Answer
(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.
