Question

Two billiard balls are at rest and touching each other on a pool table. The cue ball travels at 3.8 m/s along the line of symmetry between these balls and strikes them simultaneously. If the collision is elastic, what is the velocity of the three balls after the collision?

a) 3.8 m/s each for the cue ball and one of the other balls, 0 m/s for the third ball
b) 1.9 m/s each for all three balls
c) 0 m/s for all three balls
d) 3.8 m/s for the cue ball, 0 m/s for the other two balls

Answer 1

In an elastic collision between two billiard balls and a cue ball, the velocity of all three balls after the collision will be 3.8 m/s each for the cue ball and one of the other balls, and 0 m/s for the third ball.

Step-by-step explanation:

In an elastic collision, the total kinetic energy of the system is conserved. In this case, the cue ball strikes the two other balls simultaneously, so the momentum is also conserved. In an elastic collision between two billiard balls and a cue ball, the velocity of all three balls after the collision will be 3.8 m/s each for the cue ball and one of the other balls, and 0 m/s for the third ball.

Since the two balls are at rest, their velocities after the collision will be the same as the initial velocity of the cue ball, which is 3.8 m/s. Therefore, the correct answer is a) 3.8 m/s each for the cue ball and one of the other balls, 0 m/s for the third ball.