Test: Motion of Centre of Mass - JEE MCQ

The only force acting on the balls is force of gravity, so the acceleration of centre of mass is equal to g.

The internal forces have no effect on the trajectory of centre of mass, and the forces due to explosion are the internal forces. So the centre of mass will follow the same parabolic path even after explosion.

The velocity of the centre of mass will be zero. The initial velocity of the particles were zero because both A and B were at rest so momentum will be zero. and it is given that there is no external force hence no change in momentum of the centre of mass and it will remain zero.
By the law of conservation of momentum, the momentum is zero, and mass is not zero, velocity must be zero

As momentum is p = mv and a vector quantity, if both are moving in opposite direction then the net momentum will be mv –mv. The minus sign shows that it is in opposite direction.
So p(net) is 0.

If, r₁ and r₂ are position vectors of the centres of the positional vector of their centre of mass is given by R=(m₁r₁+m₂r₂)/ (m₁ + m₂)
R= (m₁r₂+m₂r₂)/m₁+m₂)
The acceleration of the centre of mass is given by:
A=d²R/dt
= [m₁ d₂r/dt+m₂ d²r/dt² ]/(m₁+m₂)
But, d²r₁/dt² and d²r₂/dt² are the accelerations of masses m1 and m2
Have the same magnitude (m₁-m₂)/(m₁+m₂) g
If we take acceleration of [m₁(m₁-m₂)/ (m₁+m₂)g – m2(m₁-m₂)/ (m₁+m₂)/]/ (m₁+m₂)
On simplifying that we get,
a=[(m₁-m₂)/ (m₁+m₂)]² g

When a fire cracker initially at rest, explodes into a number of fragments, the centre of mass remains stationary. This is analogous to the explosion of a ball while it is flying in a trajectory.

The momentum of the initial particle remains conserved. Hence the emitted particles’ COM will remain along +ve x-axis.

Dot product of two different unit vectors is 0 and dot product of two same unit vectors is 1. Cross product of two different unit vectors taken according to right hand thumb rule is the other vector. Cross product of two same unit vectors is 0.

Their collision will occur at their center of mass.
If we consider body of mass 'm' at origin & '2m' at distance "d" from 1st body on X-axis then, center of mass,
X = m₁(x₁)+m₂(x₂)/m₁+m₂
X = m(0)+2m(d)/m+2m
X = 2md/3m
X = 2d/3
Hence A is the correct answer.

We require an external force to move the centre of mass of a system of particles. The mutual force of electrical attraction is an internal force which can not affect the position of the centre of mass of the system.