Question

A man is standing on a cart of mass double the mass of the man. Initially, the cart is at rest. Now, man jumps horizontally with velocity u relative to the cart. Then, what is the work done by the man during the process of jumping?
A. $\dfrac{{m{u^2}}}{2}$
B. $\dfrac{{3m{u^2}}}{4}$
C. $m{u^2}$
D. None of the above

Answer 1

Hint
In a system of isolated objects (like this man and the cart), the total momentum always remains constant. And the work done will be equivalent to the kinetic energy of the system, as motion of the man is involved.
$\Rightarrow KE = \dfrac{{m{u^2}}}{2}$, where m is the mass of the object, and u is the velocity. KE is the kinetic energy due to the motion possessed by the object.

Complete step by step answer
We have a two body system consisting of a man and a cart. The man moves relative to the speed of the cart, and we are asked to find the total work done or the kinetic energy produced in this scenario. The information provided to us includes:
Mass of the man: m
Mass of the cart: 2m
Let us assume the following variables:
Velocity of the man: ${v_m}$
Velocity of the cart: v
Thus, according to the question:
$\Rightarrow {v_m} + v = u$ when in motion.
This is equivalent to:
$\Rightarrow {v_m} = u - v$
We know that the momentum of the system will remain constant:
$\Rightarrow m{v_m} = 2mv$
$\Rightarrow m(u - v) = 2mv$
Solving for the velocity of the cart, we get:
$\Rightarrow u - v = 2v$
$\Rightarrow v = \dfrac{u}{3}$
From this, the velocity of the man will be:
$\Rightarrow {v_m} = u - v = u - \dfrac{u}{3} = \dfrac{{2u}}{3}$
To calculate the total kinetic energy, we know that:
$\Rightarrow KE = \dfrac{{m{u^2}}}{2}$
For our case, this will transform to:
$\Rightarrow KE = \dfrac{{m{v_m}^2}}{2} + \dfrac{{2m{v^2}}}{2}$
$\Rightarrow KE = \dfrac{{m{{\left( {\dfrac{{2u}}{3}} \right)}^2}}}{2} + m{\left( {\dfrac{u}{3}} \right)^2}$
Opening the brackets and solving further gives us:
$\Rightarrow KE = \dfrac{{m \times 4{u^2}}}{{2 \times 9}} + m\dfrac{{{u^2}}}{9} = m{u^2}\left( {\dfrac{2}{9} + \dfrac{1}{9}} \right)$
$\Rightarrow KE = m{u^2} \times \dfrac{3}{9} = \dfrac{{m{u^2}}}{3}$
Since this answer matches none of the given options, the correct answer is option (D): None of these.

Note
Relative velocities are widely used in Physics to have an intuitive idea about the speed of the object under observation. For example, when two objects are moving independently, it is difficult to see how the motion of one is dependent on the other. But once we make one of the objects stationary and choose it as the frame of reference, the speed of the other object relative to it tells us about how much faster or slower this object is as compared to the frame of reference.