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.
Note that with |x|, the value x will always be positive when simplified. This means that:
B) |-3.5|, |4.2|, |-9| is your answer, because it becomes, 3.5, 4.2, 9
hope this helps
82.5820063 if you do that with straight math
Answer:
46, 39, 32
Explanation:
you just have to subtract 7 each time :)
