
A boat of length 10 meters and mass 450 kg is floating without motion in still water. A man of mass 50 kg standing at one end of it walks to the other end of it and stops. The magnitude of the displacement of the boat in meter relative to the ground is
A. Zero
B. 1 m
C. 2 m
D. 5 m
Answer
586.2k+ views
Hint: If the net external force on a system of the particle is zero, the center of mass of the system will not accelerate. The net force on a system of particles equals the mass of the system times the acceleration of the system’s center of mass, and the system behaves as if all of its mass were located at the system’s center of mass. Where the center of mass for n number of particles is given as
\[{x_{cm}} = \dfrac{{{x_1}{m_1} + {x_2}{m_2} + {x_3}{m_3} + ..... + {x_n}{m_n}}}{{{m_1} + {m_2} + {m_3} + ..... + {m_n}}}\]
Complete step by step answer:
Length of boat \[l = 10m\]
Mass of boat\[{m_1} = 450g\]
Mass of man\[{m_2} = 50Kg\]
If we consider the boat and the man without the motion to be a system with respect to the water, then we can say \[{f_{ext}} = 0\]since the body is in rest and hence we can also say that there will be no movement in the center of mass.
We can also say that if a man moves from one side of the boat to the other then there will be no change in center of mass since there is no horizontal force on the boat.
\[m\vartriangle {x_{CM}} = 0\]
This can be written as
\[{m_1}\vartriangle {x_1} + {m_2}\vartriangle {x_2} + .......{m_n}\vartriangle {x_n} = 0\]
Let
\[{x_1} = x\]
Hence, \[{x_2} = - \left( {x - 10} \right)\]since the length of the boat is 10cm and the man has moved in the opposite direction.
Hence the magnitude of displacement will be
\[
{m_1}\vartriangle {x_1} + {m_2}\vartriangle {x_2}_n = 0 \\
450 \times x - 50 \times \left( {10 - x} \right) = 0 \\
450x - 500 + 50x = 0 \\
500x - 500 = 0 \\
\]
By solving, we get
\[
500x = 500 \\
x = 1 \\
\]
Therefore the magnitude of the displacement of the boat is 1 meter.Option B is correct.
Note: Students must note that if a system experiences no external force, then the center of mass of the system will be at rest, and if there is any external force, then the center of mass accelerates with \[F = ma\].
\[{x_{cm}} = \dfrac{{{x_1}{m_1} + {x_2}{m_2} + {x_3}{m_3} + ..... + {x_n}{m_n}}}{{{m_1} + {m_2} + {m_3} + ..... + {m_n}}}\]
Complete step by step answer:
Length of boat \[l = 10m\]
Mass of boat\[{m_1} = 450g\]
Mass of man\[{m_2} = 50Kg\]
If we consider the boat and the man without the motion to be a system with respect to the water, then we can say \[{f_{ext}} = 0\]since the body is in rest and hence we can also say that there will be no movement in the center of mass.
We can also say that if a man moves from one side of the boat to the other then there will be no change in center of mass since there is no horizontal force on the boat.
\[m\vartriangle {x_{CM}} = 0\]
This can be written as
\[{m_1}\vartriangle {x_1} + {m_2}\vartriangle {x_2} + .......{m_n}\vartriangle {x_n} = 0\]
Let
\[{x_1} = x\]
Hence, \[{x_2} = - \left( {x - 10} \right)\]since the length of the boat is 10cm and the man has moved in the opposite direction.
Hence the magnitude of displacement will be
\[
{m_1}\vartriangle {x_1} + {m_2}\vartriangle {x_2}_n = 0 \\
450 \times x - 50 \times \left( {10 - x} \right) = 0 \\
450x - 500 + 50x = 0 \\
500x - 500 = 0 \\
\]
By solving, we get
\[
500x = 500 \\
x = 1 \\
\]
Therefore the magnitude of the displacement of the boat is 1 meter.Option B is correct.
Note: Students must note that if a system experiences no external force, then the center of mass of the system will be at rest, and if there is any external force, then the center of mass accelerates with \[F = ma\].
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