1. Momentum and Collisions Momentum and Collisions Dr. Robert MacKay Clark College, Physics

2. Introduction Review Newtons laws of motion Define Momentum Define Impulse Conservation of Momentum Collisions Explosions Elastic Collisions

3. Introduction Newtons 3 laws of motion 1. Law of inertia 2. Net Force = mass x acceleration ( F = M A ) 3. Action Reaction

4. 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

5. Newtons 2nd Law Net Force = Mass x Acceleration F = M A

6. Newtons 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

7. Newtons Law of Action Reaction (3rd Law) For every action force there is and equal and oppositely directed reaction force Ball 1 Ball 2 F 1,2 F 2,1 F 1,2 = - F 2,1

8. Momentum, p Momentum = mass x velocity is a Vector has units of kg m/s

9. Momentum, p (a vector) Momentum = mass x velocity p = m v p = ? 8.0 kg 6.0 m/s

10. Momentum, p Momentum = mass x velocity p = m v p = 160.0 kg m/s 8.0 kg V= ?

11. Momentum, p Momentum is a Vector p = m v p1 = ? p2 = ? m2= 10.0 kg V= -6.0 m/s m1= 7.5 kg V= +8.0 m/s

12. Momentum, p Momentum is a Vector p = m v p1 = +60 kg m/s p2 = - 60 kg m/s m2= 10.0 kg V= -6.0 m/s m1= 7.5 kg V= +8.0 m/s

13. Momentum, p Momentum is a Vector p = m v p1 = +60 kg m/s p2 = - 60 kg m/s the system momentum is zero., m2= 10.0 kg V= -6.0 m/s m1= 7.5 kg V= +8.0 m/s

14. Momentum, p Momentum is a Vector p = m v Total momentum of a system is a vector sum: p1+p2+p3+…….. p1p1 p2p2 p3p3 p total

15. Newtons 2nd Law Net Force = Mass x Acceleration F = M a F = M (V/t) F t = M V F t = M (V 1 -V 2 ) F t = M V 1 - M V 2 F t = p Impulse= Ft The Impulse = the change in momentum

16. Newtons 2nd Law Net Force = Mass x Acceleration F = M a or F = p/ t

17. Newtons 2nd Law Net Force = Mass x Acceleration F t = p Impulse= F t The Impulse = the change in momentum

18. If M=1500 kg and t=0.4 sec, Find p and Favg

19. 30° 50°

20. Impulse The Impulse = the change in momentum F t = p

21. Impulse The Impulse = the change in momentum F t = p

22. Newtons Law of Action Reaction (3rd Law) For every action force there is and equal and oppositely directed reaction force Ball 1 Ball 2 F 1,2 F 2,1 F 1,2 = - F 2,1

23. Newtons Law of Action Reaction (3rd Law) Ball 1 Ball 2 F 1,2 F 2,1 F 1,2 = - F 2,1 F 1,2 t = - F 2,1 t p 2 = - p 1

24. Conservation of momentum Ball 1 Ball 2 F 1,2 F 2,1 If there are no external forces acting on a system (i.e. only internal action reaction pairs), then the systems total momentum is conserved.

25. 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? M=100.0 kg after before V=? V=8.0 m/s

26. 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? M=100.0 kg after before V=? V=8.0 m/s 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

27. Explosions If V red =8.0 m/s V blue =?

28. 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? M=100.0 kg afterbefore V=? V=8.0 m/s

29. 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? M=100.0 kg afterbefore V=? V=8.0 m/s p before = p after 30kg(8.0 m/s) = 130 kg V 240 kg m/s = 130 kg V V = 1.85 m/s

30. Stick together 2 objects have same speed after colliding This 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? M=100.0 kg afterbefore V=? V=8.0 m/s

31. 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?

32. mv = (m+M) V Before and After Collision 1 / 2 (m+M)V 2 =(m+M)gh After collision but Before and After moving up

34. 2-D Collisions X axis m 1 V 10 = m 1 v 1 cos(50) + m 2 v 2 cos(40) Y axis 0 = m 1 v 1 sin(50) - m 2 v 2 sin(40)

35. 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.

36. 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.

37. 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. V V1 V2 P before =P after P Bx =P Ay & P By =P Ay 2000Kg(50 mi/hr)=3000KgV x & 1000kg(30mi/hr)=3000kgV y V x =33.3 mi/h & V y =10 mi/hr Or V= 34.8mi/hr = (sqrt(V x 2 +V x 2 ) & =16.7° = tan -1 (V y /V x ) 2000Kg 1000Kg 3000Kg

38. Elastic Collisions Bounce off without loss of energy p before = p after & KE before = KE after v1v1 m1m1 m1m1 m2m2 m2m2 v 1,f v 2,f

39. Elastic Collisions Bounce off without loss of energy p before = p after & KE before = KE after

40. Elastic Collisions Bounce off without loss of energy p before = p after & KE before = KE after + or

41. Elastic Collisions Bounce off without loss of energy p before = p after & KE before = KE after - or

42. Elastic Collisions Bounce off without loss of energy p before = p after & KE before = KE after & v1v1 m1m1 m1m1 m2m2 m2m2 v 1,f v 2,f

43. Elastic Collisions Bounce off without loss of energy p before = p after & KE before = KE after & v1v1 m m m m v 1,f = 0.0 v 2,f = v 1 if m1 = m2 = m, then v 1,f = 0.0 & v 2,f = v 1

44. Elastic Collisions Bounce off without loss of energy p before = p after & KE before = KE after & v1v1 m m M v 1,f =- v 1 v 2,f 0.0 if m 1 <<< m 2, then m 1 +m 2 m 2 & m 1 -m 2 -m 2 v 1,f = - v 1 & v 2,f 0.0 M

45. Elastic Collisions Bounce off without loss of energy v1v1 m m M v 1,f =- (v 1 +v 2 +v 2 ) v 2,f v 2 if m 1 <<< m 2 and v 2 is NOT 0.0 M Speed of Approach = Speed of separation (True of all elastic collisions) v2v2

46. Elastic Collisions m m M v 1,f =? if m 1 <<< m 2 and v 2 is NOT 0.0 v 2,f ? M 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? v 1 = 40 km/s v 2 =20 km/s

47. Elastic Collisions m m v 1,f =? if m 1 <<< m 2 and v 2 is NOT 0.0 v 2,f ? 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 trucks speed is 40 m/s and the collision is an elastic head on collision, with what speed does the ball bounce off the truck? v 1 = 20 m/s v 2 =40 m/s M M

48. Elastic Collisions Bounce off without loss of energy p before = p after & KE before = KE after v1v1 m m mm if m1 = m2 = m, then v 1,f = 0.0 & v 2,f = v 1 90°

49. Elastic Collisions Bounce off without loss of energy p before = p after & KE before = KE after v1v1 m m m m if m1 = m2 = m, then v 1,f = 0.0 & v 2,f = v 1 90° p 1 =p 1f +p 2f p1p1 p 2f p 1f 90°

50. Center of Mass The average position of the mass When we use F=ma We really mean F = m a cm The motion of an object is the combination of –The translational motion of the CM –Rotation about the CM

51. Center of mass The average position of the mass

52. CM=? M1=6.0 kg and M2=8.0 kg Ycm= Xcm=

53. CM Center of Gravity

56. 4 kg at (-2,0) 8 kg at (0,3) Where must a 10 kg mass be placed so the center of mass of the three mass system is at (0,0) ?

58. (M+ m)v=M(v+ v)+ m(v-v e ) M v= mv e Thrust=v e ( dM / dt )

59. dM=2.5 kg/s and dt =1.0 sec 5678956789 A B C D

60. The End