A man of mass M stands at one end of a plank of length L which lies at rest on a frictionless surface.The man walks to the other end of the plank. If the mass of the plank is M3, the distance that the manmoves relative to the ground is

A
$\frac{3 L}{4}$
B
$\frac{L}{4}$
C
$\frac{4 L}{5}$
D
$\frac{L}{3}$

Solution:

Let plank moves x distance in opposite direction. Then, displacement of man relative to ground will be (L $-$ x). Applying law of conservation of momentum,

$m_{R} \frac{x_{R}}{t} = m_{L} \frac{x_{L}}{t}$ or $M (L - x) = \frac{M}{3} x$

Solving this equation, we get $x = \frac{3 L}{4}$

$∴$ Displacement of man relative to ground $= L - \frac{3 L}{4} = \frac{L}{4}$

A man of mass M stands at one end of a plank of length L which lies at rest on a frictionless surface.The man walks to the other end of the plank. If the mass of the plank is $\frac{M}{3},$ the distance that the man
moves relative to the ground is

Solution:

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