Question

A man of mass $60kg$ jumps from a trolley of mass $20kg$ standing on the smooth surface with absolute velocity $3m/s$. Find the velocity of a trolley and total energy produced by man:
A. ${\text{90m/s}}$, $1.08KJ$
B. ${\text{6m/s}}$, $1.02KJ$
C. ${\text{7m/s}}$, $1.02KJ$
D. ${\text{9m/s}}$, $1.08KJ$

Answer 1

Hint: In this question, we need to determine the velocity of a trolley and total energy produced by man such that the man jumps from the trolley while standing on a smooth surface. For this, we will use the concept of momentum to determine the velocity of the trolley and then, using the relation between the energy, mass and the velocity to find the energy produced by the man.

Formula used:
${M_1}{v_1} + {M_2}{v_2} = 0$ and $E = \dfrac{1}{2}{M_1}v_1^2 + \dfrac{1}{2}{M_2}v_2^2$ where, ${M_1}$ and ${M_2}$ are the mass of the man and the mass of the trolley respectively and ${v_1}$ and ${v_2}$ are the velocity of the man and the trolley respectively.

Complete step by step answer:
We have been given the information that a man of mass $60kg$ jumps from a trolley.
The mass of the trolley is $20kg$ and is present on a smooth surface.
The man jumps with an absolute velocity of ${\text{3m/s}}$.
Now, we need to find the velocity of the trolley. Also, we have to find the total energy produced by the man.
Let ${M_1}$ and ${M_2}$ be the mass of the man and the mass of the trolley respectively.
Therefore, we now have that ${M_1} = 60kg$ and ${M_2} = 20kg$.
We already have the velocity of the man ${v_1} = {\text{3m/s}}$.
Therefore by using the formula ${M_1}{v_1} + {M_2}{v_2} = 0$, let us find the velocity of the trolley.
The velocity of the trolley can be given as
$
  {M_1}{v_1} + {M_2}{v_2} = 0 \\
  60 \times 3 + 20 \times {v_2} = 0 \\
  {v_2} = - 9{\text{m/s}} \\
 $
The negative sign indicates that the trolley was moving in the opposite direction. Hence, the velocity of the trolley is ${\text{9m/s}}$.
Now, we have to find the energy produced by the man. This can be done by using the formula $E = \dfrac{1}{2}{M_1}v_1^2 + \dfrac{1}{2}{M_2}v_2^2$.
\[
  E = \dfrac{1}{2}{M_1}v_1^2 + \dfrac{1}{2}{M_2}v_2^2 \\
   = \dfrac{1}{2} \times 60 \times {3^2} + \dfrac{1}{2} \times 20 \times {9^2} \\
   = 1.08KJ \\
 \]
Hence, the energy produced by the man is $1.08KJ$.
Therefore, we can conclude that the velocity of a trolley and total energy produced by man is ${\text{9m/s}}$ and $1.08KJ$ respectively.

So, the correct answer is “Option D”.

Note:
Whenever, two bodies are involved in terms of collision or touching then, always there will be a role of momentum. Students must know the formulas used in this solution. Then only they can answer the question. The most important formula used in this solution is $E = \dfrac{1}{2}{M_1}v_1^2 + \dfrac{1}{2}{M_2}v_2^2$ which gives the relation between the two bodies involving different masses and velocities.