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A man of \[50kg\] is standing at one end on a boat of length $25m$ and mass $200kg$. If he starts running and when he reaches the other end, he has a velocity $2m{s^{ - 1}}$ with respect to the boat. The final velocity of the boat is (in $m{s^{ - 1}}$).
A) $\dfrac{2}{5}$
B) $\dfrac{2}{3}$
C) $\dfrac{8}{5}$
D) $\dfrac{8}{3}$

Last updated date: 25th Jun 2024
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Hint: The given question is based on conservation of momentum. So, in order to get the correct solution for the given question, we need to apply the conservation of momentum for the man and the boat after reaching the other end. After that we need to solve the equation to conclude with the correct solution.

Complete step by step solution:
The mass of the man in the question is given as, $m = 50kg$
The mass of the boat in the question is given as, $M = 200kg$
The velocity of the man after reaching the other end is given as, $v = 2m{s^{ - 1}}$
And let us assume the velocity of the boat after the man reaches the other end to be, $V$
According to the question initially both the man and the boat were at rest.
Now, applying the conservation of momentum, we get,
${P_1} = {P_2}$
We know that momentum can be written as, $P = mv$
Now, let us write the equation for the conservation of momentum for the given case.
So, we get,
$ \Rightarrow mv + (m + M)V = 0$
$ \Rightarrow 50 \times 2 + (50 + 200)V = 0$
$ \Rightarrow 100 + 250V = 0$
$ \Rightarrow 250V = 100$
$\therefore V = - \dfrac{{100}}{{250}} = - \dfrac{2}{5}m{s^{ - 1}}$
Here, a negative sign shows that the velocity of the man is in the opposite direction of the boat.
Therefore, the final velocity of the boat is $\dfrac{2}{5}m{s^{ - 1}}$.

Hence, option (A) is the correct choice for the given question.

Note: According to the conservation of momentum, the momentum before collision is equal to the momentum after collision. We define momentum as the product of mass and the velocity of a body. When the bodies are moving in a straight path, the momentum is said to be linear momentum.