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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

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## Momentum and Collisions

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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 8.0 kg V= ?**Momentum , p**• Momentum is a Vector • p = m v • p1 = ? p2 = ? 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 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**Momentum , p**• Momentum is a Vector • p = m v • Total momentum of a system is a vector sum: • p1+p2+p3+…….. p3 p2 ptotal p1**Newton’s 2nd Law**• Net Force = Mass x Acceleration • F = M a • F = M (∆V/∆t) • F ∆t = M ∆V • F ∆t = M (V1-V2) • F ∆t = M V1 - M V2 • F ∆t = ∆p Impulse= F∆t • The Impulse = the change in momentum**Newton’s 2nd Law**• Net Force = Mass x Acceleration • F = M a • or • F = ∆p/ ∆t**Newton’s 2nd Law**• Net Force = Mass x Acceleration • F ∆t = ∆p Impulse= F ∆t • The Impulse = the change in momentum**If M=1500 kg and Dt=0.4 sec,**Find Dp and Favg**30°**50°**Impulse**• The Impulse = the change in momentum • F ∆t = ∆p**Impulse**• The Impulse = the change in momentum • F ∆t = ∆p**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**V=8.0 m/s • 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? after V=? M=100.0 kg M=100.0 kg before p before = p after 0 = 30kg(8.0 m/s) - 100 kg V 100 kg V = 240 kg m/s V = 2.4 m/s**Explosions**If Vred=8.0 m/s Vblue=?**“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 = 1.85 m/s**“Stick together”2 objects have same speed after**collidingThis is a perfectly inelastic collision • 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**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?**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 Collisions**X axis m1V10 = m1v1cos(50) + m2v2cos(40) Y axis 0 = m1v1sin(50) - m2v2sin(40)**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)**• 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.**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. 2-D Stick together (Inelastic) 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 CollisionsBounce off without loss of energy**• p before = p after • & • KE before = KE after m1 m2 m2 m1 v1 v1,f v2,f**Elastic CollisionsBounce off without loss of energy**• p before = p after & KE before = KE after**Elastic CollisionsBounce off without loss of energy**• p before = p after & KE before = KE after + or**Elastic CollisionsBounce off without loss of energy**• p before = p after & KE before = KE after - or**Elastic CollisionsBounce off without loss of energy**• p before = p after & KE before = KE after & m1 m2 m2 m1 v1 v1,f v2,f**Elastic CollisionsBounce off without loss of energy**• p before = p after & KE before = KE after & if m1 = m2 = m, then v1,f = 0.0 & v2,f = v1 m m m m v1 v2,f = v1 v1,f= 0.0**Elastic CollisionsBounce off without loss of energy**• p before = p after & KE before = KE after & if m1 <<< m2 , then m1+m2 ≈m2 & m1-m2 ≈ -m2 v1,f = - v1 & v2,f ≈ 0.0 M m M m v1 v2,f ≈ 0.0 v1,f=- v1**Elastic CollisionsBounce 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**if m1 <<< m2 and v 2 is NOT 0.0**Elastic Collisions 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=? M v2=20 km/s M m v2,f ≈ ? v1 = 40 km/s**if m1 <<< m2 and v 2 is NOT 0.0**Elastic Collisions 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 CollisionsBounce off without loss of energy**• p before = p after & KE before = KE after if m1 = m2 = m, then v1,f = 0.0 & v2,f = v1 m m m 90° m v1**Elastic CollisionsBounce off without loss of energy**• p before = p after & KE before = KE after if m1 = m2 = m, then v1,f = 0.0 & v2,f = v1 p1f p2f 90° m m m 90° p1 m v1 p1 =p1f +p2f**Center of Mass**• The average position of the mass • When we use F=ma • We really mean F = m acm • The motion of an object is the combination of • The translational motion of the CM • Rotation about the CM

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