Question

A man standing a cart of mass doubles the mass of man. Initially a cart is at rest. Now, man jumps horizontally with velocity relative to the cart. Then work done by man during the process of jumping will be

Answer 1

Hint: here a man of mass $m$ is standing on the cart of mass $M$ . Now, it is given in the question that the mass of cart is double the mass of man, therefore, the mass of cart will become $2m$ . here we add the kinetic of the cart and the man to calculate the work done by the man.

Complete step by step answer:
Consider a man standing on a cart of mass $M$ . The mass of the cart is double the mass of a mass. Now, if $m$ is the mass of an object, then the mass of the cart will become
$M = 2m$
Now, when the cart is at rest will be at rest, then the initial velocity of the cart and the man is zero.
$\therefore \,{u_i} = 0$
Now, when the cart starts moving it attains some velocity, let this velocity is ${v_c}$ . Also, the velocity of the man moving with the cart is ${v_m}$ . Now, the man jumps from the cart with velocity $u$ in the horizontal direction, therefore, the total velocity will become
${v_c}\hat i + {v_m}\hat i = u\hat i$
Because the man jumps in a horizontal direction and the cart is also moving in a horizontal direction that is X-axis that is why we have used $\hat i$
$${v_m}\hat i = u\hat i - {v_c}\hat i$$
$ \Rightarrow \,{v_m}\hat i = \left( {u - {v_c}} \right)\hat i$
$ \Rightarrow \,{v_m} = u - {v_c}$
This is the velocity of the man when he jumps from the cart.
Now, using the law of conservation of momentum, we get
$m{v_m} + M{v_c} = 0$
$ \Rightarrow \,m\left( {u - {v_c}} \right) + 2m\left( { - {v_c}} \right) = 0$
$ \Rightarrow \,mu - m{v_c} - 2m{v_c} = 0$
$ \Rightarrow \,mu - 3m{v_c} = 0$
$ \Rightarrow \,{v_c} = \dfrac{u}{3}$
This is the velocity of the cart.
Now, the work done by the man to jump from the cart can be calculated by adding the kinetic energy of the man and the kinetic energy of the cart and is given by
$W = \Delta K.E.$
$ \Rightarrow \,W = \dfrac{1}{2} \times m \times {\left( {u - {v_c}} \right)^2} + \dfrac{1}{2} \times M \times v_c^2$
$ \Rightarrow \,W = \dfrac{1}{2} \times m \times {\left( {u - \dfrac{u}{3}} \right)^2} + \dfrac{1}{2} \times 2m \times {\left( {\dfrac{u}{3}} \right)^2}$
$ \Rightarrow \,W = \dfrac{1}{2} \times m \times \dfrac{{4{u^2}}}{9} + m \times \dfrac{{{u^2}}}{9}$
$ \Rightarrow \,W = \dfrac{{2m{u^2}}}{9} + \dfrac{{m{u^2}}}{9}$
$ \Rightarrow \,W = \dfrac{{3m{u^2}}}{9}$
$ \Rightarrow \,W = \dfrac{{m{u^2}}}{3}$
Therefore, work done by the man during the process of jumping will be $\dfrac{{m{u^2}}}{3}$ .

Note:
Here, to calculate the work done by the man, we have taken the velocity of the cart as negative because when the man jumps in the horizontal direction, he will push the cart in the backward direction.
That is why we have taken the velocity of the cart as $ - {v_c}$ . But the velocity of the man will be positive because he jumps in the forward direction.