How to Calculate the Speed of Billiard Balls After an Elastic Collision

Question:

Two billiard balls collide in an elastic head-on collision. If the first ball, initially traveling at 3.4 m/s, comes to rest after the collision, what is the speed of the second ball?

Answer:

The speed of the second ball after the elastic collision is 3.4 m/s.

When two billiard balls collide in an elastic head-on collision, the total momentum of the system remains constant. In this scenario, the first ball is initially traveling at 3.4 m/s and comes to rest after the collision. As a result, the speed of the second ball will be equal to the initial speed of the first ball.

By conserving momentum, we can calculate the speed of the second ball after the collision. The initial momentum of the system is equal to the final momentum:

Initial momentum: m * 3.4 m/s + m * 0 m/s = 0

Final momentum: 2m * v, where v is the final velocity of the second ball

Setting the initial momentum equal to the final momentum, we get:

m * 3.4 = 2m * v

v = 3.4 m/s

Therefore, the speed of the second ball after the elastic collision is also 3.4 m/s, as both momentum and kinetic energy are conserved in elastic collisions.