Question

On a smooth, frictionless table, a billiard ball of velocity v is moving toward two other aligned billiard balls in contact. What will be the velocity of each ball after impact? Assume that all balls have the same mass and that the collisions are elastic. Ignore any rotation of the balls (Hint: Treat this as two successive collisions.)

Answer 1

Successive collision implies that the balls react instantly when the first ball collides with them. Thus we can apply momentum conservation principle: the momentum of the first ball

$$mv$$

before the interaction transforms to very small motion of the second ball. This ball gives its momentum to the third ball. Since the balls have equal masses, for the first collision of first two balls we have

$$mv1+0=0+mv2, v1=v2,$$

which means that the first ball stops and the second starts moving towards the third (but its motion is very short).

For the second collision between the second and third balls we have

$$mv2+0=0+mv3, v2=v3,$$

which means that the second ball stops and the third starts to move.

As a result we have balls 1 and 2 resting in contact with each other, and the ball 3 moving with speed v. This effect can be easily seen on Newton's cradle.