Center of Mass

A block of mass 1.9 kg is at rest at the edge of a table, of height 1 m. A bullet of mass 0.1 kg collides with the block and sticks to it. If the velocity of the bullet is 20 m/s in the horizontal direction just before the collision then the kinetic energy just before the combined system strikes the floor, is [Take g = 10 m/s² . Assume there is no rotational motion and loss of energy after the collision is negligable.]

Two bodies of mass $4 \mathrm{~g}$ and $25 \mathrm{~g}$ are moving with equal kinetic energies. The ratio of magnitude of their linear momentum is :

A circular disc of radius $R$ is removed from a bigger circular disc of radius $2R$ such that the circumferences of the discs coincide. The center of mass of the new disc is $\alpha R$ form the center of the bigger disc. The value of $\alpha$ is

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

Distance of the center of mass of a solid uniform cone from its vertex is $z{}_0$. If the radius of its base is $R$ and its height is $h$ then $z{}_0$ is equal to :

A rod of length L has non-uniform linear mass density given by $\rho$(x) = $a + b{\left( {{x \over L}} \right)^2}$ , where a and b are constants and 0 $\le$ x $\le$ L. The value of x for the centre of mass of the rod is at :

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

Two blocks of masses 10 kg and 30 kg are placed on the same straight line with coordinates (0, 0) cm and (x, 0) cm respectively. The block of 10 kg is moved on the same line through a distance of 6 cm towards the other block. The distance through which the block of 30 kg must be moved to keep the position of centre of mass of the system unchanged is :

Two bodies of mass $1 \mathrm{~kg}$ and $3 \mathrm{~kg}$ have position vectors $\hat{i}+2 \hat{j}+\hat{k}$ and $-3 \hat{i}-2 \hat{j}+\hat{k}$ respectively. The magnitude of position vector of centre of mass of this system will be similar to the magnitude of vector :

A block of mass $0.50$ $kg$ is moving with a speed of $2.00$ $m{s^{ - 1}}$ on a smooth surface. It strike another mass of $1.0$ $kg$ and then they move together as a single body. The energy loss during the collision is :

Statement - 1 : Two particles moving in the same direction do not lose all their energy in a completely inelastic collision. Statement - 2 : Principle of conservation of momentum holds true for all kinds of collisions.

This question has statement ${\rm I}$ and statement ${\rm I}$${\rm I}$. Of the four choices given after the statements, choose the one that best describes the two statements. Statement - ${\rm I}$: A point particle of mass $m$ moving with speed $\upsilon$ collides with stationary point particle of mass $M.$ If the maximum energy loss possible is given as $f\left( {{1 \over 2}m{v^2}} \right)$, then $f = \left( {{m \over {M + m}}} \right).$ Statement - ${\rm II}$: Maximum energy loss occurs when the particles get stuck together as a result of the collision.

A particle of mass $m$ moving in the $x$ direction with speed $2v$ is hit by another particle of mass $2m$ moving in the $y$ direction with speed $v.$ If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to:

A neutron moving with a speed ‘v’ makes a head on collision with a stationary hydrogen atom in ground state. The minimum kinetic energy of the neutron for which inelastic collision will take place is :

The mass of a hydrogen molecule is 3.32 $\times$ 10⁻²⁷ kg. If 10²³ hydrogen molecules strike, per second, a fixed wall of area 2 cm² at an angle of 45^o to the normal, and rebound elastically with a speed of 10³ m/s, then the pressure on the wall is nearly:

A proton of mass m collides elastically with a particle of unknown mass at rest. After the collision, the proton and the unknown particle are seen moving at an angle of 90^o with respect to each other. The mass of unknown particle is :

In a collinear collision, a particle with an initial speed v₀ strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is :

If 10²² gas molecules each of mass 10^–26 kg collide with a surface (perpendicular to it) elastically per second over an area 1 m² with a speed 10⁴ m/s, the pressure exerted by the gas molecules will be of the order of :

A particle of mass 'm' is moving with speed '2v' and collides with a mass '2m' moving with speed 'v' in the same direction. After collision, the first mass is stopped completely while the second one splits into two particles each of mass 'm', which move at angle 45° with respect to the origianl direction. The speed of each of the moving particle will be :-