Bumper car \( 1(281 \mathrm{~kg}) \) moves east at \( 2.82 \mathrm{~m} / \mathrm{s} \). Bumper car \( 2(209 \mathrm{~kg}) \) moves west at \( 1.72 \mathrm{~m} / \mathrm{s} \). If the cars have an elastic collision, what would the velocity of car 2 be afterwards? \[ \mathrm{v}_{2 \mathrm{f}}=[?] \mathrm{m} / \mathrm{s} \] Remember: East is positive ( + ); west is negative \( (-) \).

In an elastic collision, both momentum and kinetic energy are conserved. To find the final velocity of car 2, we can use the formulas for conservation of momentum and conservation of kinetic energy. Using the conservation of momentum: \[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \] Where: - \( m_1 = 281 \, \text{kg} \) (mass of car 1) - \( v_{1i} = 2.82 \, \text{m/s} \) (initial velocity of car 1) - \( m_2 = 209 \, \text{kg} \) (mass of car 2) - \( v_{2i} = -1.72 \, \text{m/s} \) (initial velocity of car 2, negative because it's moving west) Let’s denote the final velocities as \( v_{1f} \) for car 1 and \( v_{2f} \) for car 2. For simplicity, we can substitute the known values and express \( v_{1f} \) in terms of \( v_{2f} \) using the second equation for conservation of kinetic energy: \[ \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 \] Instead of diving into extensive algebra here, let's solve it directly. Using known equations for velocities after elastic collisions: \[ v_{1f} = \frac{(m_1 - m_2)v_{1i} + 2m_2v_{2i}}{m_1 + m_2} \] \[ v_{2f} = \frac{(m_2 - m_1)v_{2i} + 2m_1v_{1i}}{m_1 + m_2} \] Substituting our values into the equation for \( v_{2f} \): \[ v_{2f} = \frac{(209 - 281)(-1.72) + 2(281)(2.82)}{281 + 209} \] Calculating gives: \[ v_{2f} = \frac{(-72)(-1.72) + 1584.96}{490} = \frac{123.84 + 1584.96}{490} = \frac{1708.8}{490} \approx 3.49 \, \text{m/s} \] Thus, the final velocity of car 2 after the collision is approximately: \[ \mathrm{v}_{2 \mathrm{f}} \approx 3.49 \, \mathrm{m/s} \]