Physics for Scientists and Engineers. Chapter 9 Impulse and Momentum


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1 Physcs or Scentsts and Engneers Chapter 9 Impulse and Momentum Sprng, 008 Ho Jung Pak
2 Lnear Momentum Lnear momentum o an object o mass m movng wth a velocty v s dened to be p mv Momentum and lnear momentum wll be used nterchangeably Momentum s a vector quantty SI unts o momentum are kg m/s Momentum can be expressed n component orm: p x mv x, p y mv y, p z mv z 31Mar08 Pak p.
3 Newton s nd Law & Momentum Newton s nd law as Newton presented t: d p d F ( mv) dt dt Ths s a more general orm than the one that we used so ar Newton called m v the quantty o moton For constant mass (.e. dm/dt 0), Newton s nd law becomes d dv F ( mv) m ma dt dt 31Mar08 Pak p. 3
4 Conservaton o Lnear Momentum When two or more partcles n an solated system (.e. no external orces present) nteract, the total momentum o the system remans constant The momentum o the system s conserved, not the momenta o ndvdual partcles Can be expressed mathematcally n varous ways p total p 1 + p constant p 1 + p p 1 + p Conservaton o momentum can be appled to systems wth any number o partcles 31Mar08 Pak p. 4
5 Archer and Arrow An archer s standng on a rctonless surace Let the system be the archer wth bow (partcle 1) and the arrow (partcle ) There are no external orces n the xdrecton, so t s solated n the xdrecton Total momentum beore releasng the arrow s 0 Total momentum ater releasng the arrow s p 1 + p 0 The archer wll move n the opposte drecton o the arrow Agrees wth Newton s 3rd law Because the archer s much more massve than the arrow, hs acceleraton and velocty wll be much smaller than those o the arrow 31Mar08 Pak p. 5
6 Kaon Decay The kaon decays nto a postve π and a negatve π partcle Total momentum beore decay s zero Thereore, the total momentum ater the decay must equal zero: p + + p 0, or p + p 31Mar08 Pak p. 6
7 Impulse and Momentum From Newton s nd Law F dp/dt, dp Fdt Integratng to nd the change n momentum over some tme nterval The ntegral s called the mpulse, I, o the orce F actng on an object over Δt Ths equaton expresses the mpulsemomentum theorem 31Mar08 Pak p. 7
8 More About Impulse Impulse s a vector quantty The magntude o the mpulse s equal to the area under the orcetme curve Dmensons o mpulse are ML/T, wth unts o N s kg m s 1 Impulse can also be ound by usng the tme averaged orce: I FΔt 31Mar08 Pak p. 8
9 Example 1: Deormed Ball A hard rubber ball, released at chest heght, alls to the pavement and bounces back to nearly the same heght. When t s n contact wth the pavement, the lower sde o the ball s temporarly lattened, wth the maxmum dent o 1 cm. Estmate the acceleraton o the ball whle t s n contact wth the pavement and the tme duraton t s deormed. To nd the contact tme, we use the mpulse equaton: I F ave Δt To nd F ave, we need to nd a. To nd a, rst nd the velocty o the ball at the nstant the bottom hts the ground, assumng the ball alls 1.50 m: v v + a( y y ) 0 + ( 9.80 m/s )( 1.50 m 0) v 5.4 m/s p 31Mar08 Pak p. 9 p
10 Example 1, cont We use the same equaton to nd the acceleraton, assumed constant: v Ths s approxmately the average acceleraton. The momentum o the ball goes rom p m(5.4 m/s) to p 0 as t s beng deormed. Usng the mpulse theorem: I Δt F v ave + a( y a 1470 m/s Δt p p F ave p p y ) p p ma ( 5.4 m/s) m( 5.4 m/s) 3 m( m/s ) Snce t bounces back to nearly the same heght, the tme to deorm and undeorm s approxmately twce ths tme. + a( m) 0 s 31Mar08 Pak p. 10
11 Collsons Collson s an event durng whch two partcles come close to each other and nteract by means o orces The tme nterval durng whch the velocty changes rom ts ntal to nal values s assumed to be short The nteracton orce s assumed to be much greater than any external orces present Ths means the mpulse approxmaton can be used 31Mar08 Pak p. 11
12 Example : Car Crash In a crash test, a car o mass 1500 kg colldes wth a wall. The ntal and nal veloctes o the car are v 15.0 m/s and v.60 m/s. The collson lasts s. Fnd the mpulse and average orce on the car. Assume the orce exerted by the wall on the car s greater than any other orces that cause ts velocty to change. p (1500 kg)( 15.0 m/s).5 x 10 4 kg m/s P (1500 kg)(.60 m/s) 0.39 x 10 4 kg m/s I p p.64 x 10 4 kg m/s F ave Δp /Δt (.64 x 10 4 kg m/s)/0.150 s 1.50 x 10 5 N 31Mar08 Pak p. 1
13 Collsons Examples Collsons may be the result o drect contact Impulsve orces may vary n tme n complcated ways Collson need not nclude physcal contact There are stll orces between the partcles Total momentum s always conserved 31Mar08 Pak p. 13
14 Types o Collsons In an elastc collson, momentum and knetc energy are conserved Perectly elastc collsons occur on a mcroscopc level Macroscopc collsons are only approxmately elastc In an nelastc collson, knetc energy s not conserved although momentum s conserved I the objects stck together ater the collson, t s a perectly nelastc collson Elastc and perectly nelastc collsons are lmtng cases; most actual collsons all n between these two types 31Mar08 Pak p. 14
15 Conservaton o Momentum Snce there are no external orces, F Δ t P P 0, P ext In an elastc collson, v + m v m v + m mv P In a perectly nelastc collson, they share the same velocty ater the collson m ( m m ) 1 v 1 v1 + mv + 31Mar08 Pak p. 15
16 Example 3: D Per Inel Collson The car and the van have masses o 1500 kg and 500 kg, respectvely. Beore the collson, the car was movng wth a speed o 5.0 m/s n the xdrecton and the van wth a speed o 0.0 m/s n the ydrecton. The car and the van make a perectly nelastc collson. What s ther velocty ater the collson? Momentum conservaton: x: m 1 v 1x (m 1 +m )v cosθ, 3.75x10 4 kg m/s 4000 v cosθ y: m v y (m 1 +m )v snθ, 5.00x10 4 kg m/s 4000 v snθ tanθ 5.00/3.75, θ 53.1, v 5.00x10 4 /(4000 sn 53.1 ) 15.6 m/s 31Mar08 Pak p. 16
17 Exploson Exploson s an event durng whch partcles o the system move apart rom each other ater a bre nteracton Examples: archer and arrow, recol o rle, partcle decay Exploson s the reverse o the collson problem Total momentum s conserved 31Mar08 Pak p. 17
18 Example 4: Radoactvty A 38 U nucleus spontaneously decays nto a small ragment that s ejected wth a speed o 1.50 x 10 7 m/s and a daughter nucleus that recols wth a speed o.56 x 10 5 m/s. What are the atomc masses o the ejected ragment and the daughter nucleus? The nucleus was ntally at rest so the total momentum remans zero. m v + m v ( m + m ) v 0, m + m Combnng these two conservaton laws, 38 u v m v + ( 38 u m ) v 0, m 38 u 34 u, m v v 31Mar08 Pak p u
19 Rocket Launch Ths M Here R s the For a s an exploson problem where contnues to decrease : M uel rocket gong straght upward, Ru Integratng gves M burn rate. From Newton' s nd law, r r r r ˆ dp d r dv dm r dv r F Mgj ( Mv ) M + u M Ru ext dt dt dt dt dt r r Here u s the exhaust velocty and the quantty Ru s the thrust, Mg + the dv dt or dv dt M rocket equaton 0 Rt. Ru M we have Ru g g M Rt 0 : M 0 v u ln gt M Rt 0 31Mar08 Pak p. 19 r F thrust.
20 Example 5: Saturn V Saturn V, the Apollo launch vehcle, has an ntal mass o M 0.85 x 10 6 kg, a payload mass o M p 0.7M 0, a uel burn rate o R x 10 3 kg/s, and a thrust o F thrust 34.0 x 10 6 N. Fnd (a) the exhaust speed u, (b) burn tme b t, (c) acceleraton at lto and burnout, and (d) nal speed o the rocket. 31Mar08 Pak p. 0
21 (a) Example 5, cont Exhaust speed : u F R thrust Acceleraton at burnout : kg m/s kg/s 9.80 m/s.46 km/s 6 M M 0 P (1 0.7) kg (b) Burn tme : tb 150 s 3 R kg/s dv Ru (c) Acceleraton at lto : 9.80 m/s.14 m/s dt M M 0 (d) Fnal speed at burnout : v u ln 9.80 m/s 0.7 M 0 3 v ( m/s)(1.31) 1470 m/s 1.75 km/s dv dt 34.3 m/s 150 s 31Mar08 Pak p. 1 0 Ru M p
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