Final answer:
The horizontal component of the billiard ball's velocity after the collision is 4.063 m/s.
Step-by-step explanation:
The collision between the cue ball and the billiard ball can be treated as an elastic collision since no external forces act on the system. In an elastic collision, both the momentum and kinetic energy are conserved.
Using the conservation of momentum, we can write the equation:
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
Where m₁ is the mass of the cue ball, v₁ is the initial velocity of the cue ball, m₂ is the mass of the billiard ball, v₂ is the initial velocity of the billiard ball, v₁' is the final velocity of the cue ball, and v₂' is the final velocity of the billiard ball.
Since the billiard ball is initially at rest, its initial velocity v₂ is 0 m/s. Plugging in the given values, we have:
0.355 kg * 6.5 m/s + 0.57 kg * 0 m/s = 0.355 kg * v₁' + 0.57 kg * v₂'
From this equation, we can solve for v₂' which is the horizontal component of the billiard ball's velocity after the collision.
Let's calculate:
v₂' = (0.355 kg * 6.5 m/s) / 0.57 kg = 4.063 m/s
Therefore, the horizontal component of the billiard ball's velocity after the collision is 4.063 m/s.