Answer:
[tex]v_1 = 7.96 m/s[/tex]
[tex]v_2 = 17.8 m/s[/tex]
Explanation:
Let the mass of the other car is "m" and its kinetic energy is
[tex]K = \frac{1}{2}mv^2[/tex]
now the mass of the first car is two and half times and its kinetic energy is half that of other car
so we will have
[tex]\frac{1}{2}(2.5m)v_1^2 = \frac{1}{2}(\frac{1}{2}mv^2)[/tex]
[tex]2.5 v_1^2 = 0.5 v^2[/tex]
[tex]v_1 = 0.447 v[/tex]
now speed of both cars is increased by value of 9 m/s
so now we will have same kinetic energy for both cars
[tex]\frac{1}{2}(2.5 m)(0.447v + 9)^2 = \frac{1}{2}m(v + 9)^2[/tex]
[tex]2.5(0.447 v + 9)^2 = (v + 9)^2[/tex]
[tex]1.58(0.447v + 9) = v + 9[/tex]
[tex]0.293v = 5.22[/tex]
[tex]v = 17.8 m/s[/tex]
so speed of first car is
[tex]v_1 = 0.447 v = 7.96 m/s[/tex]
[tex]v_2 = 17.8 m/s[/tex]
