First slide

Location of centre on mass for multiple point masses

Question

A boy of mass m stands on one end of a wooden plank of length I and mass M. The plank is floating on water. If
the boy walks from one end of the plank to the other end at a constant speed, the resulting displacement of the plank is given by

Moderate

A

$\frac{ML}{m}$
B

$\frac{mL}{M}$
C

$(\frac{mL}{M + m})$
D

$(\frac{ML}{M - m})$

Solution

. Initially, when the body stands on the plank, the momentum o f boy-plank system is zero because they are at rest. Let the boy walks with a speed v on the plank. As a result, let the speed of the plank in the opposite direction be y. Now the total momentum of the system is $mv - (M + m) V$ According to the law of consecrations of momentum $mv - (M + m) V = 0$
$∴ \frac{V}{V} = \frac{m}{(M + m)} As the distance moved is proportional to the speed, the displacement L' of the plank is givea by \begin{aligned} \frac{L^{'}}{L} = \frac{V}{V} = \frac{m}{(M + m)} \\ L^{'} = (\frac{mL}{M + m}) \end{aligned}$

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