# Momentum and Collisions

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Momentum and Collisions Momentum and Collisions Dr. Robert MacKay Clark College, Physics.

Momentum and Collisions Momentum and Collisions Dr. Robert MacKay Clark College, Physics.

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Momentum and Collisions
Dr. Robert MacKay Clark College, Physics

Introduction Review Newtons laws of motion Define Momentum
Define Impulse Conservation of Momentum Collisions Explosions Elastic Collisions

Introduction Newtons 3 laws of motion 1. Law of inertia
2. Net Force = mass x acceleration ( F = M A ) 3. Action Reaction

Law of interia (1st Law) Every object continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to change that state by forces impressed upon it. acceleration = 0.0 unless the objected is acted on by an unbalanced force

Newton’s 2nd Law Net Force = Mass x Acceleration F = M A

Newton’s Law of Action Reaction (3rd Law)
You can not touch without being touched For every action force there is and equal and oppositely directed reaction force

Newton’s Law of Action Reaction (3rd Law)
Ball 1 Ball 2 F2,1 F1,2 F1,2 = - F2,1 For every action force there is and equal and oppositely directed reaction force

Momentum , p Momentum = mass x velocity is a Vector
has units of kg m/s

Momentum , p (a vector) Momentum = mass x velocity p = m v p = ?
8.0 kg 6.0 m/s

Momentum , p Momentum = mass x velocity p = m v p = 160.0 kg m/s

Momentum , p Momentum is a Vector p = m v p1 = ? p2 = ? V= +8.0 m/s
7.5 kg m2= 10.0 kg V= -6.0 m/s

Momentum , p Momentum is a Vector p = m v
p1 = +60 kg m/s p2 = - 60 kg m/s V= +8.0 m/s m1= 7.5 kg m2= 10.0 kg V= -6.0 m/s

Momentum , p Momentum is a Vector p = m v
p1 = +60 kg m/s p2 = - 60 kg m/s the system momentum is zero., V= +8.0 m/s m1= 7.5 kg m2= 10.0 kg V= -6.0 m/s

Newton’s 2nd Law Net Force = Mass x Acceleration F = M a F = M (∆V/∆t)
F ∆t = M ∆V F ∆t = M (VF-V0) F ∆t = M VF- M V0 F ∆t = ∆p Impulse= F∆t The Impulse = the change in momentum

Newton’s 2nd Law Net Force = Mass x Acceleration
F ∆t = ∆p Impulse= F ∆t The Impulse = the change in momentum

Newton’s Law of Action Reaction (3rd Law)
Ball 1 Ball 2 F2,1 F1,2 F1,2 = - F2,1 For every action force there is and equal and oppositely directed reaction force

Newton’s Law of Action Reaction (3rd Law)
Ball 1 Ball 2 F2,1 F1,2 F1,2 = - F2,1 F1,2∆t = - F2,1 ∆t ∆p2 = - ∆p1

Conservation of momentum
Ball 1 Ball 2 F2,1 F1,2 If there are no external forces acting on a system (i.e. only internal action reaction pairs), then the system’s total momentum is conserved.

“Explosions” 2 objects initially at rest
A 30 kg boy is standing on a stationary 100 kg raft in the middle of a lake. He then runs and jumps off the raft with a speed of 8.0 m/s. With what speed does the raft recoil? V=8.0 m/s after V=? M=100.0 kg M=100.0 kg before

“Explosions” 2 objects initially at rest
A 30 kg boy is standing on a stationary 100 kg raft in the middle of a lake. He then runs and jumps off the raft with a speed of 8.0 m/s. With what speed does the raft recoil? V=8.0 m/s after V=? M=100.0 kg M=100.0 kg before p before = p after = 30kg(8.0 m/s) kg V 100 kg V = 240 kg m/s V = m/s

Explosions If Vred=9.0 m/s Vblue=? 9.0 m/s

Explosions If Vred=9.0 m/s Vblue=3.0 m/s 9.0 m/s 3.0 m/s

“Stick together” 2 objects have same speed after colliding
A 30 kg boy runs and jumps onto a stationary 100 kg raft with a speed of 8.0 m/s. How fast does he and the raft move immediately after the collision? V=8.0 m/s V=? M=100.0 kg M=100.0 kg before after

“Stick together” 2 objects have same speed after colliding
A 30 kg boy runs and jumps onto a stationary 100 kg raft with a speed of 8.0 m/s. How fast does he and the raft move immediately after the collision? V=8.0 m/s V=? M=100.0 kg M=100.0 kg before after p before = p after 30kg(8.0 m/s) = 130 kg V 240 kg m/s = 130 kg V V = m/s

“Stick together” 2 objects have same speed after colliding This is a perfectly inelastic collision
V=8.0 m/s V=? M=100.0 kg M=100.0 kg before after A 30 kg boy runs and jumps onto a stationary 100 kg raft with a speed of 8.0 m/s. How fast does he and the raft move immediately after the collision?

A 20 g bullet lodges in a 300 g Pendulum
A 20 g bullet lodges in a 300 g Pendulum. The pendulum and bullet then swing up to a maximum height of 14 cm. What is the initial speed of the bullet?

Before and After Collision Before and After moving up
mv = (m+M) V Before and After Collision 1/2(m+M)V2=(m+M)gh After collision but Before and After moving up

2-D Stick together (Inelastic)
Momentum Before = Momentum After P before= P after For both the x & y components of P. A 2000 kg truck traveling 50 mi/hr East on McLoughlin Blvd collides and sticks to a 1000 kg car traveling 30 mi/hr North on Main St. What is the final velocity of the wreck? Give both the magnitude and direction OR X and Y components.

2-D Stick together (Inelastic)
A 2000 kg truck traveling 50 mi/hr East (V1) on McLoughlin Blvd collides and sticks to a 1000 kg car traveling 30 mi/hr North (V2) on Main St. What is the final velocity (V) of the wreck? Give both magnitude and direction OR X and Y components. Pbefore=Pafter PBx=PAy & PBy=PAy 2000Kg(50 mi/hr)=3000KgVx & 1000kg(30mi/hr)=3000kgVy Vx=33.3 mi/h & Vy=10 mi/hr Or V= 34.8mi/hr = (sqrt(Vx2+Vx2) & q =16.7° = tan-1(Vy/Vx) V V1 3000Kg 2000Kg V2 1000Kg

Elastic Collisions Bounce off without loss of energy
p before = p after & KE before = KE after m1 m2 m2 m1 v1 v1,f v2,f

Elastic Collisions Bounce off without loss of energy
p before = p after & KE before = KE after & if m1 = m2 = m, then v1,f = & v2,f = v1 m m m m v1 v2,f = v1 v1,f= 0.0

Elastic Collisions Bounce off without loss of energy
p before = p after & KE before = KE after & if m1 <<< m2 , then m1+m2 ≈m & m1-m2 ≈ -m2 v1,f = - v & v2,f ≈ 0.0 M M m m v1 v2,f ≈ 0.0 v1,f=- v1

Elastic Collisions Bounce off without loss of energy
if m1 <<< m2 and v 2 is NOT 0.0 Speed of Approach = Speed of separation (True of all elastic collisions) M M m m v1 v2 v1,f=- (v1 +v2 +v2) v2,f ≈ v2

M M Elastic Collisions Speed of Approach = Speed of separation m
if m1 <<< m2 and v 2 is NOT 0.0 Speed of Approach = Speed of separation (True of all elastic collisions) A space ship of mass 10,000 kg swings by Jupiter in a psuedo elastic head-on collision. If the incoming speed of the ship is 40 km/sec and that of Jupiter is 20 km/sec, with what speed does the space ship exit the gravitational field of Jupiter? m v1,f=? v2=20 km/s M M m v2,f ≈ ? v1 = 40 km/s

M M Elastic Collisions Speed of Approach = Speed of seperation v1,f=?
if m1 <<< m2 and v 2 is NOT 0.0 Speed of Approach = Speed of seperation (True of all elastic collisions) A little boy throws a ball straight at an oncoming truck with a speed of 20 m/s. If truck’s speed is 40 m/s and the collision is an elastic head on collision, with what speed does the ball bounce off the truck? v1,f=? M m v2=40 m/s M m v1 = 20 m/s v2,f ≈ ?

Elastic Collisions Bounce off without loss of energy
p before = p after & KE before = KE after if m1 = m2 = m, then v1,f = & v2,f = v1 m m m 90° m v1

Elastic Collisions Bounce off without loss of energy
p before = p after & KE before = KE after if m1 = m2 = m, then v1,f = & v2,f = v1 p1f p2f 90° m m m 90° p1 m v1 p1 =p1f +p2f

The End