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Momentum in Physics: Conservation, Impulse, and Collisions

Explore the concepts of linear momentum, conservation of momentum, impulse, and collisions in physics. Understand how momentum is conserved and how impulse is related to the change in momentum of a system.

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Momentum in Physics: Conservation, Impulse, and Collisions

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  1. Physics 1710—Chapter 9 Momentum • Solution: Electricity: 1 kw-hr = (1 x103 J/s)(3.6 x103 s) = 3.6 x106 J E/$ = 3.6 x106/$0.1 = 3.6 x107 J/$ Gasoline: 1 gal = 1.3 x108 J E/$ = 1.3 x108/$2 = 6.5 x107 J/$ Cost (gas/elec) ~ 180% for same efficiency.

  2. Physics 1710—Chapter 9 Momentum • Consider this: How high will the tennis ball bounce relative to its starting height when they are dropped as shown below?

  3. Physics 1710—Chapter 9 Momentum 1’ Lecture Linear momentum is the product of the velocity and the mass and is, therefore, a vector quantity. p = m v Momentum is always conserved. Impulse is the time integrated force.

  4. Physics 1710—Chapter 9 Momentum Linear Momentum (Latin: “movement”) p ≡ m v px = m vx py = m vy pz = m vz

  5. Physics 1710—Chapter 9 Momentum Newton’s Second Law of Motion (What Newton actually said:) ∑F = d p/dtThe net external force is equal to the time rate of change in the linear momentum. First law:In the absence of any net external force (∑F = 0)the linear momentum p of a system is conserved (d p/dt = 0).

  6. Physics 1710—Chapter 9 Momentum The Collision of Two Bodies ①⇒ ⇐② F12= - F21 F21+ F12= 0 d p1/dt +d p2/dt = 0 d( p1+ p2)/dt = 0 Thus, the total momentum is conserved in a collision.

  7. Physics 1710—Chapter 9 Momentum Conservation of Linear Momentum p1i + p2i = p1f + p2f ∑pix =∑pfx ∑piy =∑pfy ∑piz = ∑pfz

  8. Physics 1710—Chapter 9 Momentum Big Ball – Little Ball Collision Demonstration!!! ??? ⃘⃗ ⃖⃝???

  9. Physics 1710—Chapter 9 Momentum Elastic CollisionP conserved p1i+ p2i = p1f+ p2f m1v1i + m2v2i = m1v1f + m2v2f (Assume) E conserved K1i + K2i = K1f + K2f ½ m1v1i2 +½ m2v2i2= ½ m1v1f2+ ½ m2v2f2 Solve equations: v1f = [(m1 -m2)/ (m1 +m2)]v1i +[(2 m2)/ (m1 +m2)]v2i v2f = [(m2 –m1)/ (m1 +m2)]v2i +[(2 m1)/ (m1 +m2)]v1i

  10. Physics 1710—Chapter 9 Momentum v1f = [(m1 -m2)/ (m1 +m2)]v1i +[(2 m2)/ (m1 +m2)]v2i v2f = [(m2 –m1)/ (m1 +m2)]v2i +[(2 m1)/ (m1 +m2)]v1i Let m2 >> m1 ; v1i = -v2i v1f = [(0-m2)/ (0+m2)]v1i -[(2 m2)/ (0 +m2)](-v1i ) v1f = (-v1i -2 v1i ) = - 3 v1i !! K1f /K1i = v1f 2 /v1i 2 = 9 = h1i/h2f Elastic Collisionv1f = [(m1 -m2)/ (m1 +m2)]v1i +[(2 m2)/ (m1 +m2)]v2i v2f = [(m2 –m1)/ (m1 +m2)]v2i +[(2 m1)/ (m1 +m2)]v1i

  11. Physics 1710—Chapter 9 Momentum What about a ball that bounces off of the earth? What is the change in the momentum of the earth when a 0.60 kg ball falls from a height of 1.0 meter and bounces up? What is the change in velocity of the earth? (mass of earth = 6.0x1024 kg)

  12. Physics 1710—Chapter 9 Momentum Σpi = Pi +pi = 0 = M Vi + mvi Vi = - (m/M)vivi = √2gh = √2(9.8 m/s2)(1.0 m h ) = 4.4 m/sVi = - (0.6 kg/ 6.0x1024 kg )vi = - 4.4 x 10 – 25m/s If ball rebounds to same height v1f = v1i; V1f = V1i; ΔV = 8.8 x 10 – 25m/s (~1proton/36yr) What is the change in the momentum of the earth when a 0.60 kg ball falls from a height of 1.0 meter and bounces up? What is the change in velocity of the earth? (mass of earth = 6.0x1024 kg) Peer Instruction Time

  13. Physics 1710—Chapter 9 Momentum Elastic Collision Special Case m1 = m2 v1f = [(2 m2)/ (m1 +m2)]v2i v2f = [(2 m1)/ (m1 +m2)]v1i K1f = [(4 m1 m2)/ (m1 +m2) 2 ] K2i K2f = [(4 m1 m2)/ (m1 +m2) 2 ] K1i v1f = v2i K1f = K2i v2f = v1i K2f = K1i

  14. Physics 1710—Chapter 9 Momentum Elastic Collision Special Case m1 = m2 Newton’s Cradle

  15. Physics 1710—Chapter 9 Momentum Totally Inelastic Collisions (no bounce) [The objects stick together after collision.] v1f = vf v2f = vf m1 v1i + m2 v2i = (m1 + m2 ) vf vf = (m1 v1i + m2 v2i )/ (m1 + m2 ) Ki = ½ m1v1i2 +½ m2v2i2 Kf = ½ (m1 + m2)vf2

  16. Physics 1710—Chapter 9 Momentum Impulse and Momentum d p = Fdt ∆p = ∫d p = ∫Fdt = Impulse The impulse on a body equals the change in momentum.

  17. Physics 1710—Chapter 9 Momentum Impulse and “Follow Through” Demonstration

  18. Physics 1710—Chapter 9 Momentum ∆p = ∫Fdt ∆p=Fave∆t For a given force the longer it is applied the greater will be the impulse and the change in momentum.

  19. Physics 1710—Chapter 9 Momentum Impulse and Seat Belts Seat Belts ( and air bags and crumple zones) increase the stopping time ∆t. If ∆p is the same in two instants the impulse will be the same. The case with the longer ∆t will exhibit the smaller average force.

  20. Physics 1710—Chapter 9 Momentum Stopping Force ∆p = mv ∆t = s/vave = s/(v/2) Fave = ∆p/ ∆t = mv 2/(2s) Speed kills? : v 2 What about the sudden stop? :1/s

  21. Physics 1710Chapter 9 Linear Momentum and Collisions Center of Mass RCM = ∑mir/ M Or RCM = ∫r dm / M

  22. Physics 1710Chapter 9 Linear Momentum and Collisions Summary: Linear momentum is the product of the mass and velocity. Linear momentum is conserved. Impulse is the time integral of the momentum. The impulse is equal to the change in momentum of a system.

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