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 1 be afterwards? \[ v_{1 f}=[?] \mathrm{m} / \mathrm{s} \] Remember: East is positive ( + ): west is negative \( (-) \).

To solve for the final velocity of bumper car \( 1 \) after an elastic collision, we can use the conservation of momentum and the fact that kinetic energy is conserved in an elastic collision. First, we can use 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, positive because it's moving east) - \( m_2 = 209 \, \text{kg} \) (mass of car 2) - \( v_{2i} = -1.72 \, \text{m/s} \) (initial velocity of car 2, negative because it's moving west) For elastic collisions, we can also use the following equation to find the final velocities: \[ v_{1f} = \frac{(m_1 - m_2)v_{1i} + 2m_2v_{2i}}{m_1 + m_2} \] Plugging in the values, we have: \[ v_{1f} = \frac{(281 \, \text{kg} - 209 \, \text{kg})(2.82 \, \text{m/s}) + 2(209 \, \text{kg})(-1.72 \, \text{m/s})}{281 \, \text{kg} + 209 \, \text{kg}} \] Calculating that will give: \[ v_{1f} = \frac{(72\, \text{kg})(2.82\, \text{m/s}) + 2(209\, \text{kg})(-1.72\, \text{m/s})}{490\, \text{kg}} \] \[ = \frac{203.04 - 718.24}{490} \] \[ = \frac{-515.20}{490} \approx -1.05\, \text{m/s} \] So, the final velocity of car \( 1 \) after the collision would be approximately \( v_{1f} = -1.05\, \text{m/s} \), indicating that it is moving to the west.