Chapter 9. Linear Momentum

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1 Chapter 9 Linear Momentum

2 Linear Momentum Conservation of Linear Momentum Kinetic Energy of a System Collisions Collisions in Center of Mass Reference Frame MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01

3 Momentum From Newton s Laws Consider two interacting bodies with m > m 1 : F 1 F 1 m 1 m If we know the net force on each body then v = a t = Fnet m t The velocity change for each mass will be different if the masses are different. MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 3

4 Momentum Rewrite the previous result for each body as: m v = t 1 1 F1 m v = F1 t = 1 t F From the 3rd Law Combine the two results: m m 1 1 v 1 ( v v ) = m ( v v ) 1 f = m 1 i v f i MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 4

5 Momentum Simplified notation: Before : v, v 1 After : u, u 1 Rewrite the previous results with new notation: m ( v ) ( ) f v i = m v f v i m u v = m u v ( ) ( ) MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 5

6 Define Momentum The quantity (mv) is called momentum (p). p = mv and is a vector. The unit of momentum is kg m/s; there is no derived unit for momentum. MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 6

7 Momentum From a previous slide: m v = m 1 p 1 1 = p v The change in momentum of the two bodies is equal and opposite. Total momentum is conserved during the interaction; the momentum lost by one body is gained by the other. p 1 + p = 0 MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 7

8 Momentum With Calculus p = mv dp d ( mv ) dv = = m = ma dt dt dt dp F net = dt The above is not true, in this form, if the mass is changing, i.e. a rocket using up its fuel. MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 8

9 Center of Mass Momentum P = m v = p = MV sys i i i cm i i dp sys dvcm = M = MA dt dt dp sys F ext = F net ext = i dt If F = 0 then i ext P = m v = MV = Constant sys i i cm i cm MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 9

10 Impulse P = P - P = M V - V = F t ( ) f i f i Avg V i V = 0 m s ; V = -15 m s ; M = 3kg i ( ) f i P = M V - V = 3(-15-0) f ( ) P = 3-35 = -105 kg m s V f Since we are applying the change of momentum over a time interval consisting of the instant before the impact to the instant after the impact, we ignore the effect of gravity over that time period MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 10

11 Impulse and Momentum The force in impulse problems changes rapidly so we usually deal with the average force MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 11

12 Work and Impulse Comparison Work changes E T Work = 0 E T is conserved Work = Force x Distance Impulse changes P Impulse = 0 is conserved Impulse = Force x Time P MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 1

13 Typical Momentum Problems MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 13

14 Generic Zero Initial Momentum Problem p = p + p = 0 i P A p' = p' + p' = 0 i P A These problems occur whenever all the objects in the system are initially at rest and then separate. MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 14

15 Changing the Mass While Conserving the Momentum MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 15

16 Changing the Mass While Conserving the Momentum MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 16

17 Changing the Mass While Conserving the Momentum MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 17

18 Astronaut Tossing How Long Will the Game Last? Assume all astronauts are the same size and mass. Also assume all astronauts have the same strength and push with equal speed. MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 18

19 How Long Will the Game Last? -V0 V0 V 3V + V0 = 0 0 V 0 -V 0 V V 0 V = 0 0 V 3V + V0 = 0 0 MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 19

20 The Ballistic Pendulum Conservation of Energy Conservation of Momentum Bullet remains in the block MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 0

21 The High Velocity Ballistic Pendulum Bullet exits the block MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 1

22 Relative Speed Relationship The relative speed relationship applies to elastic collisions. In elastic collisions the total kinetic energy is also conserved. If a second equation is needed to solve a momentum problem, the relative speed relationship is easier to deal with than the KE equation. MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01

23 Relative Speed Relationship MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 3

24 Relative Speed in Problem Solving Need a second equation when solving for two variables in an elastic collision problem? The relative speed relationship is preferred over the conservation of kinetic energy relationship In the simpler notation: v v = u u ( ) 1 1 MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 4

25 Elastic Collision - Unequal Masses In the later slides, the initially moving mass, m n, will called be m 1 while the initially stationary mass, m c, will be m. MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 5

26 Elastic Collision Velocity Ratios Velocity Ratio Elastic Velocity Ratios U1 U Mass Ratio u v m - 1 m = - m 1+ m 1 u = v m 0 1+ m 1 MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 6

27 Elastic Collision - Fractional KE Loss For equal masses all of the KE is transferred from m 1 to m The curve is like an energy transfer function from m 1 to m. For small m the mass m 1 just rolls on through. For large m the smaller m 1 just bounces off without transferring much energy. Efficient energy transfer requires equal masses. MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 7

28 Elastic Collision - KE Transfer Fractional KE Loss Fractional KE Loss Fractional KE Loss U Velocity Ratio u = v m 0 1+ m f = m 4 m 1 m 1+ m 1 1 Mass Ratio For m > m 1 the velocity of the nd mass gets a smaller velocity and even the larger mass can t overcome the shrinking v factor in KE MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 8

29 Inelastic Collision - Equal Masses v 1 v = 0 m 1 m Before u m 1 + m After Momentum m m v = m + m u; u = v ( ) m 1 + m Kinetic Energy ( ) KE = m v ; KE = m + m u 1 1 i 1 1 f 1 ( ) m 1 1 KE f = m 1 + m v1 m 1 + m 1 m1 m1 KE = m v = KE m + m m + m KE KE f 1 1 i 1 1 m m + m f 1 = i 1 MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 9

30 Elastic and Inelastic Collisions Elastic Collision P Total is conserved Total KE is conserved P Total Inelastic Collision is conserved Total KE is NOT conserved MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 30

31 Collisions in Two and Three Dimensions MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 31

32 Totally Inelastic Collision MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 3

33 Totally Inelastic Collision If the objects stick together after the collision, then the collision is INELASTIC. Just because the objects DON T stick together after the collision, doesn t mean the collision is ELASTIC. MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 33

34 General Glancing Collision Initially only one object is moving MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 34

35 General Glancing Collision Initially only one object is moving MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 35

36 Another 3-Vector Problem Special Case Equal Masses Momentum problems where one of the objects is initially at rest yield 3-vector problems that can be solved using triangles, instead of simultaneous equations. MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 36

37 Center of Mass Reference System in Momentum Problems MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 37

38 Center of Mass Reference System Original reference frame Center-of-mass reference frame MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 38

39 Center of Mass Treatment MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 39

40 Center of Mass Treatment MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 40

41 Center of Mass Treatment MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 41

42 Center of Mass Treatment MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 4

43 Center of Mass Reference System r 1 m 1 R r r m Center of Mass System m r + m r R = m 1 + m r = r - r Laboratory System m r 1 = R - m 1 + m m1 r = R + m + m 1 r r MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 43

44 Combining Conservation of Energy, Momentum, and KE Transfer in Equal Mass Inelastic Collisions MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 44

45 h 1 h Conservation of Energy mg h = v = 1 1 g h mv 1 Conservation of Momentum mv = (m + m)u u = 1 v Conservation of Energy 1 (m)u = mg h u = g h u v g h1 h1 h = = = = g 8g 8g 4 MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 45

46 KE Transfer in Equal Mass Inelastic Collisions MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 46

47 Conservation of Momentum Two objects of identical mass have a collision. Initially object 1 is traveling to the right with velocity v 1 = v 0. Initially object is at rest v = 0. After the collision: Case 1: v 1 = 0, v = v 0 (Elastic) Case : v 1 = ½ v 0, v = ½ v 0 (Inelastic) In both cases momentum is conserved. Case 1 Case MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 47

48 Conservation of Kinetic Energy? Case 1 Case KE After = KE Before (Conserved) KE After = ½ KE Before (Not Conserved) KE = Kinetic Energy = ½mv o MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 48

49 Where Does the KE Go? Case 1 Case KE After = KE Before (Conserved) KE After = ½ KE Before (Not Conserved) In Case each object shares the Total KE equally. Therefore each object has KE = 5% of the original KE Before MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 49

50 50% of the KE is Missing Each rectangle represents KE = ¼ KE Before MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 50

51 Turn Case 1 into Case Each rectangle represents KE = ¼ KE Before Take away 50% of KE. Now the total system KE is correct Use one half of the 50% taken to speed up object 1. Now it has 5% of the initial KE But object has all the KE and object 1 has none Use the other half of the 50% taken to slow down object. Now it has only 5% of the the initial KE Now they share the KE equally and we see where the missing 50% was spent. MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 51

52 Extra Slides MFMcGraw-PHY 45 Chap09Ha-Momentum-Revised-10//01 5

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