Two cars collide at an icy intersection and stick together afterward. The 1st car has a mass of 1400kg and was approaching at 7.00m/s due south. The 2nd car has a mass of 900kg and was approaching at 23.0 m/s due west. (a) Calculate the final velocity (magnitude in m/s and direction in degrees counterdockwise from the west) of the cars(B) how much kinetic energy in J is lost in the collision?

[tex]\begin{gathered} p_{iy}=p_{fy} \\ m_1v_{iy}=(m_1+m_2)v_{fy} \\ (1400\text{ kg\rparen\lparen7 m/s\rparen= \lparen1400 kg+900 kg\rparen v}_{fy} \end{gathered}[/tex]

Solving for vfy, we get:

[tex]\begin{gathered} 9800\text{ kg m/s = 2300 kg v}_{fy} \\ \\ v_{fy}=\frac{9800\text{ kg m/s}}{2300\text{ kg}} \\ \\ v_{fy}=4.26\text{ m/s due to south} \end{gathered}[/tex]

For the horizontal direction, we get:

[tex]\begin{gathered} p_{ix}=p_{fx} \\ m_2v_{ix}=(m_1+m_2)v_{fx} \\ (900\text{ kg\rparen\lparen23 m/s\rparen = \lparen1400 kg + 900 kg\rparen}v_{fx} \\ 20700\text{ kg m/s = \lparen2300 kg\rparen}v_{fy} \\ \\ v_{fx}=\frac{20700\text{ kg m/s}}{2300\text{ kg}} \\ \\ v_{fx}=9\text{ m/s} \end{gathered}[/tex]

Now, we can calculate the magnitude and direction of the final velocity as follows

[tex]\begin{gathered} v_f=\sqrt{v_{fx}^2+v_{fy}^2} \\ \\ v_f=\sqrt{4.26^2+9^2} \\ v_f=9.96\text{ m/s} \\ \\ \\ \theta=\tan^{-1}(\frac{v_{fy}}{v_{fx}}) \\ \\ \theta=\tan^{-1}(\frac{9}{4.26}) \\ \\ \theta=64.67° \end{gathered}[/tex]

Part (b)

Then, the loss in kinetic energy can be calculated as:

[tex]\begin{gathered} \text{ Loss in KE = }KE_i-KE_f \\ \text{ Loss in KE = \lparen}\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2)-(\frac{1}{2}(m_1+m_2)v_f^2) \\ \\ \text{ Loss in KE = \lparen}\frac{1}{2}(1400)(7)^2+\frac{1}{2}(900)(23)^2)-(\frac{1}{2}(1400+900)(9.96)^2) \\ \\ \text{ Loss in KE = 272350 J - 114081.84 J} \\ \text{ Loss in KE = 158,268.16 J} \end{gathered}[/tex]

Therefore, the loss in kinetic energy was 158,268.16 Joules.