A billiard ball traveling at 3.8 m/s has an elastic head-on collision with a billiard ball of equal mass that is initially at rest. The first ball is at rest after the collision. What is the speed of the second ball after the collision?

To solve this problem, we can use the principle of conservation of momentum. In an elastic collision, the total momentum of the system before the collision is equal to the total momentum after the collision.

Let's denote the masses of the two billiard balls as m1 and m2, and their initial velocities as v1 and v2, respectively. According to the problem, m1 = m2, v1 = 3.8 m/s, and v2 = 0 m/s.

The conservation of momentum equation is:

m1 * v1 + m2 * v2 (before collision) = m1 * v1' + m2 * v2' (after collision)

Since m1 = m2, we can simplify the equation:

m1 * v1 + m1 * v2 (before collision) = m1 * v1' + m1 * v2' (after collision)

Now, plug in the known values into the equation:

m1 * 3.8 m/s + m1 * 0 m/s = m1 * v1' + m1 * v2'

Since m1 appears on both sides of the equation, we can cancel them out:

3.8 m/s + 0 m/s = v1' + v2'

After simplifying, we get:

3.8 m/s = v1' + v2'

Since v1' is given as 0 m/s (the first ball comes to rest after the collision), we can substitute it into the equation:

3.8 m/s = 0 m/s + v2'

Therefore, the speed of the second ball after the collision is 3.8 m/s.

The only way to conserve both kinetic energy and momentum for the second ball to acquire the velocity of the first ball, if the first ball stops.

That is what happens in a head-on elastic collision.