A billiard ball traveling at 12 m/s has an elastic head-on collision with a billiard ball of equal mass that is initially at rest. The first ball is at rest after the collision.

Question

What is the speed of the second ball after the collision?

Answer 1

Since the billiard balls are of equal mass, when this situation occurs in an elastic collision all of the momentum from the first ball will be transferred to the second ball giving a final velocity of the second ball of 12 m/s.

We can also prove this using the conservation of momentum equation.

m₁v_1i + m₂v_2i = m₁v_1f + m₂v_2f

since the masses are all the same, I can cancel them out since every term is multiplied by mass.

This then simplifies to:

v_1i + v_2i = v_1f + v_2f

Plugging in our values:

12 + 0 = 0 +v_2f

This gives v_2f a velocity of 12 m/s.

Hope this helps!