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.
Control of the upper classes
