General Physics (PHY 2130)

Transcription

1 General Physcs (PHY 2130) Lecture 17 Momentum Momentum and mpulse Conservaton of momentum

2 Lghtnng Revew Last lecture: 1. Work and energy: elastc potental energy momentum Revew Problem: An 8.0 kg crate, ntally at rest, sldes down a slope as n the pcture. The work of the force of frcton s 300 J. What s the speed of the crate at the bottom of the slope? 15 m

3 Revew problem An 8.0 kg crate, ntally at rest, sldes down a slope as n the pcture. The work of the force of frcton s 300 J. What s the speed of the crate at the bottom of the slope? Idea: use energy consderatons wth nonconservatve forces present W nc ΔE E f E 15 m At the top: At the bottom: E mgh + 0 E f 0 + mv2 2 Thus: W nc mv2 2 mgh or v 2 ( m mgh +W nc) 19.2 m / s

4 Recall: Momentum From Newton s laws: force must be present to change an object s velocty (speed and/or drecton) Wsh to consder effects of collsons and correspondng change n velocty Golf ball ntally at rest, so some of the KE of club transferred to provde moton of golf ball and ts change n velocty Method to descrbe s to use concept of lnear momentum Lnear momentum product of mass velocty scalar vector

5 Momentum p mv Vector quantty, the drecton of the momentum s the same as the velocty s Apples to two-dmensonal moton as well p mv and p x x y mv y Sze of momentum: depends upon mass depends upon velocty

6 Impulse In order to change the momentum of an object (say, golf ball), a force must be appled The tme rate of change of momentum of an object s equal to the net force actng on t F net ma m(v f v ) Δt Δ p Δt or : Δ p F netδt Gves an alternatve statement of Newton s second law (F Δt) s called the mpulse Impulse s a vector quantty, the drecton s the same as the drecton of the force

7 Graphcal Interpretaton of Impulse Usually force s not constant, but tme-dependent mpulse F Δ t area under F() t curve Δt If the force s not constant, use the average force appled The average force can be thought of as the constant force that would gve the same mpulse to the object n the tme nterval as the actual tme-varyng force gves n the nterval If force s constant: mpulse F Δt

8 Example: Impulse Appled to Auto Collsons The most mportant factor s the collson tme or the tme t takes the person to come to a rest Ths wll reduce the chance of dyng n a car crash Ways to ncrease the tme Seat belts Ar bags The ar bag ncreases the tme of the collson and absorbs some of the energy from the body

9 Example: A car of mass 1500 kg colldes wth a wall and rebounds as shown. If the collson lasts for s, Fnd (a) the mpulse delvered to the car due to the collson and (b) the magntude and drecton of the average force exerted on the car. Assume force exerted by wall s large compared wth other forces Gravtatonal and normal forces are perpendcular and so do not effect the horzontal momentum Can apply mpulse approxmaton 9 (a) Impulse delvered to the car Δp p p f 4 (1500kg 2.60m / s) (1500kg kg. m / s ( 15.0m / s)) (b) The average force exerted on the car Δp Δt kg. m / s 0.150s 5 F av N

10 ConcepTest Suppose a png-pong ball and a bowlng ball are rollng toward you. Both have the same momentum, and you exert the same force to stop each. How do the tme ntervals to stop them compare? 1. It takes less tme to stop the png-pong ball. 2. Both take the same tme. 3. It takes more tme to stop the png-pong ball.

11 ConcepTest Suppose a png-pong ball and a bowlng ball are rollng toward you. Both have the same momentum, and you exert the same force to stop each. How do the tme ntervals to stop them compare? 1. It takes less tme to stop the png-pong ball. 2. Both take the same tme. 3. It takes more tme to stop the png-pong ball. Note: Because force equals the tme rate of change of momentum, the two balls loose momentum at the same rate. If both balls ntally had the same momenta, t takes the same amount of tme to stop them.

12 Problem: Teeng Off A 50-g golf ball at rest s ht by Bg Bertha club wth 500-g mass. After the collson, golf leaves wth velocty of 50 m/s. a) Fnd mpulse mparted to ball b) Assumng club n contact wth ball for 0.5 ms, fnd average force actng on golf ball

13 Problem: teeng off Gven: mass: m50 g kg velocty: v50 m/s Fnd: 1. Use mpulse-momentum relaton: mpulse Δp mv mv ( kg )( 50 m s) 2.50 kg m 2. Havng found mpulse, fnd the average force from the defnton of mpulse: f s 0 mpulse? F average? Δp F Δt, thus F Δp Δt kg m 3 10 N s s Note: accordng to Newton s 3 rd law, that s also a reacton force to club httng the ball: F Δt F mv mv f f mv + MV R f Δt, or ( ) MV f MV, mv + MV or of club CONSERVATION OF MOMENTUM

14 Conservaton of Momentum Defnton: an solated system s the one that has no external forces actng on t Momentum n an solated system n whch a collson occurs s conserved (regardless of the nature of the forces between the objects) A collson may be the result of physcal contact between two objects Contact may also arse from the electrostatc nteractons of the electrons n the surface atoms of the bodes

15 Conservaton of Momentum The prncple of conservaton of momentum states when no external forces act on a system consstng of two objects that collde wth each other, the total momentum of the system before the collson s equal to the total momentum of the system after the collson

16 Conservaton of Momentum Mathematcally: m1 v1 + m2 v2 m1 v1 f + m2 v2 f Momentum s conserved for the system of objects The system ncludes all the objects nteractng wth each other Assumes only nternal forces are actng durng the collson Can be generalzed to any number of objects

17 Problem: Teeng Off (cont.) Let s go back to our golf ball and club problem: Ball : Δp 2.50 kg m s, m 50 gramm Club : m v Δv ( ) f v 2.50 kg m s, so ( ) 2.50 kg m s v f v 5m s 0.5 kg 50 m s factor of 10 tmes smaller

18 18 More on Conservaton of Momentum v 1 v 2 m 1 >m 2 m 1 m 2 A short tme later the masses collde. m 1 m 2 What happens?

19 19 Durng the nteracton: N 1 N 2 y F 21 F 12 x w 1 w 2 F y N 1 w1 0 F y N 2 w2 0 F x F 21 m a 1 1 F x F 12 m a 2 2 There s no net external force on ether mass.

20 20 Conservaton of Momentum f f f f p p p p p p p p p p t F t F F F Δ Δ Δ Δ If net external force actng on a system s zero, then the momentum of the system s conserved f ext p p F 0, If

21 Example: A rfle has a mass of 4.5 kg and t fres a bullet of 10.0 grams at a muzzle speed of 820 m/s. What s the recol speed of the rfle as the bullet leaves the barrel? 21 Gven: m r 4.5 kg m b 10.0 g 0.01 kg v b 820 m/s v r v b 0 m/s Fnd: v r? Idea: as long as the rfle s horzontal, there wll be no net external force actng on the rfle-bullet system and momentum wll be conserved. p p f 0 m b v b + m r v r v r m " b v b $ m r # 1.82 m/s 0.01 kg 4.5 kg % '820 m/s &

22 ConcepTest Suppose a person jumps on the surface of Earth. The Earth 1. wll not move at all 2. wll recol n the opposte drecton wth tny velocty 3. mght recol, but there s not enough nformaton provded to see f that could happened

23 ConcepTest Suppose a person jumps on the surface of Earth. The Earth 1. wll not move at all 2. wll recol n the opposte drecton wth tny velocty 3. mght recol, but there s not enough nformaton provded to see f that could happened Note: momentum s conserved. Let s estmate Earth s velocty after a jump by a 80-kg person. Suppose that ntal speed of the jump s 4 m/s, then: Person : Δp 320 kg m Earth : V Δp M Earth Earth V Earth 320 kg m 6 10 kg s 320 kg m s, so s m s tny neglgble velocty, n opposte drecton

24 24 Do you know why polce use skd marks to fnd veloctes of vehcles before collson??? See page 246

25 Rocket Propulson The basc equaton for rocket propulson s: v f v v e M ln M M s the ntal mass of the rocket plus fuel M f s the fnal mass of the rocket plus any remanng fuel The speed of the rocket s proportonal to the exhaust speed f

26 Thrust of a Rocket The thrust s the force exerted on the rocket by the ejected exhaust gases The nstantaneous thrust s gven by Δv ΔM Ma M v Δt e Δt The thrust ncreases as the exhaust speed ncreases and as the burn rate (ΔM/Δt) ncreases