11. A toy yacht and a toy boat are moving toward each other on a frictionless track The yacht has a mass of 4 kg and is moving to the right at 3m/s while the boat has a mass of 2 kg and is moving to the left at 3m/s After the collision, the yacht bounces back at 1m/s a. Draw & label a picture and use that picture to solve for the final velocity of the toy boat after the collision.Be sure to include units & directions. b. Calculate the change in kinetic energy during the collision. c.Explain which type of collision is shown in this problem.

a. To solve for the final velocity of the toy boat after the collision, we can use the conservation of momentum. The total momentum before the collision must be equal to the total momentum after the collision.<br /><br />Let's denote the final velocity of the toy boat as $v_b$.<br /><br />The total momentum before the collision is given by:<br />$P_{\text{initial}} = m_{\text{yacht}} \cdot v_{\text{yacht}} + m_{\text{boat}} \cdot v_{\text{boat}}$<br /><br />Substituting the given values:<br />$P_{\text{initial}} = 4 \, \text{kg} \cdot 3 \, \text{m/s} + 2 \, \text{kg} \cdot (-3 \, \text{m/s})$<br />$P_{\text{initial}} = 12 \, \text{kg} \cdot \text{m/s} - 6 \, \text{kg} \cdot \text{m/s}$<br />$P_{\text{initial}} = 6 \, \text{kg} \cdot \text{m/s}$<br /><br />The total momentum after the collision is given by:<br />$P_{\text{final}} = m_{\text{yacht}} \cdot v_{\text{yacht}}' + m_{\text{boat}} \cdot v_{\text{boat}}'$<br /><br />Substituting the given values:<br />$P_{\text{final}} = 4 \, \text{kg} \cdot (-1 \, \text{m/s}) + 2 \, \text{kg} \cdot v_{\text{boat}}'$<br />$P_{\text{final}} = -4 \, \text{kg} \cdot \text{m/s} + 2 \, \text{kg} \cdot v_{\text{boat}}'$<br />$P_{\text{final}} = 2 \, \text{kg} \cdot v_{\text{boat}}' - 4 \, \text{kg} \cdot \text{m/s}$<br /><br />Since the total momentum before and after the collision must be equal, we can set $P_{\text{initial}}$ equal to $P_{\text{final}}$ and solve for $v_{\text{boat}}'$:<br />$6 \, \text{kg} \cdot \text{m/s} = 2 \, \text{kg} \cdot v_{\text{boat}}' - 4 \, \text{kg} \cdot \text{m/s}$<br />$2 \, \text{kg} \cdot v_{\text{boat}}' = 10 \, \text{kg} \cdot \text{m/s}$<br />$v_{\text{boat}}' = 5 \, \text{m/s}$<br /><br />Therefore, the final velocity of the toy boat after the collision is $5 \, \text{m/s}$ to the right.<br /><br />b. To calculate the change in kinetic energy during the collision, we need to find the initial and final kinetic energies and then subtract the final kinetic energy from the initial kinetic energy.<br /><br />The initial kinetic energy is given by:<br />$KE_{\text{initial}} = \frac{1}{2} m_{\text{yacht}} v_{\text{yacht}}^2 + \frac{1}{2} m_{\text{boat}} v_{\text{boat}}^2$<br /><br />Substituting the given values:<br />$KE_{\text{initial}} = \frac{1}{2} \cdot 4 \, \text{kg} \cdot (3 \, \text{m/s})^2 + \frac{1}{2} \cdot 2 \, \text{kg} \cdot (-3 \, \text{m/s})^2$<br />$KE_{\text{initial}} = \frac{1}{2} \cdot 4 \, \text{kg} \cdot 9 \, \text{m}^2/\text{s}^2 + \frac{1}{2} \cdot 2 \, \text{kg} \cdot 9 \, \text{m}^2/\text{s}^2$<br />$KE_{\text{initial}} = 18 \, \text{J} + 9 \, \text{J}$<br />$KE_{\text{initial}} = 27 \, \text{J}$<br /><br />The final kinetic energy is given by:<br />$KE_{\text{final}} = \frac{1}{2} m_{\text{yacht}} v_{\text{yacht}}'^2 + \frac{1}{2} m_{\text{boat}} v_{\text{boat