Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1200 kg and is approaching at 8.00m/s due south. The second car has a mass of 850 kg and is approaching at 17.0m/s due west. (a) Calculate the final velocity (magnitude and direction) of the cars. (b) How much kinetic energy is lost in the collision

Since the collision is an inelastic collision, kinetic energy will be reduced and the momentum will be conserved. To calculate the common velocity after collision we use the formula below

[tex]m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{1})v\\m_{1}=1200kg, m_{2}=850kg\\v_{1}=8m/s,v_{2}=17m/s\\[/tex]

If we substitute value into the equation we arrive at

[tex](1200*8)+(850*17)=(1200+850)v\\v=\frac{9600+14450}{2050} \\v=11.73m/s[/tex]

b. the change in kinetic energy is expressed as

[tex]K.E=K.E_{final}- K.E_{initial}\\[/tex]

[tex]K.E_{initial}=1/2m_{1}v_{1}^{2}+1/2m_{2}v_{2}^{2}\\K.E_{initial}=1/2[(1200*8^{2} )+(850*17^{2} )]\\K.E_{initial}=1/2(76800+245650)\\K.E_{initial}=161225m/s[/tex]

[tex]K.E_{final}=1/2(m_{1}+m_{2})v\\K.E_{final}=1/2(1200+850)11.73\\K.E_{final}=12,023m/s[/tex]

The lost kinetic energy is calculated as

[tex]K.E_{lost}=K.E_{final}- K.E_{initial}\\K.E_{lost}=12023-161225\\K.E_{lost}=149,202m/s[/tex]