Question

If two balls of equal momentum are moving towards each other, then find the velocity of the centre of mass:

1.

Half of the sum of the velocities of the two balls

2.

Zero

3.

Equal to the velocity of the heavy ball

4.

Can't determined

Answer 1

Correct Answer - Option 2 :

Zero

CONCEPT:

Centre of mass:

• The centre of mass of a body or system of a particle is defined as, a point at which the whole of the mass of the body or all the masses of a system of particle appeared to be concentrated.

The motion of the centre of mass:

• Let there are n particles of masses m1, m2,..., mn.
• If all the masses are moving then,

⇒ Mv = m1v1 + m2v2 + ... + mnvn

⇒ Ma = m1a1 + m2a2 + ... + mnan

⇒ \(M\vec{a}=\vec{F_1}+\vec{F_2}+...+\vec{F_n}\)

⇒ M = m1 + m2 + ... + mn

• Thus, the total mass of a system of particles times the acceleration of its centre of mass is the vector sum of all the forces acting on the system of particles.
• The internal forces contribute nothing to the motion of the centre of mass.

EXPLANATION:

Given P₁ = P, and P₂ = -P

Where P₁ = momentum of the ball 1, and P2 = momentum of the ball 2

• Since both the balls are moving in the opposite direction, therefore if the P₁ is taken as positive then P₂ will be taken negatively.
• Let m1 and m2 are the masses of ball 1 and ball 2 respectively, and v1 and v2 are the velocities of ball 1 and ball 2 respectively.

We know that,

⇒ P₁ = m₁v₁     -----(1)

⇒ P2 = m2v2     -----(2)

• We know that if a system of particles have n particles and all are moving with some velocity, then the velocity of the centre of mass is given as,

⇒ \(V=\frac{m_1v_1+m_2v_2+...+m_nv_n}{m_1+m_2+...+m_n}\)     -----(3)

Therefore if the two balls of masses m1, and m2 are moving with velocity v1, and v2 respectively, then the velocity of the center of mass is given as,

⇒ \(V=\frac{m_1v_1+m_2v_2}{m_1+m_2}\)     -----(4)

By equation 1, equation 2, and equation 4,

⇒ \(V=\frac{P_1-P_2}{m_1+m_2}\)

⇒ \(V=\frac{P-P}{m_1+m_2}\)

⇒ V = 0 m/sec

• Hence, option 2 is correct.