This is solved in essentially the same way as in your other question [26141465] with some slight differences.
By conservation of momentum,
(281 kg) (2.82 ms⁻¹) + (209 kg) (-1.72 ms⁻¹) = (281 kg) v₁ + (209 kg) v₂
and by conservation of energy,
(281 kg) (2.82 ms⁻¹)² + (209 kg) (-1.72 ms⁻¹)² = (281 kg) v₁² + (209 kg) v₂²
The momentum equation reduces to
432.94 kg•ms⁻¹ = (281 kg) v₁ + (209 kg) v₂
energy equation reduces just as before to
2852.93 J = (281 kg) v₁² + (209 kg) v₂²
Solve for v₁ (same technique as before); you should end up with
v₁ ≈ -1.05 ms⁻¹
and solving for v₂,
v₂ ≈ 3.49 ms⁻¹
