A 900-kg car traveling east at 15. 0 m/s suddenly collides with a 750-kg car traveling north at 20. 0 m/s. The cars stick together after the collision. In what direction does the wreckage move just after the collision?.

By conservation of momentum,

(900 kg) (15.0 m/s) i + (750 kg) (20.0 m/s) j = (900 kg + 750 kg) v

where v is the velocity vector of the combined cars. Solve for v :

(13,500 i + 15,000 j) kg•m/s = (1650 kg) v

v ≈ (8.18 i + 9.09 j) m/s

Then the direction of v relative to East is θ such that

tan(θ) ≈ 9.09/8.18   ⇒   θ ≈ 48.0°

so the wreckages would move at approximately 48.0° north of east.

answer:

Pink?

Explanation

i guessed that becuz ur pfp sorry if im wrong