Chapter 7: Linear Momentum

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

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

4. Motion and Its Conservation Not Conserved Not Conserved Not Conserved Conserved Conserved stops smooth and level rough

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

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 • mDv = Ft = -3000 kg·m/s • Dv = (-3000 kg·m/s)/1000kg = -3m/s • vf = vi + Dv • = 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)

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

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

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

14. Ex: 2 objects, complete inelastic • (m1v1)initial + (m2v2)initial = (m1v1)final + (m2v2)final • Each car has mass 1000kg • (1000)(10) + (1000)(0) = (1000)v + (1000)v • 10,000 = 2000v v = 5 m/s

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 = ½mv2. • K-final = ½(m)(v/2)2 + ½(m)(v/2)2 = ¼ mv2. • 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

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. 2 dimensional p conservation

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

20. 1 2 Initial momentum Collision-impulse p + impulse Final momentum

21. 1 2 Initial momentum Collision-impulse p + impulse Final momentum

22. 1 2 Initial momentum Collision-impulse p + impulse Final momentum

23. 1 2 Initial momentum Collision-impulse p + impulse Final momentum

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

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

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. 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. (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. 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. 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. Example Elastic Collisions

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

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

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