You have a system of two equations involving $v_1$, $v_2$, $v_{1f}$, and $v_{2f}$. Let's solve them step by step.

The system of equations:

1. $v_1 + v_2 = v_{1f} + v_{2f}$ (Equation 1)
2. $v_1^2 + v_2^2 = v_{1f}^2 + v_{2f}^2$ (Equation 2)

Step 1: Simplify using algebraic identities

We'll subtract the two velocities in both equations.

From Equation 1: $v_1 - v_{1f} = v_{2f} - v_2$

This suggests that the difference between initial and final velocities of $v_1$ is equal to the opposite difference between initial and final velocities of $v_2$.

Step 2: Relate $v_1$ and $v_2$ using the second equation

Rewrite Equation 2 as: $v_1^2 - v_{1f}^2 = v_{2f}^2 - v_2^2$

Factor both sides: $(v_1 - v_{1f})(v_1 + v_{1f}) = (v_{2f} - v_2)(v_{2f} + v_2)$

From Step 1, we already know that $v_1 - v_{1f} = v_{2f} - v_2$, so we can substitute: $(v_{2f} - v_2)(v_1 + v_{1f}) = (v_{2f} - v_2)(v_{2f} + v_2)$

Divide both sides by $v_{2f} - v_2$ (assuming it's not zero): $v_1 + v_{1f} = v_{2f} + v_2$

This shows that the sum of initial and final velocities for $v_1$ equals the sum of final and initial velocities for $v_2$, suggesting that the velocities are exchanged.

Conclusion:

The equations describe a perfectly elastic collision, where the velocities of the two objects are swapped: $v_{1f} = v_2 \quad \text{and} \quad v_{2f} = v_1$

Would you like more details on the derivation, or any further clarifications?

Follow-up questions:

1. What are elastic and inelastic collisions, and how do they differ?
2. How does the conservation of kinetic energy apply to these equations?
3. Can this system be solved for any general form, or are there constraints?
4. How do we interpret the physical meaning of the velocity exchange?
5. How does this result apply to a real-world scenario, like billiard balls?

Tip: In physics problems involving energy, always check whether kinetic energy is conserved to determine if the collision is elastic.