Two carts collide on a frictionless air track. Cart 1, with mass 0.350 kg and a compressed spring, moves at 2.00 m/s and recoils at -4.00 m/s after releasing its spring's energy. Cart 2, with mass 0.500 kg, starts at -0.500 m/s. What is cart 2's final velocity?

1. Momentum conservation gives m1v1 + m2v2 equal to m1v1 prime + m2v2 prime. 2. Substituting known values, (0.350)(2.00) + (0.500)(-0.500) equal to (0.350)(-4.00) + (0.500)v2 prime. 3. The left side gives 0.700 minus 0.250, which is 0.450, and the first term on the right gives -1.40. 4. Solving for v2 prime gives (0.450 minus (-1.40)) divided by 0.500, which is 1.85 divided by 0.500, or 3.70 m/s. 5. This is an inelastic collision, but one where the spring adds internal kinetic energy rather than removing it, which is why cart 2 ends up moving so much faster than either cart started. _Source: OpenStax College Physics (CC BY 4.0), Ch 8 "Linear Momentum and Collisions", section 8.5 Inelastic Collisions in One Dimension_