# Chapter 8 - Momentum, Impulse, and Collision - Problems - Exercises - Page 264: 8.22

(a) The speed of the lighter car is 0.409 m/s to the right. (b) The change in kinetic energy during the collision is -2670 J.

#### Work Step by Step

(a) Let $m_A$ be the mass of the heavier car. Let $m_B$ be the mass of the lighter car. $m_Av_{A2}+m_Bv_{B2}=m_Av_{A1}+m_Bv_{B1}$ $m_Bv_{B2}=m_Av_{A1}+m_Bv_{B1}-m_Av_{A2}$ $v_{B2}=\frac{m_Av_{A1}+m_Bv_{B1}-m_Av_{A2}}{m_B}$ $v_{B2}=\frac{(1750~kg)(1.50~m/s)+(1450~kg)(-1.10~m/s)-(1750~kg)(0.250~m/s)}{1450~kg}$ $v_{B2} = 0.409~m/s$ The speed of the lighter car is 0.409 m/s to the right. (b) $K_1 = \frac{1}{2}m_Av_{A1}^2+\frac{1}{2}m_Bv_{B1}^2$ $K_1 = \frac{1}{2}(1750~kg)(1.50~m/s)^2+\frac{1}{2}(1450~kg)(1.10~m/s)^2$ $K_1 = 2846~J$ $K_2 = \frac{1}{2}m_Av_{A2}^2+\frac{1}{2}m_Bv_{B2}^2$ $K_2 = \frac{1}{2}(1750~kg)(0.25~m/s)^2+\frac{1}{2}(1450~kg)(0.409~m/s)^2$ $K_2 = 176~J$ We can find the change in kinetic energy during the collision. $\Delta K = K_2-K_1$ $\Delta K = 176~J-2846~J$ $\Delta K = -2670~J$ The change in kinetic energy during the collision is -2670 J.

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