1. Chapter 7: Linear Momentum CQ: 2 Problems: 1, 7, 22, 41, 45, 47. Momentum & Impulse Conservation of Momentum Types of Collisions 1

2. Type of Collision Kinetic Energy Conserved Momentum Conserved Complete Inelastic X yes Inelastic X yes Elastic yes

3. Conserved Quantities Energy – has many forms Motion – has only one form How does Nature conserve motion? Speed? Velocity? Mass x Velocity? 3

4. 4 stops Conserved Not Conserved smooth and levelrough Motion and Its Conservation

5. 5 Momentum Momentum = mv Symbol: p, P. SI Unit: kg·m/s Ex. 1000kg car moves at 8m/s. mv = (1000kg)(8m/s) = 8,000 kg·m/s

6. 6 Impulse impulse = Ft force x time = change in momentum SI Unit: N·s = kg·m/s Ex. A 4000N force acts for 0.010s. impulse = Ft = (4000N)(0.010s) = 40 N·s

7. 7 Impulse and Momentum Impulse = Ft = change-in-momentum consequence of Newton’s 2 nd law that Ft = change in momentum F = ma Ft = mat Ft = (m)(at) Ft = (m)(  v) N·s = kg·m/s

8. 8 Impulse Example A braking force of 4000N acts for 0.75s on a 1000kg car moving at 5.0m/s. impulse = Ft = (-4000N)(0.75s) = -3000 N·s m  v = Ft = -3000 kg·m/s  v = (-3000 kg·m/s)/1000kg = -3m/s v f = v i +  v = 5m/s + (-3m/s) = 2m/s

9. Collisions “Brief” interaction between objects Objects share the collision force (N3L) One object gets +Ft, the other gets –Ft. As a whole, they receive no impulse, thus no change in total momentum of this system (due to the collision) 9

10. 10

11. 11 Conservation of Momentum (i.e. Motion): Initial Momentum (Motion): Final Momentum (Motion): (Since )

12. 12 Collisions elastic: total KE stays same inelastic: total KE decreases complete inelastic: objects move at same velocity after collision, total KE decreases

13. 13 Conservation of Linear Momentum If net-external force = 0 (e.g. level frictionless surface)

14. 14 Ex: 2 objects, complete inelastic (m 1 v 1 ) initial + (m 2 v 2 ) initial = (m 1 v 1 ) final + (m 2 v 2 ) final Each car has mass 1000kg (1000)(10) + (1000)(0) = (1000)v + (1000)v 10,000 = 2000v v = 5 m/s

15. 15 Conservation of Momentum may occur when KE is lost Ex: Mass m, Speed v, hits & sticks to another mass m, speed = 0. Final speed of each object is v/2. K-initial = ½(m)(v) 2 + 0 = ½mv 2. K-final = ½(m)(v/2) 2 + ½(m)(v/2) 2 = ¼ mv 2. Half the original KE converted to heat, sound, etc.

16. Elastic Collisions Approach velocity = Separation velocity Equal mass-collisions: exchange velocity Ex: straight pool shot Ex: car +5m/s bumps car at +3m/s (bumping car +3, bumped +5) Ex: (4kg)  10m/s (1kg)  5m/s Result: (4kg)  8m/s (1kg)  13m/s 16

17. 17 Ex: collision type 2kg @ +6m/s hits 1kg @ +3m/s After) 2kg @ +4.8ms, 1kg @ ???? mv-before = mv-after (2)(6)+(1)(3) = (2)(4.8)+(1)(v) 15 = 9.6 + v v = 5.4 Is this collision elastic or inelastic?

18. 18 2 dimensional p conservation

19. 19 Example: m1 = 0.010kg, m2 = 1.0kg, v1i = 200m/s. Calculate vf (mom. cons.), Then calculate h (using energy).

20. Initial momentum Collision-impulse p + impulse Final momentum 12

21. Initial momentum Collision-impulse p + impulse Final momentum 12

22. Initial momentum Collision-impulse p + impulse Final momentum 12

23. Initial momentum Collision-impulse p + impulse Final momentum 12

24. Summary momentum = mass x velocity Ft = change-in-momentum Momentum is conserved when net external forces are negligible Momentum conservation may occur for elastic & inelastic collisions Elastic: approach & separation vel. equal 24

25. 25 complete inelastic (e.g. 62) 2000kg truck vi = +10m/s hits and locks with 1000kg car vi = -4m/s. mv-before = mv-after (2000)(10) + (1000)(-4) = (3000)v 20,000 – 4,000 = 3000v 16,000 = 3000v v = 5.33 m/s

26. 26

27. 27

28. 28 dropped ball (e.g. 21) 1kg ball hits ground at -5m/s and bounces off with +4m/s. Contact time t = 0.33s. (Fnet)t = mvf –mvi = m(vf – vi) (Fc – mg)t = m(vf – vi) (Fc – 9.8)(0.33) = (1)(4 –(-5)) = 9 (Fc – 9.8) = 27 Fc = 37 newtons

29. 29 Two masses move on a frictionless horizontal surface. M1 = 1kg, v1i = 4m/s. M2 = 2kg, v2i = 1m/s. The masses collide along a straight line. Find v1f, if v2f = 2.3 m/s and no other external forces act.

30. 30 (cont) Calculate the initial and final kinetic energies. It is possible for kinetic energy to decrease due to the production of thermal energy in a collision. In this case 2.73J of Thermal Energy were created by the collision.

31. 31 08-4. A 0.0149kg bullet moves horizontally at 830 feet per second and strikes a 10lb wood block lying at rest on a horizontal surface. The bullet takes 1.0 millisecond to stop inside the block. a) Convert the data to SI units. b) Calculate the speed the block moves just after the bullet stops in the block. System momentum conserved when external impulse is negligible.

32. 32 Calculate the kinetic energy of the bullet before the collision and of the moving block + bullet after the collision. What percent of the original kinetic energy is converted to other energies? What percent is retained as kinetic?

33. 33 Example Elastic Collisions

34. 34 Elastic Collisions Where One Object is at Rest Before Collision.

35. 35 Elastic Collisions Where One Object is at Rest Before Collision.

36. 36 Elastic Collisions Where One Object is at Rest Before Collision.

37. 1a) (2000)(10)+(1000)(0) = (3000)vf vf = 6.67m/s Kf = ½ (3000)(6.67)(6.67) = 66,700J

38. 1b) (2000)(10) = (2000)(5) + (1000)(vf) 20,000 = 10,000 + 1,000vf 10,000 = 1,000vf vf = 10m/s Kf-sys = ½ (2000)(5)(5) + ½ (1000)(10)(10) 75,000J (Not elastic, but more elastic than previous)

39. 2 (2000)(5)+(500)(0) = (2500)vf vf = 4m/s E2 = ½ (2500)(4)(4) = E3 = (fk)(12) 20,000 = fk(12) fk = 1,667N frict.coeff. = fk/FN = 1,667/(500)(9.8) = 0.34