
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 walked to the other end of the plank. If the mass of the plank is \[\dfrac{M}{3}\], the distance that the man moves relative to the ground is-
A. \[\dfrac{{3L}}{4}\]
B. \[\dfrac{L}{4}\]
C. \[\dfrac{{4L}}{5}\]
D. \[\dfrac{L}{3}\]
Answer
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Hint:Centre of mass is an imaginary point where the whole mass of an object may be concentrated. We can solve this problem by using the concept of centre of mass and centre of momentum by considering the initial and final positions.
Formula Used:
\[{X_{CM}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2}}}{{{m_1} + {m_2}}}\]
Here ${m_1}$ and $ {m_2}$ are the mass of the body and ${X_{CM}}$ is the centre of mass.
Complete step by step answer:
Mass of man=\[M\]
Mass of plank= \[M/3\]
Let man is standing at the starting of plank, at \[X = 0\], end of plank, at \[X = L\]
Centre of mass of man is at \[X = 0\]
Centre of mass of man is at \[X = L/2\]
Position of centre of mass for the system
\[ \dfrac{{M \times 0 + (M/3) \times (L/2)}}{{M + (M/3)}} = \dfrac{L}{8}\]
Now, consider man is walking from starting to end point, travels a distance $Z$
Now, the centre of mass of man is at $x = z$
Now, the centre of mass of plank will shift $(L-Z)$ towards left,
Centre of mass of plank is at
\[\left[ {\dfrac{L}{2} - (L - z)} \right] = \left( {Z - \dfrac{L}{2}} \right)\]
We know there is no external force, centre of mass will not change its position.
\[ \dfrac{{Mz + (Mz/3) - ML/6 \times }}{{(4M/3)}} = \dfrac{L}{8}\]
After solving this equation,
We get-\[\dfrac{L}{8} = z - \dfrac{L}{8}\]
\[\therefore z = \dfrac{L}{4}\]
So, option B is correct.
Note: Centre of mass is a distribution of mass in space where the weighted relative position of the distributed mass equals to zero. If no external force is acting then the centre of mass of a body will not change.When we are studying the dynamics of the motion of the system of a particle as a whole, then we need not bother about the dynamics of individual particles of the system. But only focus on the dynamic of a unique point corresponding to that system.
Formula Used:
\[{X_{CM}} = \dfrac{{{m_1}{x_1} + {m_2}{x_2}}}{{{m_1} + {m_2}}}\]
Here ${m_1}$ and $ {m_2}$ are the mass of the body and ${X_{CM}}$ is the centre of mass.
Complete step by step answer:
Mass of man=\[M\]
Mass of plank= \[M/3\]
Let man is standing at the starting of plank, at \[X = 0\], end of plank, at \[X = L\]
Centre of mass of man is at \[X = 0\]
Centre of mass of man is at \[X = L/2\]
Position of centre of mass for the system
\[ \dfrac{{M \times 0 + (M/3) \times (L/2)}}{{M + (M/3)}} = \dfrac{L}{8}\]
Now, consider man is walking from starting to end point, travels a distance $Z$
Now, the centre of mass of man is at $x = z$
Now, the centre of mass of plank will shift $(L-Z)$ towards left,
Centre of mass of plank is at
\[\left[ {\dfrac{L}{2} - (L - z)} \right] = \left( {Z - \dfrac{L}{2}} \right)\]
We know there is no external force, centre of mass will not change its position.
\[ \dfrac{{Mz + (Mz/3) - ML/6 \times }}{{(4M/3)}} = \dfrac{L}{8}\]
After solving this equation,
We get-\[\dfrac{L}{8} = z - \dfrac{L}{8}\]
\[\therefore z = \dfrac{L}{4}\]
So, option B is correct.
Note: Centre of mass is a distribution of mass in space where the weighted relative position of the distributed mass equals to zero. If no external force is acting then the centre of mass of a body will not change.When we are studying the dynamics of the motion of the system of a particle as a whole, then we need not bother about the dynamics of individual particles of the system. But only focus on the dynamic of a unique point corresponding to that system.
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