Two identical billiard balls traveling at the same speed have a head-on collision and rebound. If the balls had twice the mass, but maintained the same size and speed, how would the rebound be different?

a. No difference.

b. At a higher speed.

c. At slower speed.

a) No difference

Explanation:

Since the billiard balls are identical, they have the same mass. Also they have the same speed

Since the angular momentum is conserved and the total energy is conserved (if we assume elastic collision)

1/2 m1 * v i1² + 1/2 m2 * v i1² = 1/2 m1 * v f1² + 1/2 m2 * v f2²

where m = mass, vi = initial velocity, vf = final velocity

since m1=m2=m, vi1=vi2=vi

1/2 m1 * v i1² + 1/2 m2 * v i1² = 1/2 m1 * v f1² + 1/2 m2 * v f2²

m * v i² = 1/2 m (v f1² + v f2²)

vi² = 1/2 (v f1² + v f2²)

since the 2 balls are indistinguishable from each other (they have identical initial mass and velocity) there is no reason for a preferential speed for one of the balls and therefore its velocities must be equal. Thus vf1=vf2=vf

therefore

v i² = 1/2 (v f1² + v f2²) = v i1² = 1/2 * 2vf² = vf²

and thus

vi = vf

in conclusion, there is no difference in speed after the rebound