Question

A white pool ball (m₁ = 0.3 kg) moving at a speed of v_o1 = +3 m/s collides head-on with a red pool ball (m₂ = 0.4 kg) initially moving at a speed of v_o2 = - 2 m/s. Neglecting friction and assuming the collision is perfectly elastic, what is the velocity of each ball after the collision?

Answer 1

The sketches below represent the system before and after the collision

Since momentum is conserved in this collision, we can set the total momentum of the system before the collision equal to the total momentum after the collision:

But here we have two unknowns, v_f1 and v_f2, and only one equation. We can generate another equation containing the same variables if we remember that both momentum and kinetic energy are conserved in an elastic collision. 

Substituting the equation solved above for v_f1 into the equation for conservation of kinetic energy, we get one equation with only one unknown, namely v_f2. Substituting the known values into this equation and solving for v_f2 gives v_f2 = +2.3 m/s. Solving for v_f1 in the equation above gives v_f1 = - 2.7 m/s.

It can be shown that if we solve the equations for conservation of momentum and conservation of kinetic energy simultaneously, it always turns out that the relative speeds of the two masses remains the same (except for a negative sign) before and after a perfectly elastic collision regardless of the masses of the two objects. That is, 

v₀₁ − v₀₂ = − (v_f1 − v_f2) 

Collisions in Two Dimensions