Linear Momentum Isolated Systems Nonisolated Systems: Impulse

Linear Momentum Isolated Systems Nonisolated Systems: Impulse Lana Sheridan De Anza College Nov 7, 2016

Last time introduced momentum Newton s Second Law: more general form relation to force relation to Newton s third law conservation of momentum the rocket equation

Overview applying the rocket equation conservation of momentum in isolated systems nonisolated systems impulse average force

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 )

The Rocket Equation For a rocket burning fuel at a constant rate: ( ) mi v f = v i + v e ln m f 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 dt

Example 9.17 A rocket moving in space, far from all other objects, has a speed of 3.0 10 3 m/s relative to the Earth. Its engines are turned on, and fuel is ejected in a direction opposite the rocket s motion at a speed of 5.0 10 3 m/s relative to the rocket. 1 (a) What is the speed of the rocket relative to the Earth once the rocket s mass is reduced to half its mass before ignition? (b) What is the thrust on the rocket if it burns fuel at the rate of 50 kg/s? 3 Serway & Jewett, page 279.

Example 9.17 (a) Speed with half of mass gone? Let initial mass be m i ( ) mi v f = v i + v e ln m f

Example 9.17 (a) Speed with half of mass gone? Let initial mass be m i ( ) mi v f = v i + v e ln m f ( ) v f = (3.0 10 3 ) + (5.0 10 3 mi ) ln (1/2)m i v f = (3.0 10 3 ) + (5.0 10 3 ) ln (2)

Example 9.17 (a) Speed with half of mass gone? Let initial mass be m i ( ) mi v f = v i + v e ln m f ( ) v f = (3.0 10 3 ) + (5.0 10 3 mi ) ln (1/2)m i (b) Thrust if dm dt v f = (3.0 10 3 ) + (5.0 10 3 ) ln (2) v f = 6.5 10 3 m/s = 50 kg/s? F Thrust = v e dm dt

Isolated Systems and Linear Momentum Last time we talked about systems that interact internally, but do not experience external forces. In that case momentum is conserved.

Isolated Systems and Linear Momentum Last time we talked about systems that interact internally, but do not experience external forces. In that case momentum is conserved. Example 9.2 When discussing energy, we ignored the kinetic energy of the Earth when considering the energy of a system consisting of the Earth and a dropped ball. ( K ball + U g = 0.) Verify this is a reasonable thing to do.

Isolated Systems and Linear Momentum Example 9.2 When discussing energy, we ignored the kinetic energy of the Earth when considering the energy of a system consisting of the Earth and a dropped ball. ( K ball + U g = 0.) Verify this is a reasonable thing to do. Need to argue that K Earth << K ball. Momentum is conserved when we drop a ball, if we include the Earth in our system: p b + p E = 0. p b = p E

Isolated Systems and Linear Momentum Need to argue that K Earth << K ball.

Isolated Systems and Linear Momentum Need to argue that K Earth << K ball. This is equivalent to saying K Earth K ball << 1.

Isolated Systems and Linear Momentum Need to argue that K Earth << K ball. This is equivalent to saying K Earth K ball << 1. K Earth K ball = 1 2 m E v 2 E,f 1 2 m bv 2 b,f = m ( ) 2 E ve,f m b v b,f = m ( ) 2 E mb m b m E

Conservation of Linear Momentum alysis Review Model: from Isolated lastsystem time: (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.

Nonisolated Systems What happens when the system is not isolated: when external forces act? The momentum of the system will change.

Nonisolated Systems What happens when the system is not isolated: when external forces act? The momentum of the system will change. F ext = dp sys dt

Nonisolated Systems What happens when the system is not isolated: when external forces act? The momentum of the system will change. How will it change? F ext = dp sys dt

Impulse & Nonisolated Systems Generalized Newton s second law: where F net = i F i. How will it change? F net = dp dt

Impulse & Nonisolated Systems Generalized Newton s second law: where F net = i F i. How will it change? F net = dp dt p f = F net dt +p i This change in momentum is called the impulse: I = p = F net dt where F net is the net external force

Nonisolated Systems ed a system s nonisolated. ual to the um theorem: (9.13) System boundary Impulse Momentum The change in the total momentum of the system is equal to the total impulse on the system.

the same set of choices. (iii) If the system system Connection from the same to set of Energy choices. case (Momentum) yzed and have defined a system system, the system is nonisolated. of the Work system is equal Heat to the e impulse momentum theorem: al Kinetic energy Potential energy Internal energy Mechanical waves System boundary (9.13) Impulse Momentum e m. Matter transfer Electrical Electromagnetic transmission radiation The change in the total momentum of the system is equal to the total impulse on the system. Energy transfers cross the boundary W = F dr umpers? AM Impulse crosses the boundary I = F dt = p

Impulse Since impulse is simply the change in momentum, it has the same units as momentum, kg m s 1. For a force applied over a time interval t = t f t i : I = tf t i F net dt

Impulse Since impulse is simply the change in momentum, it has the same units as momentum, kg m s 1. For a force applied over a time interval t = t f t i : I = tf t i F net dt We can also relate the impulse to the average net force applied on this time interval. I = F net,avg t and we can find the average force using the definition of the average value of a function:

same as the direction of the change in momentum. Impulse has the dimensions of momentum, that is, ML/T. Impulse is not a property of a particle; rather, it is a measure of the degree to which an external force changes the particle s momentum. Because the net force imparting an impulse F net,avg = 1 to a particle can generally vary in time, it is convenient to define a time-averaged tf net force: F net dt t 1 S t i a F 2 avg ; 1 t f Dt 3 a FS dt (9.10) t i Impulse The impulse imparted to the particle by the force is the area under the curve. F The time-averaged net force gives the same impulse to a particle as does the timevarying force in (a). F t i a tf ( F ) avg t t t f t i t f b F net dt = I = F net,avg t 2 Here we are integrating force with respect to time. Compare this strategy with our efforts in Chapter 7, where we t integrated force with respect to position to i find the work done by the force. Fig ing (b) for lin (o sam in (

Example 9.3.3 How Good Are the Bumpers? lar crash In a particular test, a car crash of mass test, 1 a car 500 of kg mass colwall 1500 as shown kg collides in Figure with a 9.4. wall. The The initial initial locities and final of the velocities car are of S v the car are: i 5215.0 i^ m/s 0 i^ m/s, respectively. If the collision lasts the impulse caused v i = 15.0 by the i m/s collision and et force exerted and von f = the 2.60 car. i m/s. If the collision lasts 0.150 s, find the impulse caused by the collision and the ze The average collision net force time exerted is short, on the so we car. can car being brought to rest very rapidly ving back in the opposite direction with eed. Let us assume the net force exerted on 1 Serway & Jewett, page 255. AM a 15.0 m/s +2.60 m/s Before After Hyundai Motors/HO/Landov Figure 9.4 (Example 9.3) (a) Th result of its collision with the wall.

Example 9.3 Impulse? I = p

Example 9.3 Impulse? I = p = m(v f v f ) = (1500 kg)(2.60 ( 15.0) m/s) i = 2.64 10 4 i kg m/s

Example 9.3 Impulse? I = p = m(v f v f ) = (1500 kg)(2.60 ( 15.0) m/s) i = 2.64 10 4 i kg m/s Average net force?

Example 9.3 Impulse? I = p Average net force? = m(v f v f ) = (1500 kg)(2.60 ( 15.0) m/s) i = 2.64 10 4 i kg m/s F net,avg = I t

Example 9.3 Impulse? I = p Average net force? = m(v f v f ) = (1500 kg)(2.60 ( 15.0) m/s) i = 2.64 10 4 i kg m/s F net,avg = I t = 2.46 104 i kg m/s 0.150 s = 1.76 10 5 i N

Question and (g) there is no motion beforehand and plenty of motion afterward? Section 9.3 Analysis Model: Nonisolated System (Momentum) 12. A man claims that he can hold onto a 12.0-kg child in a Q/C head-on collision as long as he has his seat belt on. Consider this man in a collision in which he is in one of two identical cars each traveling toward the other at 60.0 mi/h relative to the ground. The car in which he rides is brought to rest in 0.10 s. (a) Find the magnitude of the average force needed to hold onto the child. (b) Based on your result to part (a), is the man s claim valid? (c) What does the answer to this problem say about laws requiring the use of proper safety devices such as seat belts and special toddler seats? 13. An estimated force W time curve for a baseball struck by a bat is shown in Figure P9.13. From F (N) 20 000 15 000 10 000 F max = 18 000 N w op to on th co su 19. T fo ti va Fi im th (b p n ve it th

Question (a) Average force to hold child:

Question (a) Average force to hold child: F net,avg = I t

Question (a) Average force to hold child: F net,avg = I t = m v f v i t (12)(60 mi / h)(1609 m/mi) = (0.10 s)(3600 s/h) = 3.2 10 3 N

Question (a) Average force to hold child: F net,avg = I t = m v f v i t (12)(60 mi / h)(1609 m/mi) = (0.10 s)(3600 s/h) = 3.2 10 3 N (b) man s claim?

Question (a) Average force to hold child: F net,avg = I t = m v f v i t (12)(60 mi / h)(1609 m/mi) = (0.10 s)(3600 s/h) = 3.2 10 3 N (b) man s claim? It seems unlikely that he will be able to exert 3200 N of force on the child.

Summary using the rocket equation nonisolated systems impulse average force (Uncollected) Homework Serway & Jewett, Read along in Chapter 9. PREV: Ch 9, onward from page 283. Probs: 1, 3, 5, 7, 61, 63 Ch 9, onward from page 284. Probs: 13, 15, 17, 19