Question

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.

Answer 1

Answer:

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