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**Momentum and Collisions**

Dr. Robert MacKay Clark College, Physics

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**Introduction Review Newtons laws of motion Define Momentum**

Define Impulse Conservation of Momentum Collisions Explosions Elastic Collisions

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**Introduction Newtons 3 laws of motion 1. Law of inertia**

2. Net Force = mass x acceleration ( F = M A ) 3. Action Reaction

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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

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

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**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

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**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

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**Momentum , p Momentum = mass x velocity is a Vector**

has units of kg m/s

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**Momentum , p (a vector) Momentum = mass x velocity p = m v p = ?**

8.0 kg 6.0 m/s

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

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**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

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**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

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**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

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**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

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**Newton’s 2nd Law Net Force = Mass x Acceleration**

F ∆t = ∆p Impulse= F ∆t The Impulse = the change in momentum

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**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

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**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

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**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.

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**“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

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**“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

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

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

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**“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

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**“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

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**“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?

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**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?

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**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

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**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.

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**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

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**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

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**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

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**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

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**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

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**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

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**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 ≈ ?

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**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

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**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

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Chapter 2 Newton’s Laws of Motion

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