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1 Chapter 9 Linear Momentum. Announcements Assignments due Saturday Midterm Exam II: October 17 (chapters 6-9) Formula sheet will be posted Practice problems.

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Presentation on theme: "1 Chapter 9 Linear Momentum. Announcements Assignments due Saturday Midterm Exam II: October 17 (chapters 6-9) Formula sheet will be posted Practice problems."— Presentation transcript:

1 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 Impulse is a vector, in the same direction as the average force. 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

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 Going Bowling II p p 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 longer time to bring to rest?

9 Going Bowling II We know: Here, F and  p are the same for both balls! It will take the same amount of time to stop them. p p so  p = F av  t 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 longer time to bring to rest? av  t pp F 

10 Going Bowling III p p 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? a) the bowling ball b) same distance for both c) the Ping-Pong ball d) impossible to say

11 Going Bowling III p p Use the work-energy theorem: W =  KE. The ball with less mass has the greater speed, and thus the greater KE. In order to remove that KE, work must be done, where W = Fd. Because the force is the same in both cases, the distance needed to stop the less massive ball must be bigger. 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? a) the bowling ball b) same distance for both c) the Ping-Pong ball d) impossible to say

12 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 With no net force:

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 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: With no net external force:

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

17 Lecture 11 Momentum, Energy, and Collisions

18 Linear Momentum Impulse With no net external force:

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

20 1 2 Nuclear Fission I A uranium nucleus (at rest) undergoes fission and splits into two fragments, one heavy and the other light. Which fragment has the greater momentum? a) the heavy one b) the light one c) both have the same momentum d) impossible to say 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 Nuclear Fission II a) the heavy one b) the light one c) both have the same speed d) impossible to say 1 2 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 Nuclear Fission II 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 the light fragment has the greater speed. a) the heavy one b) the light one c) both have the same speed d) impossible to say 1 2 A uranium nucleus (at rest) undergoes fission and splits into two fragments, one heavy and the other light. Which fragment has the greater speed?

23 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. We know that p x =0, p y = 0 in initial state and no external forces act in the horizontal looking from above: x y v1v1 v2v2 v3v3

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.

25

26 Center of Mass

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

29 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: With no net external force: RECALL:

30 Motion about the Center of Mass The center of mass of 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 Motion of the center of mass Action/Reaction pairs inside the system cancel out

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 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. p cannonball = (10 kg)(50 m/s) = 500 kg-m/s p flatcar = 500 kg-m/s = (1000 kg)(0.5 m/s) 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?

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 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. 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.)

40 Two objects collide... and stick A completely inelastic collision: no “bounce back” No external forces... so momentum of system is conserved initialp x = mv 0 finalp x = (2m)v f mv 0 = (2m)v f v f = v 0 / 2 mass m

41 Inelastic collision: What about energy? mass m v f = v 0 / 2 Kinetic energy is lost! KE final = 1/2 KE initial 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 Elastic vs. Inelastic Inelastic collision: momentum is conserved but kinetic energy is not Completely inelastic collision: colliding objects stick together, maximal loss of kinetic energy Elastic collision: momentum and kinetic energy is conserved.

44 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) KE final < KE initial Momentum Conservation:

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 In case I, the solid wall clearly stops the car. In cases II and III, because p tot = 0 before the collision, then p tot must also be zero after the collision, which means that the car comes to a halt in all three cases. 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?

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

48 Crash Cars II a) I b) II c) III d) II and III e) all three The car on the left loses the same KE in all three cases, but in case III, the car on the right loses the most KE because KE = mv 2 and the car in case III has the largest velocity. If all three collisions below are totally inelastic, which one(s) will cause the most damage (in terms of lost energy)?

49 Ballistic pendulum: the height h can be found using conservation of mechanical energy after the object is embedded in the block. v f = m v 0 / (m+M) momentum conservation in inelastic collision KE = 1/2 (mv 0 ) 2 / (m+M) energy conservation afterwards PE = (m+M) g h h max = (mv 0 ) 2 / [2 g (m+M) 2 ]

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

51 Inelastic Collisions in 2 Dimensions Energy is not a vector equation: there is only 1 conservation of energy equation Momentum is a vector equation: there is 1 conservation of momentum equation per dimension For collisions in two dimensions, conservation of momentum is applied separately along each axis:

52 Elastic Collisions In elastic collisions, both kinetic energy and momentum are conserved. One-dimensional elastic collision:

53 Elastic Collisions in 1-dimension solving for the final speeds: Note: relative speed is conserved for head-on (1-D) elastic collision We have two equations: conservation of momentumconservation of energy and two unknowns (the final speeds). For special case of v 2i = 0

54 Limiting cases of elastic collisions note: relative speed conserved

55 Limiting cases note: relative speed conserved

56 Limiting cases note: relative speed conserved

57 Toy Pendulum Could two balls recoil and conserve both momentum and energy? Incompatible!

58 Elastic Collisions II v v m M Carefully place a small rubber ball (mass m) on top of a much bigger basketball (mass M) and drop these from the same height h so they arrive at the ground with the speed v. What is the velocity of the smaller ball after the basketball hits the ground, reverses direction, and then collides with the small rubber ball? a) zero b) v c) 2v d)3v e) 4v

59 Remember that relative velocity has to be equal before and after collision! Before the collision, the basketball bounces up with v and the rubber ball is coming down with v, so their relative velocity is –2v. After the collision, it therefore has to be +2v!! Elastic Collisions II v v v v 3v v (a)(b)(c) m M Carefully place a small rubber ball (mass m) on top of a much bigger basketball (mass M) and drop these from the same height h so they arrive at the ground with the speed v. What is the velocity of the smaller ball after the basketball hits the ground, reverses direction, and then collides with the small rubber ball? a) zero b) v c) 2v d)3v e) 4v

60 Elastic Collisions I v 2 v 1 at rest a) situation 1 b) situation 2 c) both the same Consider two elastic collisions: 1) a golf ball with speed v hits a stationary bowling ball head-on. 2) a bowling ball with speed v hits a stationary golf ball head-on. In which case does the golf ball have the greater speed after the collision?

61 Remember that the magnitude of the relative velocity has to be equal before and after the collision! Elastic Collisions I v 1 In case 1 the bowling ball will almost remain at rest, and the golf ball will bounce back with speed close to v. v 2 2v2v In case 2 the bowling ball will keep going with speed close to v, hence the golf ball will rebound with speed close to 2v. a) situation 1 b) situation 2 c) both the same Consider two elastic collisions: 1) a golf ball with speed v hits a stationary bowling ball head-on. 2) a bowling ball with speed v hits a stationary golf ball head-on. In which case does the golf ball have the greater speed after the collision?


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