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 1 be afterwards? \[ \mathrm{v}_{1 \mathrm{~F}}=[?] \mathrm{m} / \mathrm{s} \] Remember East is positive \( (-) \) : west is negative \( (-) \).

To find the final velocity of bumper car 1 after an elastic collision, we can use the conservation of momentum and the principle of conservation of kinetic energy. Assuming both cars collide elastically and along a straight line (in the east direction), we can set up two equations. 1. **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) - \( m_2 = 209 \, \text{kg} \) (mass of car 2) - \( v_{1i} = 2.82 \, \text{m/s} \) (initial velocity of car 1) - \( v_{2i} = 1.72 \, \text{m/s} \) (initial velocity of car 2) - \( v_{1f} \) and \( v_{2f} \) are the final velocities of car 1 and car 2 after the collision. 2. **Conservation of Kinetic Energy**: \[ \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 \] We can simplify the problem using formulas for the final velocities after an elastic collision: \[ v_{1f} = \frac{(m_1 - m_2)v_{1i} + 2m_2 v_{2i}}{m_1 + m_2} \] \[ v_{2f} = \frac{(m_2 - m_1)v_{2i} + 2m_1 v_{1i}}{m_1 + m_2} \] Plugging in the numbers for \( v_{1f} \): \[ v_{1f} = \frac{(281 \, \text{kg} - 209 \, \text{kg}) \times 2.82 \, \text{m/s} + 2 \times 209 \, \text{kg} \times 1.72 \, \text{m/s}}{281 \, \text{kg} + 209 \, \text{kg}} \] \[ v_{1f} = \frac{(72 \cdot 2.82) + (717.76)}{490} \] \[ v_{1f} = \frac{203.04 + 717.76}{490} \] \[ v_{1f} \approx \frac{920.8}{490} \approx 1.88 \, \text{m/s} \] Thus, the final velocity of car 1 is: \[ \mathrm{v}_{1 \mathrm{~F}} \approx 1.88 \, \mathrm{m/s} \]