Head-on Elastic Collision

Let two balls A and B having masses m_A and m_B respectively collide head-on. Let v_Ai and v_Bi be the velocities of the two balls before collision and v_Af and v_Bf be their velocities after the collision.

Now applying the laws of conservation of momentum and kinetic energy,

For conservation of momentum

m_Av_Ai + m_Av_Bi = m_Av_Af + m_Bv_Bf

For conservation of kinetic energy

½m_Av_Ai² + ½m_Bv_Bi² = ½m_Av_Af² + ½m_Bv_Bf²

There are only two unknown quantities (velocities of the balls after collision) and there are two independent equations.

Case I: Collision of two particles of identical masses

Suppose that the two balls colliding with each other are identical i.e. m_A = m_B = m.

Then, the second term in the equations will drop out resulting in

v_Af = v_Bi

v_Bf = v_Ai

If two identical balls collide head-on, their velocities after collision get interchanged. After collision, the velocity of A is same as that of B before collision and the velocity of B is same as that of A before collision.

Case II: Collision of two particles of unequal masses

Assume that m_B is very large compared to m_A and particle B is initially at rest.

m_B >> m_A and v_Bi = 0

Then, the mass m_A can be neglected in comparison to m_B.

v_Af ≈ –v_Ai

v_Bf ≈ 0

After collision, the heavy particle continues to be at rest. The light particle returns back on its path with a velocity equal to its the initial velocity. This is what happens when a child hits a wall with a ball.