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 45º Pc θ P'

1 2 Px = Px' -(900)(25)sin45º = -(900 + 1900)v'sinθ v'sinθ = 5.68 Py = Py' -(900)(25)cos45º -(1900)(14) = -(900 + 1900)v'cosθ 2 v'cosθ = 15.18 θ = 20.5º Now divide 1 by 2: v'sin(20.5º) = 5.68 v'sinθ = 5.68 v' = 16.2 m/s v'cosθ = 15.18 Final velocity = 16.2 m/s [S 20.5º W] tanθ = 0.374

2. An 8 kg piece of jelloTM 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. 3.8 kg 9 m/s 12 m/s 2.2 kg 8.0 kg 30º v3' θ 2 kg

v3'sinθ = 1.6 1 2 v3'cosθ = 8.55 v3'sin10.6º = 1.6 v3'sinθ = 1.6 Px = Px' 0= (3.8)(9)cos30º – (2.2)(12) - 2v3'sinθ v3'sinθ = 1.6 1 Py = Py' 0= (3.8)(9)sin30º – 2v3'cosθ 2 v3'cosθ = 8.55 θ = 10.6º Now divide 1 by 2: v3'sin10.6º = 1.6 v3'sinθ = 1.6 v3' = 8.7 m/s v3'cosθ = 8.55 Final velocity = 8.7 m/s [S 10.6º W] tanθ = 0.187