Center of Mass

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 :