A pair of 0.20 kg billiard balls make an elastic collision. Before the collision, the 4-ball was moving 0.50 m/s to the right, and the 8-ball was moving 1.0 m/s to the left. After the collision, the 4-ball is now moving at 1.0 m/s to the left. What is the velocity (magnitude and direction) of the 8-ball after the collision?

To solve this problem, we can use the conservation of momentum and the fact that the collision is elastic to find the velocity of the 8-ball after the collision.

We are given (assuming to the right is positive):

m = 0.20 kg
v₀₄ = 0.50 m/s
v₀₈ = -1.0 m/s
v_f₄ = -1.0 m/s

The conservation of momentum before and after the collision can be expressed as:

[tex]mv_{0_{4}}+mv_{0_{8}}=mv_{f_{4}}+mv_{f_{8}}[/tex]

Given the masses are equal, we can simplify the momentum equation to:

[tex]\Longrightarrow v_{0_{4}}+v_{0_{8}}=v_{f_{4}}+v_{f_{8}}[/tex]

Plugging in the known values:

[tex]\Longrightarrow 0.50 \text{ m/s}-1.0 \text{ m/s}=-1.0 \text{ m/s}+v_{f_{8}}\\\\\\\\\Longrightarrow -0.50 \text{ m/s}=-1.0 \text{ m/s}+v_{f_{8}}\\\\\\\\\therefore v_{f_8}=\boxed{0.50 \text{ m/s}}[/tex]

Thus, the final velocity of the 8-ball after the collision is 0.50 m/s to the right.