Chapter 9 Linear Momentum

1. Chapter 9 Linear Momentum

2. Announcements • Assignments due Saturday • Midterm Exam II: October 17 (chapters 6-9) • Formula sheet will be posted • Practice problems • Practice exam next week

3. Linear Momentum Momentum is a vector; its direction is the same as the direction of the velocity.

4. Momentum and Newton’s Second Law Newton’s second law, as we wrote it before: is only valid for objects that have constant mass. Here is a more general form (also useful when the mass is changing):

5. Impulse The same change in momentum may be produced by a large force acting for a short time, or by a smaller force acting for a longer time. Impulse quantifies the overall change in momentum Impulse is a vector, in the same direction as the average force.

6. Impulse We can rewrite as So we see that The impulse is equal to the change in momentum.

7. Why we don’t dive into concrete The same change in momentum may be produced by a large force acting for a short time, or by a smaller force acting for a longer time.

8. p p Going Bowling II a) the bowling ball b) same time for both c) the Ping-Pong ball d) impossible to say A bowling ball and a Ping-Pong ball are rolling toward you with the same momentum. If you exert the same force to stop each one, which takes a longertimeto bring to rest?

9. p p = F  t av p Going Bowling II a) the bowling ball b) same time for both c) the Ping-Pong ball d) impossible to say A bowling ball and a Ping-Pong ball are rolling toward you with the same momentum. If you exert the same force to stop each one, which takes a longertimeto bring to rest? We know: sop = Fav t Here,Fandp are thesamefor both balls! It will take thesame amount of timeto stop them.

10. p p Going Bowling III a) the bowling ball b) same distance for both c) the Ping-Pong ball d) impossible to say A bowling ball and a Ping-Pong ball are rolling toward you with the same momentum. If you exert the same force to stop each one, for which is the stopping distance greater?

11. p p Going Bowling III a) the bowling ball b) same distance for both c) the Ping-Pong ball d) impossible to say A bowling ball and a Ping-Pong ball are rolling toward you with the same momentum. If you exert the same force to stop each one, for which is the stopping distance greater? Use the work-energy theorem:W = KE. The ball withless masshas thegreater speed,and thus thegreater KE.In order to remove that KE, work must be done, whereW = Fd. Because the force is thesamein both cases, the distance needed to stop theless massive ballmust bebigger.

12. With no net force: Conservation of Linear Momentum The net force acting on an object is the rate of change of its momentum: If the net force is zero, the momentum does not change! • A vector equation • Works for each coordinate separately

13. Internal Versus External Forces Internal forces act between objects within the system. As with all forces, they occur in action-reaction pairs. As all pairs act between objects in the system, the internal forces always sum to zero: Therefore, the net force acting on a system is the sum of the external forces acting on it.

14. With no net external force: Momentum of components of a system Internal forces cannot change the momentum of a system. However, the momenta of pieces of the system may change. An example of internal forces moving components of a system:

15. Kinetic Energy of a System Another example of internal forces moving components of a system: The initial momentum equals the final (total) momentum. But the final Kinetic Energy is very large

16. Birth of the neutrino Beta decay fails momentum conservation? First detection 1956 Pauli “fixes” it with a new ghost-like, undetectable particle Bohr scoffs

17. Lecture 11 Momentum, Energy, and Collisions

18. Impulse Linear Momentum With no net external force:

19. 1 2 Nuclear Fission I a) the heavy one b) the light one c) both have the same momentum d) impossible to say A uranium nucleus (at rest) undergoes fission and splits into two fragments, one heavy and the other light. Which fragment has the greater momentum?

20. 1 2 Nuclear Fission I a) the heavy one b) the light one c) both have the same momentum d) impossible to say A uranium nucleus (at rest) undergoes fission and splits into two fragments, one heavy and the other light. Which fragment has the greater momentum? The initial momentum of the uranium was zero, so the final total momentum of the two fragments must also be zero.Thus the individual momenta are equal in magnitude and opposite in direction.

21. 1 2 Nuclear Fission II a) the heavy one b) the light one c) both have the same speed d) impossible to say A uranium nucleus (at rest) undergoes fission and splits into two fragments, one heavy and the other light. Which fragment has the greater speed?

22. 1 2 Nuclear Fission II a) the heavy one b) the light one c) both have the same speed d) impossible to say A uranium nucleus (at rest) undergoes fission and splits into two fragments, one heavy and the other light. Which fragment has the greater speed? We have already seen that the individual momenta are equal and opposite. In order to keep the magnitude of momentum mv the same, the heavy fragment has the lower speed and thelight fragment has the greater speed.

23. y x v3 A plate drops onto a smooth floor and shatters into three pieces of equal mass. Two of the pieces go off with equal speeds v along the floor, but at right angles to one another. Find the speed and direction of the third piece. looking from above: We know that px=0, py = 0 in initial state and no external forces act in the horizontal v2 v1

24. An 85-kg lumberjack stands at one end of a 380-kg floating log, as shown in the figure. Both the log and the lumberjack are at rest initially. (a) If the lumberjack now trots toward the other end of the log with a speed of 2.7 m/s relative to the log, what is the lumberjack’s speed relative to the shore? Ignore friction between the log and the water. (b) If the mass of the log had been greater, would the lumberjack’s speed relative to the shore be greater than, less than, or the same as in part (a)? Explain.

26. Center of Mass

27. Center of Mass The center of mass of a system is the point where the system can be balanced in a uniform gravitational field. For two objects: The center of mass is closer to the more massive object. Think of it as the “average location of the mass”

28. The center of mass need not be within the object Center of Mass

29. With no net external force: Momentum of components of a system Internal forces cannot change the momentum of a system. RECALL: However, the momenta of pieces of the system may change. An example of internal forces moving components of a system:

30. Motion about the Center of Mass The center of massof a complex or composite object follows a trajectory as if it were a single particle - with mass equal to the complex object, and experiencing a force equal to the sum of all external forces on that complex object

31. Action/Reaction pairs inside the system cancel out Motion of the center of mass

32. The total mass multiplied by the acceleration of the center of mass is equal to the net external force The center of mass accelerates just as though it were a point particle of mass M acted on by

33. Momentum of a composite object

34. Recoil Speed a) 0 m/s b) 0.5 m/s to the right c) 1 m/s to the right d) 20 m/s to the right e) 50 m/s to the right A cannon sits on a stationary railroad flatcar with a total mass of 1000 kg. When a 10-kg cannonball is fired to the left at a speed of 50 m/s, what is the recoil speed of the flatcar?

35. Recoil Speed a) 0 m/s b) 0.5 m/s to the right c) 1 m/s to the right d) 20 m/s to the right e) 50 m/s to the right A cannon sits on a stationary railroad flatcar with a total mass of 1000 kg. When a 10-kg cannonball is fired to the left at a speed of 50 m/s, what is the recoil speed of the flatcar? Because the initial momentum of the system was zero, the final total momentum must also be zero.Thus, the final momenta of the cannonball and the flatcar must be equal and opposite. pcannonball = (10 kg)(50 m/s) = 500 kg-m/s pflatcar = 500 kg-m/s = (1000 kg)(0.5 m/s)

36. Recoil Speed a) 0 m/s b) 0.5 m/s to the right c) 1 m/s to the right d) 20 m/s to the right e) 50 m/s to the right A cannon sits on a stationary railroad flatcar with a total mass of 1000 kg. When a 10-kg cannonball is fired to the left at a speed of 50 m/s, what is the speed of the center of mass?

37. Recoil Speed a) 0 m/s b) 0.5 m/s to the right c) 1 m/s to the right d) 20 m/s to the right e) 50 m/s to the right A cannon sits on a stationary railroad flatcar with a total mass of 1000 kg. When a 10-kg cannonball is fired to the left at a speed of 50 m/s, what is the speed of the center of mass? Internal forces cannot change the motion of the center of mass. The CM was originally motionless at zero, and remains so after the gun is fired.

38. Rolling in the Rain a) speeds up b) maintains constant speed c) slows down d) stops immediately An open cart rolls along a frictionless track while it is raining. As it rolls, what happens to the speed of the cart as the rain collects in it? (Assume that the rain falls vertically into the box.)

39. Rolling in the Rain a) speeds up b) maintains constant speed c) slows down d) stops immediately An open cart rolls along a frictionless track while it is raining. As it rolls, what happens to the speed of the cart as the rain collects in it? (Assume that the rain falls vertically into the box.) Because the rain falls in vertically, it adds no momentum to the box, thus the box’s momentum is conserved. However, because the mass of the box slowly increases with the added rain, its velocity has to decrease.

40. initial final px = mv0 px = (2m)vf mass m mass m Two objects collide... and stick No external forces... so momentum of system is conserved mv0 = (2m)vf vf = v0 / 2 A completely inelastic collision: no “bounce back”

41. Kinetic energy is lost! KEfinal = 1/2 KEinitial mass m mass m Inelastic collision: What about energy? vf = v0 / 2 initial final

42. Collisions This is an example of an “inelastic collision” Collision: two objects striking one another Elastic collision⇔ “things bounce back” ⇔ energy is conserved Inelastic collision: less than perfectly bouncy ⇔ Kinetic energy is lost Time of collision is short enough that external forces may be ignored so momentum is conserved Completely inelastic collision: objects stick together afterwards. Nothing “bounces back”. Maximal energy loss

43. Completely inelastic collision: colliding objects stick together, maximal loss of kinetic energy Elastic collision: momentum and kinetic energy is conserved. Elastic vs. Inelastic Inelastic collision: momentum is conserved but kinetic energy is not

44. Momentum Conservation: Completely Inelastic Collisions in One Dimension Solving for the final momentum in terms of initial velocities and masses, for a 1-dimensional, completely inelastic collision between unequal masses: Completely inelastic only (objects stick together, so have same final velocity) KEfinal < KEinitial

45. Crash Cars I a) I b) II c) I and II d) II and III e) all three If all three collisions below are totally inelastic, which one(s) will bring the car on the left to a complete halt?

46. Crash Cars I a) I b) II c) I and II d) II and III e) all three If all three collisions below are totally inelastic, which one(s) will bring the car on the left to a complete halt? In case I, the solid wall clearly stops the car. In cases II and III, becauseptot = 0 before the collision, thenptot must also be zero after the collision, which means that the car comes to a halt in all three cases.

47. Crash Cars II a) I b) II c) III d) II and III e) all three If all three collisions below are totally inelastic, which one(s) will cause the most damage (in terms of lost energy)?

48. The car on the left loses the same KE in all three cases, but incase III, the car on the right loses the most KE becauseKE = mv2 and the car incase IIIhas thelargest velocity. Crash Cars II a) I b) II c) III d) II and III e) all three If all three collisions below are totally inelastic, which one(s) will cause the most damage (in terms of lost energy)?

49. momentum conservation in inelastic collision PE = (m+M) g h energy conservation afterwards hmax = (mv0)2 / [2 g (m+M)2] KE = 1/2 (mv0)2 / (m+M) vf = m v0 / (m+M) Ballistic pendulum: the height h can be found using conservation of mechanical energy after the object is embedded in the block.

50. approximation Velocity of the ballistic pendulum Pellet Mass (m): 2 g Pendulum Mass (M): 3.81 kg Wire length (L): 4.00 m