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?

In an elastic collision, both momentum and kinetic energy are conserved. You can use the formulas for conservation of momentum and kinetic energy to solve for the velocities after the collision. The equations are: 1. Momentum: \( m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f} \) 2. Kinetic Energy: \( \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 = \frac{1}{2} m_1 v_{1f}^2 + \frac{1}{2} m_2 v_{2f}^2 \) Where: - \( m_1 \) and \( m_2 \) are the masses of the cars, - \( v_{1i} \) and \( v_{2i} \) are the initial velocities, - \( v_{1f} \) and \( v_{2f} \) are the final velocities after collision. Given: - \( m_1 = 281 \, \text{kg} \), \( v_{1i} = 2.82 \, \text{m/s} \) - \( m_2 = 209 \, \text{kg} \), \( v_{2i} = 1.72 \, \text{m/s} \) Now let's solve for \( v_{1f} \): Set up the momentum equation: \[ 281 \times 2.82 + 209 \times 1.72 = 281 v_{1f} + 209 v_{2f} \] And set up the kinetic energy equation: \[ \frac{1}{2} \times 281 \times (2.82)^2 + \frac{1}{2} \times 209 \times (1.72)^2 = \frac{1}{2} \times 281 v_{1f}^2 + \frac{1}{2} \times 209 v_{2f}^2 \] Solving this system of equations typically involves substitution because you have two equations with two unknowns (the final velocities). If both speeds are positive post-collision (which they should be), car \( 1 \) would end up with a velocity of approximately \( 2.10 \, \text{m/s} \) after the collision, depending on how you simplify and solve for the second velocity.