Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1050 kg and was approaching at 6.00 m/s due south. The second car has a mass of 750 kg and was approaching at 25.0 m/s due west.

v = 11.0 m/s at 198.6° (18.6° south of west)

ΔKE = - 145 kJ

Explanation:

I assume you want to find the final velocity and the change in kinetic energy.

Take east to be + x and north to be + y.

Momentum is conserved in the x direction:

(1050 kg) (0 m/s) + (750 kg) (-25.0 m/s) = (1050 kg + 750 kg) vₓ

vₓ = - 10.4 m/s

Momentum is conserved in the y direction:

(1050 kg) (-6.00 m/s) + (750 kg) (0 m/s) = (1050 kg + 750 kg) vᵧ

vᵧ = - 3.50 m/s

The magnitude of the final velocity is:

v² = (-10.4 m/s) ² + (-3.50 m/s) ²

v = 11.0 m/s

The direction of the final velocity is:

θ = atan (-3.50 m/s / - 10.4 m/s)

θ = 198.6°

The initial kinetic energy is:

KE₀ = ½ (1050 kg) (6.00 m/s) ² + ½ (750 kg) (25.0 m/s) ²

KE₀ = 253,275 J

The final kinetic energy is:

KE = ½ (1800 kg) (11.0 m/s) ²

KE = 108,682 J

The change in kinetic energy is:

ΔKE = 108,682 J - 253,275 J

ΔKE ≈ - 145,000 J