Center of Mass objective question answers test

We know that the resultant of all forces, on any system of particles, is zero. Therefore their centre of mass does not depend upon the forces acting on the particles.

We have Mass of the first body (M₁) = 5Kg and mass of the second body (M₂) = 20Kg. We know that as both the bodies are moving towards each other, without any external forces, therefore their centre of mass remains stationary.

We know that when the man climbs up to the balloon using a rope, then no external force is involved.
Therefore centre of mass of the ‘man and balloon’ system will remain stationary.

We have Mass of body A (m_A) = M; Mass of part B (m_B) = M/3 and mass of part C (m_C) = 2M/3.
We know that as the body A breaks into two parts while falling vertically downwards, under gravity, therefore there is no external force acting on the bodies. Thus centre of mass of bodies B and C taken together does not shift.

We have, Mass of first particles (m₁) = m; Mass of second particle (m₂) = m; Velocity of first particle (v₁) = 2v and velocity of second particle (v₂) = -v (Minus sing due to opposite direction).
We know that velocity of their centre of mass (v_cm) = (m₁ v₁ + m₂ v₂)/(m₁ + m₂) = [(m × 2v) + m(-v)]/(m+m) = (2mv – mv)/ 2m = mv/2m = v/2.

Given: Mass of first particle = m₁; Mass of the second particle = m₂; Distance through which the first particle is pushed toward which the first particle is pushed towards mass centre (d₁) = d.
Let distance of first particle from mass centre = a and distance of second particle form mass centre = -b (Minus sign due to opposite side of second particle from mass centre)
We know that sum of moments of masses about mass centre is zero.
Therefore initial sum of moments of mass = m₁a + m₂(-b) = 0 or m₁a – m₂(b) = 0
Similarly, final sum of moments after the mass m₁ is pushed = m₁ (a-d₁) + m₂(-b-d₂) = 0
d₂ = m₁d / m₂
(Where d₂ is the distance moved by second particle towards the mass centre and negative sign indicates direction of movement of second particle).

Given that Mass of the first block (m₁) = 10 Kg; Mass of second block (m₂) = 4 Kg; Velocity of first block due to impulse (v₁) = 14 m-s^-1 and velocity of second block (v₂) = 0 (because the impulsive force acts only on the first block).
We know that the velocity of the centre of mass (v_cm) = (m₁ v₁ + m₂ v₂)/(m₁+m₂) = [(10×14) + (4×0)]/(10+4) = 140/14 = 10 m-s^-1.

Given that Mass of the first particle (m₁) = m; Mass of second particle (m₂) = m; Acceleration of first particle (a₁) = a and acceleration of second particle (a₂) = 0 (because it is at rest.)
We know that acceleration of the centre of mass (a) = (m₁ v₁ + m₂ v₂)/(m₁+m₂) = (m×a+m×0)/(m+m) = ma/2m = a/2.

We know that as all the metal balls are identical in mass and radius, therefore the center of mass of the system will be at the point of intersection of their medians.