Two cars approach along icy streets which meet at a right angle to one another. The cars collide and stick together. One car has a mass of 1200 kg and had a speed of 30 km/h in the +x direction before the collision. The second car has a mass of 1500 kg and was traveling in the +y direction before the collision. After the collision, the wreakage moved off at an angle of 64o to the x axis.

Question

Two cars approach along icy streets which meet at a right angle to one another. The cars collide and stick together. One car has a mass of 1200 kg and had a speed of 30 km/h in the +x direction before the collision. The second car has a mass of 1500 kg and was traveling in the +y direction before the collision. After the collision, the wreakage moved off at an angle of 64o to the x axis. 
a) What was the initial speed of the heavier car? 
b) What percentage KE was lost in the collision?

I need help please. Show the equation and then plug the numbers into that equation

Henry · Answer

M1 = 1200kg.
V1 = 30km/h = 30,000m/3600s = 8.33 m/s[0o].

M2 = 1500 g.
V2[90o] = ?.

a. M1*V1 + M2*V2 = M1*V + M2*V.
1200*8.33 + 1500V2[90o] = 1200V[64o] + 1500V[64o].
10,000 + 1500V2[90o] = 2700V[64o].
The hor. components before the collision = Hor. components after collision:
10,000 = 2700V*Cos64.
V = 8.45 m/s. = The velocity after the collision.

The Y before = Y after collision:
1500V2 = 2700V*sin64.
1500V2 = 2700*8.45*sin64,
1500V2 = 20,503, V2 = 13.7 m/s. = Speed of heavier car.

Henry · Answer

b. KEb = 0.5M1*V1^2 + 0.5M2*V2^2 = Kinetic energy before collision. KEa = 0.5M1*V^2 + 0.5M2*v^2 = Kinetic energy after collision. KE Lost = KEb - KEa. %KE Lost = KE Lost/KEb.