A billiard ball traveling at 8.0 m/s has an elastic collision with a second billiard ball of equal mass that is at rest. After

Question

the collision, the first ball is at rest. What is the velocity of the second ball after the collision?

Answer 1

v1=8m/s, v2=0 , m1=m2->m , v=?
m1*v1+m2*v2=(m1+m2)*v
m*8+m*0=(m+m)*v
8m=2m*v
8m/2m=v
4=v

v=4m/s

Answer 2

Anything that hits something of the same mass will transfer the velocity to the other object while coming to a rest itself so the velocity of 8 would be given to the other billboard ball. additionally within the question itself elastic collision is mention and elastic collision means a collision where both the momentum and kinetic energy is conserved meaning that the momentum doesn't change