Question

A running man $A$ has the same kinetic energy as that of a boy $B$ of half his mass. The man $A$ speeds up by $2\,m{s^{ - 1}}$ and the boy $B$ changes his speed by $x$ so that the kinetic energies of $A$ and $B$ are equal again. Then the value of $x$ is:
(A) $4\,m{s^{ - 1}}$
(B) $2\sqrt 2 \,m{s^{ - 1}}$
(C) $\dfrac{1}{{\sqrt 2 }}\,m{s^{ - 1}}$
(D) $2\,m{s^{ - 1}}$

Answer 1

The value of the $x$ can be determined by using the kinetic energy equation of the man and the boy, by equating the kinetic energy of the man and boy, the velocity equation is determined. And by using the velocity information in the kinetic energy equation, and then by equating the kinetic energy equation, the value of $x$ can be determined.

Formula used:
The kinetic energy of the body is given by,
$KE = \dfrac{1}{2}m{v^2}$
Where, $KE$ is the kinetic energy of the body, $m$ is the mass of the body and $v$ is the velocity of the body.

Complete step by step answer:
Given that,
The mass of the boy is equal to the half mass of the man.
If the mass of the boy is $m$, then the mass of the man is $2m$. So that the mass of the boy is half of the mass of the man.
Now,
The kinetic energy of the man is given by,
$KE = \dfrac{1}{2}m{v^2}$
By substituting the mass of the man in the above equation, then
$KE = \dfrac{1}{2}\left( {2m} \right){v^2}\,....................\left( 1 \right)$
Where, $v$ is the velocity of the man.
Now,
The kinetic energy of the boy is given by,
$KE = \dfrac{1}{2}m{v^2}$
By substituting the mass of the boy in the above equation, then
$KE = \dfrac{1}{2}mv{'^2}\,....................\left( 2 \right)$
Where, {v^'} is the velocity of the boy.
By equation the equation (1) and equation (2), then
$\dfrac{1}{2}\left( {2m} \right){v^2} = \dfrac{1}{2}mv{'^2}$
By cancelling the same terms in the above equation, then
$2{v^2} = v{'^2}$
By taking the square root on both side in the above equation, then
$v' = \sqrt 2 v$
Now the velocity of the both man and boy is increasing, then
The kinetic energy of the man is given by,
$KE = \dfrac{1}{2}\left( {2m} \right){\left( {v + 2} \right)^2}\,....................\left( 3 \right)$
The kinetic energy of the boy is given by,
$KE = \dfrac{1}{2}m{\left( {v' + x} \right)^2}\,....................\left( 4 \right)$
By equating the equation (3) and the equation (4), then
$\dfrac{1}{2}\left( {2m} \right){\left( {v + 2} \right)^2} = \dfrac{1}{2}m{\left( {v' + x} \right)^2}$
By cancelling the same terms in the above equation, then
$2{\left( {v + 2} \right)^2} = {\left( {v' + x} \right)^2}$
By taking the square root on both side in the above equation, then
$\sqrt 2 \left( {v + 2} \right) = \left( {v' + x} \right)$
By multiplying the terms in the above equation, then
$v' + x = \sqrt 2 v + 2\sqrt 2$
By substituting the value of $v'$ in the above equation, then
$\sqrt 2 v + x = \sqrt 2 v + 2\sqrt 2$
By rearranging the terms in the above equation, then
$x = \sqrt 2 v + 2\sqrt 2 - \sqrt 2 v$
By subtracting the terms in the above equation, then
$x = 2\sqrt 2 \,m{s^{ - 1}}$

So, the correct answer is “Option B”.

Note:
The kinetic energy of the body is directly proportional to the mass of the body and the square of the velocity of the body.
The mass of the body and the velocity of the body increases, the kinetic energy of the body also increases accordingly.