(D) The displacement of the center of mass is given by the formula: $\Delta X_{\text{COM}} = \frac{m_1 \Delta x_1 + m_2 \Delta x_2}{m_1 + m_2}$.Since

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(D) The displacement of the center of mass is given by the formula: $\Delta X_{\text{COM}} = \frac{m_1 \Delta x_1 + m_2 \Delta x_2}{m_1 + m_2}$.Since the center of mass must remain at its original position,the net displacement of the center of mass is zero,i.e.,$\Delta X_{\text{COM}} = 0$.Let the displacement of $m_1$ be $\Delta x_1 = +2 \text{ cm}$ (towards the center of mass) and the displacement of $m_2$ be $\Delta x_2 = -x$ (towards the center of mass).Substituting the values into the equation:$0 = \frac{3 \times 2 + 2 \times (-x)}{3 + 2}$$0 = \frac{6 - 2x}{5}$$6 - 2x = 0$$2x = 6$$x = 3 \text{ cm}$.Therefore,the particle of mass $m_2$ must move by $3 \text{ cm}$ towards the center of mass.

Class 11 Physics System of Particles and Rotational Motion Motion of Centre of Mass

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In a system,two particles of masses $m_1 = 3 \text{ kg}$ and $m_2 = 2 \text{ kg}$ are placed at a certain distance from each other. The particle of mass $m_1$ is moved towards the center of mass of the system through a distance of $2 \text{ cm}$. In order to keep the center of mass of the system at the original position,the particle of mass $m_2$ should move towards the center of mass by a distance of . . . . . . $\text{cm}$.

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$5$
B
$4$
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$6$
D
$3$

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System of Particles and Rotational Motion

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Motion of Centre of Mass

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$A$ ladder $AP$ of length $5\, m$ is inclined to a vertical wall and is slipping over a horizontal surface with a velocity of $2\, m/s$. When the end $A$ is at a distance of $3\, m$ from the wall,what is the velocity of the center of mass $(C.M.)$ of the ladder at this moment (in $, m/s$)?

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Three men $A, B$ and $C$ of masses $40\,kg, 50\,kg$ and $60\,kg$ are standing on a plank of mass $90\,kg$,which is kept on a smooth horizontal plane. If $A$ and $C$ exchange their positions,then mass $B$ will shift:

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$A$ 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 $3M$,the distance that the man moves relative to the ground is:

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$A$ small ball $B$ of mass $m$ is suspended with a light inelastic string of length $L$ from a block $A$ of the same mass $m$,which can move on a smooth horizontal surface as shown in the figure. The ball is displaced by an angle $\theta$ from the equilibrium position and then released. The displacement of the block when the ball reaches the equilibrium position is

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$A$ balloon of mass $M$ with a light rope and a monkey of mass $m$ are at rest in mid-air. If the monkey climbs up the rope and reaches the top of the rope,the distance by which the balloon descends will be (Total length of the rope is $L$):

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$A$ ladder $AP$ of length $5\, m$ is inclined to a vertical wall and is slipping over a horizontal surface with a velocity of $2\, m/s$. When the end $A$ is at a distance of $3\, m$ from the wall,what is the velocity of the center of mass $(C.M.)$ of the ladder at this moment (in $, m/s$)?

Three men $A, B$ and $C$ of masses $40\,kg, 50\,kg$ and $60\,kg$ are standing on a plank of mass $90\,kg$,which is kept on a smooth horizontal plane. If $A$ and $C$ exchange their positions,then mass $B$ will shift:

$A$ 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 $3M$,the distance that the man moves relative to the ground is:

$A$ balloon of mass $M$ with a light rope and a monkey of mass $m$ are at rest in mid-air. If the monkey climbs up the rope and reaches the top of the rope,the distance by which the balloon descends will be (Total length of the rope is $L$):

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