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

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.

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.

To solve this problem, we need to use the principle of conservation of momentum and the principle of conservation of kinetic energy.

a) Let's calculate the initial velocity of the heavier car (car 2). The momentum before the collision is equal to the momentum after the collision.

The momentum (p) of an object is given by the equation:

p = mass × velocity

For car 1 (with a mass of 1200 kg and speed of 30 km/h in the +x direction):
p1 = (1200 kg) × (30 km/h) = 36,000 kg·km/h

For car 2 (with a mass of 1500 kg and traveling in the +y direction):
p2 = (1500 kg) × (vy)

Since the cars stick together, the wreckage moves off at an angle of 64° to the x-axis. We need to find the y-component of velocity for car 2, vy.

Using the trigonometric relationship:

tan(64°) = vy / vx

Substituting the given values, we have:

tan(64°) = vy / (30 km/h)

Solving for vy:

vy = tan(64°) × (30 km/h)

Now, the total momentum after the collision is given by:

p_total = (mass car 1 + mass car 2) × (velocity after collision)

Using the given angle and the x and y-components of velocity, we can write:

p_total = (1200 kg + 1500 kg) × sqrt(vx^2 + vy^2)

Since the wreckage moves off at an angle of 64° to the x-axis, we have:

p_total = 2700 kg × sqrt(vx^2 + vy^2)

Setting the initial and final momenta equal to each other, we have:

36,000 kg·km/h = 2700 kg × sqrt(vx^2 + vy^2)

To find vx^2 + vy^2, we square both sides:

(36,000 kg·km/h)^2 = (2700 kg)^2 × (vx^2 + vy^2)

Now, let's convert the units to SI units:
1 km/h = 1000 m/3600 s ≈ 0.2778 m/s
So, we have:
vx = 30 km/h ≈ 8.33 m/s

Substituting the known values into the equation above, we can solve for vx^2 + vy^2.

b) Once we find the initial velocity of the heavier car (car 2), we can calculate the initial kinetic energy (KE) using the equation:

KE = (1/2) × mass × velocity^2

Substituting the known values, we can calculate the initial KE for car 2.

To find the percentage KE lost in the collision, we can use the equation:

% KE lost = ((initial KE - final KE) / initial KE) × 100

Substituting the initial and final KE, we can calculate the percentage KE lost.

Apply these equations step by step, and don't forget to check your units along the way to get the accurate solutions for a) and b).