Let m₁ and v₁ denote the mass and initial velocity of the first ball, and m₂ and v₂ the same quantities for the second ball. Momentum is conserved throughout the collision, so
m₁ v₁ + m₂ v₂ = m₁ v₁' + m₂ v₂'
where v₁' and v₂' are the balls' respective velocities after the collision.
Kinetic energy is also conserved, so
1/2 m₁ v₁² + 1/2 m₂ v₂² = 1/2 m₁ (v₁')² + 1/2 m₂ (v₂')²
or
m₁ v₁² + m₂ v₂² = m₁ (v₁')² + m₂ (v₂')²
From the momentum equation, we have
(0.600 kg) (9.00 m/s) + (0.300 kg) (-8.00 m/s) = (0.600 kg) v₁' + (0.300 kg) v₂'
which simplifies to
10.0 m/s = 2 v₁' + v₂'
so that
v₂' = 10.0 m/s - 2 v₁'
From the energy equation, we have
(0.600 kg) (9.00 m/s)² + (0.300 kg) (-8.00 m/s)² = (0.600 kg) (v₁')² + (0.300 kg) (v₂')²
which simplifies to
67.8 J = (0.600 kg) (v₁')² + (0.300 kg) (v₂')²
or
226 m²/s² = 2 (v₁')² + (v₂')²
Substituting v₂' yields
226 m²/s² = 2 (v₁')² + (10.0 m/s - 2 v₁')²
which simplifies to
3 (v₁')² - (20.0 m/s) v₁' - 63.0 m²/s² = 0
Solving for v₁' using the quadratic formula gives two solutions,
v₁' ≈ -2.33 m/s or v₁' = 9.00 m/s
but the second solution corresponds to the initial conditions, so we omit that one.
Then the second ball has velocity
v₂' = 10.0 m/s - 2 (-2.33 m/s)
v₂' ≈ 14.7 m/s
