In the first case the mass of the man was $M$, the velocity of the man was ${v_1}$ and he was having a kinetic energy which is half that of the boy with mass $\dfrac{M}{2}$ and velocity${v_2}$.In the second case when the velocity of the man becomes $v + 1$their kinetic energy becomes equal. Comparing these two cases given we have two find the velocity ${v_1}$ of the man.

Formula used
$K.E = \dfrac{1}{2}M{v^2}$,${K.E}$ where is the kinetic energy, $M$is the mass and $v$ is the velocity of the object whose kinetic energy we are calculating.

Complete step-by-step answer:
Kinetic energy is defined as the half of the product of mass and square times the velocity.
$K.E = \dfrac{1}{2}M{v^2}$,${K.E}$ where is the kinetic energy, $M$is the mass and $v$ is the velocity of the object whose kinetic energy we are calculating.
First Case: The kinetic energy of the man is half the kinetic energy of the boy with half the mass. The man has a mass$M$ and velocity ${v_1}$ whereas the boy has mass $\dfrac{M}{2}$ and velocity ${v_2}$.
$\dfrac{1}{2}Mv_1^2 = \dfrac{1}{2}\left( {\dfrac{1}{2}\dfrac{M}{2}v_2^2} \right)$
$v_1^2 = \dfrac{1}{4}\left( {v_2^2} \right)$
Second Case: The kinetic energy of the man is equal to the kinetic energy of the boy with half the mass. The man has a mass$M$ and velocity ${v_1} + 1$ whereas the boy has mass $\dfrac{M}{2}$ and velocity${v_2}$.
$\dfrac{1}{2}M{\left( {{v_1} + 1} \right)^2} = \dfrac{1}{2}\dfrac{M}{2}v_2^2$
${\left( {{v_1} + 1} \right)^2} = \dfrac{1}{2}v_2^2$
By taking the ratio both these relations we get
$\dfrac{{v_1^2}}{{{{\left( {{v_1} + 1} \right)}^2}}} = {\left( {\dfrac{{{v_1}}}{{{v_1} + 1}}} \right)^2} = \dfrac{1}{2}$
Taking the reciprocal,
${\left( {\dfrac{{{v_1} + 1}}{{{v_1}}}} \right)^2} = 2$
$\dfrac{{{v_1} + 1}}{{{v_1}}} = \sqrt 2$
$1 + \dfrac{1}{{{v_1}}} = \sqrt 2$
${v_1} = \dfrac{1}{{\left( {\sqrt 2 - 1} \right)}}m/s$
The correct answer is C

Note: Now that we have found the velocity of the man we can use any of the relations that we have found to calculate the velocity of boy. We calculated that
$v_1^2 = \dfrac{1}{4}\left( {v_2^2} \right)$
${v_1} = \dfrac{{{v_2}}}{2}$
We know ${v_1} = \dfrac{1}{{\left( {\sqrt 2 - 1} \right)}}m/s$
Therefore ${v_2} = \dfrac{1}{{2\left( {\sqrt 2 - 1} \right)}}m/s$