Conservation of Momentum in Two Dimensions Since momentum

Conservation of Momentum in Two Dimensions Since momentum is a vector, we can use conservation of momentum in the X and Y directions separately As with any two-dimensional vector problem, we can use vector components or the sine and cosine laws 1. A 900 kg car is travelling [SW] with a speed of 25 m/s when it has a completely inelastic collision with a 1900 kg truck moving South at 14 m/s. Determine the velocity with which the two vehicles are moving after they collide. Pt P' 45º Pc θ

P x = P x' -(900)(25)sin 45º = -(900 + 1900)v'sinθ = 5. 68 1 P y = P y' -(900)(25)cos 45º -(1900)(14) = -(900 + 1900)v'cosθ = 15. 18 2 Now divide 1 by 2: v'sinθ = 5. 68 v'cosθ = 15. 18 tanθ = 0. 374 θ = 20. 5º v'sin(20. 5º) = 5. 68 v' = 16. 2 m/s Final velocity = 16. 2 m/s [S 20. 5º W]

2. An 8 kg piece of jello. TM is at rest on a large kitchen table when it explodes into three pieces. A 2. 2 kg piece moves off horizontally with a velocity of 12 m/s [W] and a 3. 8 kg piece moves off with a velocity of 9 m/s [E 30º N]. Find the velocity of the third piece. 9 m/s 12 m/s 8. 0 kg 2. 2 kg 30º v 3' θ 2 kg 3. 8 kg

P x = P x' 0= (3. 8)(9)cos 30º – (2. 2)(12) - 2 v 3'sinθ = 1. 6 1 P y = P y' 0= (3. 8)(9)sin 30º – 2 v 3'cosθ = 8. 55 2 θ = 10. 6º Now divide 1 by 2: v 3'sin 10. 6º = 1. 6 v 3'sinθ = 1. 6 v 3' = 8. 7 m/s v 3'cosθ = 8. 55 tanθ = 0. 187 Final velocity = 8. 7 m/s [S 10. 6º W]