Center of Mass

If the resultant of all the external forces acting on a body be zero, the centre of mass of the system

Consider a two-particle system with particles having masses $m_1$ and $m_2$. If the first particle is pushed towards the centre of mass through a distance $d$, by what distance should the second particle be moved, so as to keep the centre of mass at the same position?

Which of the following points is the likely position of the centre of mass of the system shown in figure?

Two particles each of mass $1 \mathrm{g}$ are placed at a distance of $1 \mathrm{m}$ and $3 \mathrm{m}$ respectively from the origin along $\mathrm{X}$-axis. The centre of the mass of the system of the particles is

Two particles each of mass $1 \mathrm{g}$ are placed at a distance of $1 \mathrm{m}$ and $3 \mathrm{m}$ respectively from the origin along $x$-axis. The centre of the mass of the system of the particles is

Two particles $A$ and $B$, initially at rest, move towards each other under a mutual force of attraction. At the instant when the speed of $A$ is $v$ and the speed of $B$ is $2v$, the speed of the center of mass of the system is:

A projectile is launched from the origin with speed $v$ at an angle $\theta$ from the horizon. At the highest point in the trajectory, the projectile breaks into two pieces, $A$and $B$ of masses $m$ and $2m$ respectively. Immediately after the breakup, piece $A$ is at rest relative to the ground. Neglect air resistance. Which of the following sentences most accurately describes what happens next? 

In a $\mathrm{HCl}$ molecule, the separation between the nuclei of the two atoms is about $1.27 \mathrm{\AA } \left(1 \mathrm{\AA }={10}^{-10} \mathrm{m}\right)$. Given that a chlorine atom is about $35.5$ times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus, the position of centre of mass of the molecule is

Center of mass of three particles of masses $1 \mathrm{k}\mathrm{g}, 2\mathrm{ }\mathrm{k}\mathrm{g}$ and $3 \mathrm{k}\mathrm{g}$ lies at the point $(1, 2, 3)$ and center of mass of another system of particles $3 \mathrm{k}\mathrm{g}$ and $2 \mathrm{k}\mathrm{g}$ lies at the point $(-\mathrm{1,} 3,-2)$. The point at which a particle of mass $5 \mathrm{k}\mathrm{g}$ be placed, so that the center of mass of first system coincides with centre of mass of whole system is:

The center of mass of a system of two particles of masses $m_1$ and $m_2$ is at a distance $d_1$ from $m_1$ and at a distance $d_2$ from mass $m_2$, such that:

The $\mathrm{x}-\mathrm{y}$ coordinates of the center of mass of a uniform $\mathrm{L}$ shaped lamina of mass $3 \mathrm{k}\mathrm{g}$ is:

Three particles of masses $1\mathrm{ }\mathrm{k}\mathrm{g},\mathrm{ }\frac{3}{2} \mathrm{k}\mathrm{g}$ and $2 \mathrm{k}\mathrm{g}$ are located at the vertices of an equilateral triangle (see figure below). The coordinates of centre of mass of this system is located at:

A system of two particles is having masses ${\mathrm{m}}_1$ and ${\mathrm{m}}_2$. If the particle of mass ${\mathrm{m}}_1$ is pushed towards the center of mass of the particles through a distance$\mathrm{d}$, by what distance the particle if mass ${\mathrm{m}}_2$ should be moved so as to keep the center of mass of particles at the original position?

As shown in figure from a uniform rectangular sheet a triangular sheet is removed from one edge. The shift of center of mass is
 

A rope of length 30 cm is on a horizontal table with maximum length hanging from edge A of the table. The coefficient of friction between the rope and table is 0.5. The distance of centre of mass of the rope from A is

Two blocks of masses $10\mathrm{\hspace{0.17em}}\mathrm{kg}$ and $30\mathrm{\hspace{0.17em}}\mathrm{kg}$ are placed along a vertical line. If $10 \mathrm{kg}$ block is raised through a height of $7\mathrm{\hspace{0.17em}}\mathrm{cm}$, then the distance through which other mass should be moved to raise the center of mass of the system by $1\hspace{0.17em}\mathrm{cm}$ is

Two spherical bodies of masses $M$ and $5M$ and radii $R$ and $2R$, respectively, are released in free space, with initial separation between their centres equal to $12R$. If they attract each other due to gravitational force only, then the distance covered by the smaller body, just before collision is

Three identical spheres, each of mass $1$ $\mathrm{kg}$ are kept as shown in figure below, touching each other, with their centers on a straight line. If their centres are marked $P$, $Q$, $R$respectively, the distance of centre of mass of the system from $P$ is

Consider a two particle system with particles having masses $m_1$ and $m_2$. If the first particle is pushed towards the centre of mass through a distance $d$, by what distance should the second particle be moved, so as to keep the centre of mass at the same position?

A body $\mathit{\text{A}}$ of mass $\mathit{\text{M }}$while falling vertically downwards under gravity breaks into two parts; a body $\mathit{\text{B}}$ of mass $\frac{1}{3}\mathit{\text{M}}$ and body $\mathit{\text{C}}$ of mass $\frac{2}{3}\mathit{\text{M}}$. The center of mass of bodies  $\mathit{\text{B}}$ and $\mathit{\text{C}}$ taken together shifts compared to that of body $\mathit{\text{A}}$ towards