Linear Momentum. Lana Sheridan. Nov 6, De Anza College

Transcription

1 Linear Momentum Lana Sheridan De Anza College Nov 6, 2017

2 Last time energy practice

3 Overview introducing momentum Newton s Second Law: more general form relation to force relation to Newton s third law conservation of momentum the rocket equation

4 Linear Momentum Introduce a new quantity: For a particle, linear momentum, p, is product of the particle s mass with its velocity. p = mv

5 Linear Momentum Introduce a new quantity: For a particle, linear momentum, p, is product of the particle s mass with its velocity. p = mv It is a vector.

6 Linear Momentum Introduce a new quantity: For a particle, linear momentum, p, is product of the particle s mass with its velocity. p = mv It is a vector. Newton called it the quantity of motion.

7 Linear Momentum Introduce a new quantity: For a particle, linear momentum, p, is product of the particle s mass with its velocity. p = mv It is a vector. Newton called it the quantity of motion. Units:

8 Linear Momentum Introduce a new quantity: For a particle, linear momentum, p, is product of the particle s mass with its velocity. p = mv It is a vector. Newton called it the quantity of motion. Units: kg m/s

9 Momentum vs. Kinetic energy Momentum: p = mv Kinetic energy: K = 1 2 mv 2 = p2 2m

10 Momentum vs. Kinetic energy Momentum: p = mv Kinetic energy: K = 1 2 mv 2 = p2 2m Both depend on m and v only. K is a scalar. p is a vector. In a system with many particles, can have i p i = 0 even if i K i 0.

11 Question Quick Quiz Two objects have equal kinetic energies. How do the magnitudes of their momenta compare? (A) p 1 < p 2 (B) p1 = p2 (C) p1 > p2 (D) not enough information to tell 1 From Serway & Jewett, page 250.

12 Question Quick Quiz Two objects have equal kinetic energies. How do the magnitudes of their momenta compare? (A) p 1 < p 2 (B) p1 = p2 (C) p1 > p2 (D) not enough information to tell 1 From Serway & Jewett, page 250.

13 Question Quick Quiz Your physical education teacher throws a baseball to you at a certain speed and you catch it. The teacher is next going to throw you a medicine ball whose mass is ten times the mass of the baseball. You are given the following choices: You can have the medicine ball thrown with (i) the same speed as the baseball (ii) the same momentum, or (iii) the same kinetic energy. Rank these choices from easiest to hardest to catch, in terms of work done to catch the ball. (A) (i), (ii), (iii) (B) (iii), (ii), (i) (C) (ii), (iii), (i) (D) all are equivalent 2 From Serway & Jewett, page 250.

14 Question Quick Quiz Your physical education teacher throws a baseball to you at a certain speed and you catch it. The teacher is next going to throw you a medicine ball whose mass is ten times the mass of the baseball. You are given the following choices: You can have the medicine ball thrown with (i) the same speed as the baseball (ii) the same momentum, or (iii) the same kinetic energy. Rank these choices from easiest to hardest to catch, in terms of work done to catch the ball. (A) (i), (ii), (iii) (B) (iii), (ii), (i) (C) (ii), (iii), (i) (D) all are equivalent 2 From Serway & Jewett, page 250.

15 Conservation of Linear Momentum alysis Model: Isolated System (Momentum) 251 For an isolated system, ie. a system with no external forces, total linear momentum is conserved. efined a the system h is the served: (9.5) System boundary Momentum If no external forces act on the system, the total momentum of the system is constant. 1 Figures from Serway & Jewett.

16 Conservation of Linear Momentum For an isolated system, ie. a system with no external forces, total linear momentum is conserved: ( ) d p i = 0 i

17 Conservation of Linear Momentum For an isolated system, ie. a system with no external forces, total linear momentum is conserved: ( ) d p i = 0 i This corresponds to a translational symmetry in the equations of motion.

18 Conservation of Linear Momentum For an isolated system, ie. a system with no external forces, total linear momentum is conserved: ( ) d p i = 0 i This corresponds to a translational symmetry in the equations of motion. (Note: before when speaking of energy isolated meant not exchanging energy now, for momentum, it means, no external forces act on the system.)

19 Force and Momentum: Newton s Second Law Newton s second law for a particle whose mass might change (more general!): F = dp

20 Force and Momentum: Newton s Second Law Newton s second law for a particle whose mass might change (more general!): F = dp This is how Newton thought of his second law.

21 Force and Momentum: Newton s Second Law Newton s second law for a particle whose mass might change (more general!): F = dp This is how Newton thought of his second law. Also F avg = p t

22 Force and Momentum This is the definition to use when mass is changing with time. F = dp F = d(mv)

23 Force and Momentum This is the definition to use when mass is changing with time. F = dp F = d(mv) F = m dv dm +v

24 Force and Momentum This is the definition to use when mass is changing with time. If m is constant dm F = dp F = d(mv) F = m dv dm +v = 0, and this reduces to: F = m a

25 , you compare one particular j^ m/s. Then, a ject for 5.00 s. ity, using the late its accelerulate its accelobject s vector Find the work. (f) Find the f. (g) Find the State the result nd (c), and the Generalized Newton s second law example #94, page Q/C Sand from a stationary hopper falls onto a moving conveyor belt at the rate of 5.00 kg/s as shown in Figure P9.94. The conveyor belt is supported by fric- Sand from ationless stationary rollers hopper and moves fallsat onto a constant a moving speed conveyor of v 5 belt at the rate of kg/s m/s under as shown. the action The of conveyor a constant belt horizontal is supported by external force S Fext supplied by the motor that drives frictionless rollers the belt. and Find moves (a) the atsand s a constant rate of change speedof ofmomen- tum in ofthe a constant horizontal horizontal direction, (b) external the force force of fric- F ext v = m/s under the action tion exerted by the belt on the sand, (c) the external supplied by the motor that drives the belt. Find (a) the sand s rate force S Fext, (d) the work done by S Fext in 1 s, and (e) the of change ofkinetic momentum energy inacquired the horizontal by the falling direction, sand (b) each the force of friction exerted second by due the to belt the change on thein sand, its horizontal and (c) motion. the external (f) Why are the answers to parts (d) and (e) different? force F ext. kes a perfectly 1.00-g object. r the collision. r the collision 0.0 g. (c) Find 2.00-g particle ) and (b). In e more kinetic its initial direction. (a) Find the final speeds of the two particles in terms of v i. (b) What is the angle u at which the particle 3m is scattered? v S F ext ity of Oregon technique of Figure P On a horizontal air track, a glider of mass m carries

26 Generalized Newton s second law example (a) the sand s rate of change of momentum in the horizontal direction dp = d(mv) 0 dv = m + v dm = v dm

27 Generalized Newton s second law example (a) the sand s rate of change of momentum in the horizontal direction dp = d(mv) 0 dv = m + v dm = v dm = (0.750 m/s)(5.00 kg/s) = 3.75 N

28 Newton s Third Law and Conservation of Momentum m 1 S F 21 m 2 S v 1 S F 12 Figure 9.1 Two particles interact with each other. According to Newton s third law for two interacting particles: Newton s third law, we must have S F 12 52F S 21. F 21 = F 12 S v 2 A 60-kg archer stands at horizontally at 85 m/s. W after firing the arrow? From Newton s third law, we is paired with a force in the force causes the archer to sl problem. We cannot determ under constant acceleration eration of the archer. We ca force because we don t know are of no help because we k string back or the elastic pot Despite our inability to s this problem is very simple motion, linear momentum. To of two particles (Fig. 9.1) wit an instant of time. Because that from the other particle force) acts on particle 2, th opposite in direction that particles form a Newton s th express this condition as

29 Newton s Third Law and Conservation of Momentum m 1 S F 21 m 2 S v 1 S F 12 S v 2 Figure 9.1 Two particles interact with each other. According to Newton s third law, we must have S F 12 52F S 21. Newton s third law for two interacting particles: F 21 = F 12 dp 1 = dp 2 applying a conservation principle, conservation of energy. Let us conside situation and see if we can solve it with the models we have developed so f A 60-kg archer stands at rest on frictionless ice and fires a kg horizontally at 85 m/s. With what velocity does the archer move across after firing the arrow? From Newton s third law, we know that the force that the bow exerts on is paired with a force in the opposite direction on the bow (and the arc force causes the archer to slide backward on the ice with the speed reques problem. We cannot determine this speed using motion models such as th under constant acceleration because we don t have any information about eration of the archer. We cannot use force models such as the particle un force because we don t know anything about forces in this situation. Ener are of no help because we know nothing about the work done in pulling string back or the elastic potential energy in the system related to the taut b Despite our inability to solve the archer problem using models learn this problem is very simple to solve if we introduce a new quantity that motion, linear momentum. To generate this new quantity, consider an isolat of two particles (Fig. 9.1) with masses m 1 and m 2 moving with velocities S v 1 an instant of time. Because the system is isolated, the only force on one that from the other particle. If a force from particle 1 (for example, a gra force) acts on particle 2, there must be a second force equal in magn opposite in direction that particle 2 exerts on particle 1. That is, the forc particles form a Newton s third law action reaction pair, and S F12 52F S 2 express this condition as S F21 1 S F From a system point of view, this equation says that if we add up the forc particles in an isolated system, the sum is zero. Let us further analyze this situation by incorporating Newton s secon the instant shown in Figure 9.1, the interacting particles in the system h erations corresponding to the forces on them. Therefore, replacing the each particle with ma S for the particle gives m 1a S 1 1 m 2a S Now we replace each acceleration with its definition from Equation 4.5: m 1 d vs 1 1 m d vs If the masses m 1 and m 2 are constant, we can bring them inside the deriva ation, which gives d 1m 1 v S d 1m S 2 v d 1m 1 v S 1 1 m 2 v S Notice that the derivative of the sum m 1 v S 1 1 m 2 v S 2 with respect to tim Consequently, this sum must be constant. We learn from this discussion

30 Newton s Third Law and Conservation of Momentum m 1 S F 21 m 2 S v 1 S F 12 S v 2 Figure 9.1 Two particles interact with each other. According to Newton s third law, we must have S F 12 52F S 21. Newton s third law for two interacting particles: F 21 = F 12 dp 1 applying a conservation principle, conservation of energy. Let us conside situation and see if we can solve it with the models we have developed so f A 60-kg archer stands at rest on frictionless ice and fires a kg horizontally at 85 m/s. With what velocity does the archer move across after firing the arrow? From Newton s third law, we know that the force that the bow exerts on is paired with a force in the opposite direction on the bow (and the arc force causes the archer to slide backward on the ice with the speed reques problem. We cannot determine this speed using motion models such as th under constant acceleration because we don t have any information about eration of the archer. We cannot use force models such as the particle un force because we don t know anything about forces in this situation. Ener are of no help because we know nothing about the work done in pulling string back or the elastic potential energy in the system related to the taut b Despite our inability to solve the archer problem using models learn this problem is very simple to solve if we introduce a new quantity that motion, linear momentum. To generate this new quantity, consider an isolat of two particles (Fig. 9.1) with masses m 1 and m 2 moving with velocities S v 1 an instant of time. Because the system is isolated, the only force on one that from the other particle. If a force from particle 1 (for example, a gra force) acts on particle 2, there must be a second force equal in magn opposite in direction that particle 2 exerts on particle 1. That is, the forc particles form a Newton s third law action reaction pair, and S F12 52F S 2 express this condition as S F21 1 S F From a system point of view, this equation says that if we add up the forc particles in an isolated system, the sum is zero. Let us further analyze this situation by incorporating Newton s secon the instant shown in Figure 9.1, the interacting particles in the system h erations corresponding to the forces on them. Therefore, replacing the each particle with ma S for the particle gives = dp 2 d (p 1 + p 2 ) = 0 m 1a S 1 1 m 2a S Now we replace each acceleration with its definition from Equation 4.5: m 1 d vs 1 1 m d vs If the masses m 1 and m 2 are constant, we can bring them inside the deriva ation, which gives d 1m S 1 v 1 Implies: 2 p total = p 1 + p 2 does not change with time. Or, p total = 0. 1 d 1m S 2 v d 1m S 1 v 1 1 m S 2 v Notice that the derivative of the sum m 1 v S 1 1 m 2 v S 2 with respect to tim Consequently, this sum must be constant. We learn from this discussion

31 Newton s Third Law and Conservation of Momentum Newton s third law conservation of momentum No external forces (only internal action-reaction pairs): p total = 0

32 The Rocket Equation A case where the mass is changing. A famous example: what happens to a rocket as it burns fuel. ( The rocket equation )

33 The Rocket Equation Main idea: conserve momentum between the rocket and the ejected propellant. Suppose that the rocket burns fuel at a steady rate with respect to time, and let the initial mass of the rocket and propellent be m i (at time t i ) and the final mass be m f (at t f ). Newton s third law: F rocket = F exhaust 2 Figure by Wikipedia user Skorkmaz.

34 The Rocket Equation Consider this interaction in the frame of the rocket and suppose the force F r on the rocket is constant with time. Newton s Second Law for the rocket: F r = dp = m dv dm +v

35 The Rocket Equation Consider this interaction in the frame of the rocket and suppose the force F r on the rocket is constant with time. Newton s Second Law for the rocket: F r = dp = m dv + v dm 0 In the rocket frame, the rocket is not moving (v = 0).

36 The Rocket Equation Consider this interaction in the frame of the rocket and suppose the force F r on the rocket is constant with time. Newton s Second Law for the rocket: F r = dp = m dv + v dm 0 In the rocket frame, the rocket is not moving (v = 0). Newton s second law for the exhaust: dv e F e = m e + v dm e e

37 The Rocket Equation Consider this interaction in the frame of the rocket and suppose the force F r on the rocket is constant with time. Newton s Second Law for the rocket: F r = dp = m dv + v dm 0 In the rocket frame, the rocket is not moving (v = 0). Newton s second law for the exhaust: dv F e = m 0 e e + v dm e e Constant force constant exhaust velocity in the frame of the rocket. ( dv e = 0)

38 The Rocket Equation So, if v is the velocity of the rocket F e = F r dm e v e = m dv

39 The Rocket Equation So, if v is the velocity of the rocket F e = F r dm e v e = m dv but dm e = dm since the rate mass in ejected is equal to the rate of loss of rocket mass: v e dm = m dv

40 The Rocket Equation So, if v is the velocity of the rocket F e = F r dm e v e = m dv but dm e = dm since the rate mass in ejected is equal to the rate of loss of rocket mass: v e dm = m dv

41 The Rocket Equation So, if v is the velocity of the rocket F e = F r dm e v e = m dv but dm e = dm since the rate mass in ejected is equal to the rate of loss of rocket mass: v e dm = m dv

42 The Rocket Equation So, if v is the velocity of the rocket F e = F r dm e v e = m dv but dm e = dm since the rate mass in ejected is equal to the rate of loss of rocket mass: v e dm = m dv Consider the x-axis to drop the vector notation, noticing v e points in the opposite direction of v and dv. Therefore, dm v e = m dv

43 The Rocket Equation Rocket s change in velocity, v? dv = v e m dm

44 The Rocket Equation Rocket s change in velocity, v? dv = v e m v = tf t i dm v e dm m(t)

45 The Rocket Equation Rocket s change in velocity, v? dv = v e m v = tf t i mf = m i dm v e dm m(t) v e m dm

46 The Rocket Equation Rocket s change in velocity, v? dv = v e m v = tf t i mf = m i dm v e dm m(t) v e m dm [ ] mf = v e ln(m) m i ( ) mi v = v e ln m f

47 The Rocket Equation For a rocket burning fuel at a constant rate: ( ) mi v f = v i + v e ln m f where m i is the initial mass of the rocket and m f is the final (smaller) mass.

48 The Rocket Equation For a rocket burning fuel at a constant rate: ( ) mi v f = v i + v e ln m f where m i is the initial mass of the rocket and m f is the final (smaller) mass. The thrust on an object is the forward force on the object generated by engines / a propulsion system. For a contant velocity of the exhaust, v e : Thrust = v e dm

49 The Rocket Equation The last slide gives the most commonly stated version of the rocket equation. The vector form of these equations are: (Both for a rocket burning fuel at a constant rate, constant exhaust velocity) ( ) mi v f = v i v e ln m f The minus sign indicates that the exhaust velocity is opposite to the direction of the acceleration of the rocket. F Thrust = v e dm The direction of the exhaust velocity is opposite to the direction of the acceleration of the rocket, but the rocket s mass is decreasing, is negative. so dm

50 Summary linear momentum momentum in an isolated system conservation of momentum rocket equation (Uncollected) Homework Serway & Jewett, Read the first sections of Chapter 9, read ahead about non-isolated systems. Ch 9, onward from page 283. Probs: 1, 3, 5, 7, 61, 63