Question

A man of \[M = 58kg\;\] jumps from an aeroplane as shown in figure. He sees the hard ground below him and a lake at a distance \[d = 1m\] from the point directly below him. He immediately puts off his jacket (mass \[m = 2kg\]) and throws it in a direction directly away from the lake. If he just fails to strike the ground, find the distance (in \[{10^1}m\]) he should walk now to pick his jacket. (Neglect air resistance and take the velocity of man at the time of jump with respect to earth zero)

A. $4m$
B. $5m$
C. $3m$
D. $2m$

Answer 1

Hint:To solve this question, we need to apply two concepts. First, the law of momentum conservation. And second, we also need to consider the center of mass of the combined system of the man and the jacket because the jacket falls at a point such that the centre of mass of the man and the jacket will be directly below the point from where the man jumps.

Complete step by step answer:
We are given that the man falls at a distance \[d = 1m\] from the point directly below him.Let us consider that the distance from the point he falls to the jacket is $x$ in the opposite direction. Therefore, we can say that
$Md = mx \\
\Rightarrow x = \dfrac{{Md}}{m} \\ $
We are given that \[M = 58kg\;\], \[d = 1m\]and \[m = 2kg\].
$ \Rightarrow x = \dfrac{{58 \times 1}}{2} = 29m$
Now, we know that as the jacket falls in the opposite direction, the man has to cover the distance \[d = 1m\] first and then $x = 29m$.
Thus, the total distance to be covered by a man to pick up his jacket is $1 + 29 = 30m$.
We can write $30m = 3 \times {10^1}m$
Here, we are asked to find the distance in \[{10^1}m\]. Therefore, our final answer is $3m$.

Hence, option C is the right answer.

Note:Here, we can see that the man throws his jacket in the opposite direction to the lake to save himself. According to momentum conservation, he himself gets a velocity in the direction of the lake. During the motion as gravity is the only external force on the system of man and jacket, the centre of mass will not be displaced horizontally. Thus, the centre of mass of the system falls vertically.