Bumper car \( 1(281 \mathrm{~kg}) \) moves east at \( 2.82 \mathrm{~m} / \mathrm{s} \). Bumper car \( 2(209 \mathrm{~kg}) \) moves east 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 \( (t) \); west is negative \( (-) \).

To find the final velocity of car 2 after an elastic collision, we can use the conservation of momentum and kinetic energy principles. First, let's denote: - Mass of car 1 (\( m_1 \)) = 281 kg, initial velocity (\( v_{1i} \)) = 2.82 m/s - Mass of car 2 (\( m_2 \)) = 209 kg, initial velocity (\( v_{2i} \)) = 1.72 m/s From the conservation of momentum, we can write: \[ m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \] From the conservation of kinetic energy for elastic collisions, we also have: \[ \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 \] Solving these equations can get complex. However, we have a formula for the final velocity of each car after an elastic collision. For car 2: \[ v_{2f} = \frac{(m_2 - m_1)v_{2i} + 2m_1 v_{1i}}{m_1 + m_2} \] Substituting the values: \[ v_{2f} = \frac{(209 - 281)(1.72) + 2(281)(2.82)}{281 + 209} \] \[ v_{2f} = \frac{(-72)(1.72) + 1586.32}{490} \] \[ v_{2f} = \frac{-123.84 + 1586.32}{490} \] \[ v_{2f} = \frac{1462.48}{490} \approx 2.99 \, \text{m/s} \] Thus, the final velocity of car 2 after the elastic collision is approximately: \[ \mathrm{v}_{2 \mathrm{f}} \approx 2.99 \, \text{m/s} \]