You should be able to solve 1D and 2D Momentum problems. Linear Momentum Learning Goal: You should be able to solve 1D and 2D Momentum problems.

Linear Momentum Antoine Lavoisier proved that mass was conserved by burning iron in a closed system. (The mass of any closed system remains constant.) Descartes pondered the actions of collisions in 1644 and wondered if some quantity of motion was conserved; why do objects slow down? Where did the motion go? Newton defined a quantity of motion that was conserved and called it momentum.

Linear Momentum The momentum of an object is defined as the product of its mass and velocity (not speed). Linear momentum looks at velocities just before or after a collision. The forces between the objects and external forces can be neglected or cancel out in an isolated system. We’ll look at the effects of forces later with impulse causing a change in momentum.

Linear Momentum Momentum is a vector quantity measured in kg m/s or Ns. (or g cm/s or kg km/h, etc.) p = mv The total momenta of components before a collision must equal the total momenta of those components after the collision. pi = pf or p = 0

Colliding Objects For two colliding objects bouncing off each other m1v1 + m2v2 = m1v1’ + m2v2’ For two colliding objects that stick m1v1 + m2v2 = (m1+m2)v’ In general: pi = pf

Videos http://www.youtube.com/watch?v=OuA-znVMY3I http://www.youtube.com/watch?v=mFNe_pFZrsA&list= TL82-EFBENCmwAYWaj2Lb5rFQTWUM8PlHF

1D Problem A 75.0 kg rugby player runs at 4.0 m/s [E] and collides head-first with a 90.0 kg running at 4.0 m/s [W]. The lighter player bounces off at 1.0 m/s [W]. What is the final velocity of the heavier player right after the collision? pi = pf m1v1 + m2v2 = m1v1’ + m2v2’ Sub in values! v2’ = 0.17 m/s [E]

1D Problem #2 A 6.0 Mg train rolls right at 2.0 m/s and collides and sticks to a 3.0 Mg train rolling left at 3.0 m/s. Find the final velocity of the trains. pi = pf m1v1 + m2v2 = (m1+m2)v’  sub in values v’ = 0.33 m/s [right]

2D Problem A 1200.0 kg Smart Car travels East at 10.0 m/s and collides with a 2000.0 kg Hummer moving North at 5.0 m/s. If the two cars lock together upon impact, find their initial resulting velocity. pi = pf As this is not a linear question, we can use vectors and a vector diagram:

2D Diagram p=mv pH = (2000kg)(5m/s) = 10000kgm/s [N] pS = (1200kg)(10m/s) = 12000 kgm/s [E]

θ = tan-1[(10000kgm/s)/(12000kgm/s)] 2D Problem continued pR2 = (10000kgm/s)2+(12000 kgm/s)2 |pR| = kgm/s |vR| = pR/mT   = m/s θ = tan-1[(10000kgm/s)/(12000kgm/s)] = o N of E vR =

pix = pfx and piy = pfy Crazy 2D Problem A grenade of mass 1.20 kg is at rest and it explodes into three pieces. One 0.50 kg piece travels at 3.0 m/s [N] and a 0.30 kg piece travels at 4.0 m/s [SW]. What is the velocity of the third piece? Δp = 0 where pi = 0 pix = pfx and piy = pfy

p1’ is moving North so has no x-component. p1y’ = 0.50 kg (3.0 m/s)[N] Crazy 2D Problem contd. Object 1 p1’ p1’ is moving North so has no x-component. p1y’ = 0.50 kg (3.0 m/s)[N] = 1.5 kgm/s [N] p1x’ = 0

Crazy 2D Problem contd. p2’ p2y’ = cos45 (1.2 kgm/s) [S] p2x’ Object 2 p2y’ = cos45 (0.30kg)(4m/s) = cos45 (1.2 kgm/s) [S] p2x’ = sin45 (1.2 kgm/s) [W] p2’ p2x’ p2y’

2D Crazy Q still Momentum is conserved in each plane p1x’ + p2x’ + p3x’ = 0 0 + (- sin 45 (1.2 kgm/s)) + p3x’ = 0 p3x’ = p1y’ + p2y’ + p3y’ = 0 1.5 kgm/s + (- cos 45 (1.2 kgm/s)) + p3y’ = 0 p3y’ =