Question

A running has half the kinetic energy of a running boy of half his mass. The man speeds up by 1m/s and then has K.E. as that of the boy. What was the original speed of the man and the boy?
A) $\sqrt 2 m/s;\left( {2\sqrt 2 - 1} \right)m/s$
B) $\left( {\sqrt 2 - 1} \right)m/s;\left( {2\sqrt 2 - 1} \right)m/s$
C) $\left( {\sqrt 2 + 1} \right)m/s;\left( {2\sqrt 2 + 1} \right)m/s$
D) $\left( {\sqrt 2 + 1} \right)m/s;\sqrt 2 m/s$

Answer 1

Hint: Apply the formula of the kinetic energy which $K.E. = \dfrac{1}{2}\mathop {Mv}\nolimits^2 $ and the relation between the velocities of the man and the boy. Then find the value of the original speed of the man using the given condition for the equal kinetic energies of the man and the boy

Formula Used:
$K.E. = \dfrac{1}{2}\mathop {Mv}\nolimits^2 $
 Where $K.E.$ 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:
The value of the mass of the man and the velocity of the man is unknown and hence let us consider the mass of the man as $m$
The value of the velocity of the man be $v$

So, we get the kinetic energy of the man using the formula given as:
The kinetic energy of the man is used using the formula as :
$K.E. = \dfrac{1}{2}\mathop {Mv}\nolimits^2 $
Where $K.E.$ is the kinetic energy, $M$ is the mass and $v$ is the velocity of the object whose kinetic energy we are calculating.


The value of the kinetic energy of the man becomes:
$\mathop {K.E}\nolimits_m = \dfrac{1}{2}\mathop {Mv}\nolimits^2 $
Now the mass of a boy can be written as: $\dfrac{m}{2}$
The velocity of the boy is assumed as $\mathop v\nolimits_b $
Now the given condition is
$\implies$ $\dfrac{1}{2}\mathop {mv}\nolimits^2 = \dfrac{1}{4} \times \dfrac{m}{2}\mathop v\nolimits_b^2 $
We get,
$\implies$ $v = \dfrac{{\mathop v\nolimits_b }}{2}........eqn{\text{ }}1$
Now, the value of the kinetic energy is the same as that of the boy when the value of the velocity of the man is increased by 1 meter per second.
So, we can write the condition as:
$\implies$ $\dfrac{1}{2}\mathop {m\left( {v + 1} \right)}\nolimits^2 = \dfrac{1}{2} \times \dfrac{m}{2}\mathop {\left( {\mathop v\nolimits_b } \right)}\nolimits^2 $
Substituting the value of the $\mathop v\nolimits_b $ we get the value of the initial velocity of the man is,
$\implies$ $v = \dfrac{{\sqrt 2 - 1}}{1}m/s$

So, the value of the speed of the man initially becomes: $\left( {\sqrt 2 - 1} \right)m/s$.
Now put the value of speed of man in equation $1$
$\left( {\sqrt 2 - 1} \right) = \dfrac{{\mathop v\nolimits_b }}{2}$
\[\mathop v\nolimits_b = 2(\sqrt 2 - 1)m/s\]
So, the original speed of boy \[\mathop v\nolimits_b = 2(\sqrt 2 - 1)m/s\]
And the original speed of the man is $\left( {\sqrt 2 - 1} \right)m/s$
So, the correct answer is option B.

Note: One of the main possible mistakes that we can make in this kind of problem is finding the condition to apply for the given problem. The given is that the velocity of the man’s speed is increased by one meter per second the kinetic energy is the same. We need to carefully apply the condition to get the value of the original speed of the person.