Ground Rules. PC1221 Fundamentals of Physics I. Linear Momentum, cont. Linear Momentum. Lectures 17 and 18. Linear Momentum and Collisions


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1 PC Fundamentals of Physcs I Lectures 7 and 8 Lnear omentum and Collsons Dr Tay Seng Chuan Ground Rules Swtch off your handphone and pager Swtch off your laptop computer and keep t No talkng whle lecture s gong on No gosspng whle the lecture s gong on Rase your hand f you have queston to ask Be on tme for lecture Be on tme to come back from the recess break to contnue the lecture Brng your lecturenotes to lecture Lnear omentum The lnear momentum of a partcle or an object that can be modeled as a partcle of mass m movng wth a velocty v s defned to be the product of the mass and velocty: p = m v The terms momentum and lnear momentum wll be used nterchangeably n ths course, e, when we say momentum we also means lnear momentum (whch s n a straght lne) 3 Lnear omentum, cont Lnear momentum s a vector quantty Its drecton s the same as the drecton of v The dmensons of momentum (mass x velocty) are L/T The SI unts of momentum are kg m / s omentum can be expressed n component form (small letter p): omentum n x drecton p x = m v x p y = m v y p z = m v z omentum n y drecton omentum n z drecton 4
2 Newton and omentum Newton called the product mv the quantty of moton of the partcle Newton s Second Law can be used to relate the momentum of a partcle to the resultant force actng on t dv d( mv) dp Σ F= ma= m = = dt dt dt wth constant mass 5 ( v) dv d m dp Σ F= ma= m = = dt dt dt The tme rate of change of the lnear momentum of a partcle s equal to the net force actng on the partcle Ths s the form n whch Newton presented the Second Law It s a more general form than the one we used prevously Ths form also allows for mass changes omentum approach can be used to analyse the moton n a system of partcles 6 Conservaton of Lnear omentum Whenever two or more partcles n an solated system nteract, the total momentum of the system remans constant The momentum of the system s conserved, but the momentum of ndvdual partcle may not necessarly conserved. The total momentum of an solated system equals ts ntal momentum 7 Intal Sum Conservaton of omentum, Conservaton of momentum can be expressed mathematcally n varous ways p total = p + p = constant p + p = p f + p f fnal Sum In component form for the varous drectons, the total momentum n each drecton s ndependently conserved p x = p fx p y = p fy p z = p fz Conservaton of momentum can be appled to systems wth any number of partcles 8
3 Conservaton of omentum, Archer Example The archer s standng on a frctonless surface (ce). We know the mass of the archer (wth bow) and the mass of the arrow, and the speed of the arrow. What wll be the recol speed of the archer? Approaches to solve ths problem: Newton s Second Law no, no nformaton about F or a Energy approach no, no nformaton about work or energy Let the system be the archer wth bow (partcle ) and the arrow (partcle ) There are no external forces n the xdrecton, so t s solated n terms of momentum n the x drecton Total momentum before releasng the arrow s 0 The total momentum after releasng the arrow s p f + p f = 0 omentum yes 9 0 p f + p f = 0, or, m v f + m v f = 0 The archer wll move n the opposte drecton of the arrow after the release Agrees wth Newton s Thrd Law Because the archer s much more massve than the arrow, hs acceleraton and velocty wll be much smaller than those of the arrow Impulse and omentum From Newton s Second Law = F Solvng for dp (by cross multplyng) gves dp = Fdt By ntegraton, we can fnd the change n momentum over some tme nterval t f Δ p= p f p = Fdt = I t The ntegral s called the mpulse (I )of the force F actng on an object over the tme Δt The mpulse mparted to a partcle by a force s equal to the change n the momentum of the partcle (mpulsemomentum theorem). Ths s equvalent to Newton s Second Law.
4 f Δ = f = dt = t p p p F I Impulse s a vector quantty Force The magntude of the mpulse s equal to the area under the forcetme curve Dmensons of mpulse are (L T  ) T = L T  = L / T Impulse s not a property of the partcle, but a measure of the change n momentum of the partcle t Tme 3 If I throw an egg at you, how are you gong to catch t? 4 Force f Δ = f = dt = t p p p F I Tme t If we are to keep the mpulse to be a constant,.e., the area under the curve s to reman unchanged, but we extend the nterval of (t f  t ), what wll be the net effect? 5 Example. A garden hose s held as shown. The hose s orgnally full of motonless water. What addtonal force s necessary to hold the nozzle statonary after the water flow s turned on, f the dscharge rate s kg/s wth a speed of 5.0 m/s? Answer: The force exerted on the water by the hose s Δp mv water f mv ( kg)( 5.0 m s) 0 F = = = = Δt Δt.00 s Accordng to Newton's thrd law, the water exerts a force of equal magntude back on the hose. Thus, the gardener must apply a 5.0 N force (n the drecton of the velocty of the extng water stream) to hold the hose statonary. 5.0 N 6
5 f Δ = f = dt = t t p p p F I = F Δt Impulse Approxmaton The mpulse can also be found by usng the tme and averaged force I = FΔt Ths would gve the same mpulse as the tmevaryng force does 7 In many cases, one force actng on a partcle wll be much greater than any other force actng on the partcle When usng the Impulse Approxmaton, we wll assume ths s true The force wll be called the mpulse force p and p f represent the momenta (momentums) mmedately before and after the collson respectvely The partcle s assumed to move very lttle durng the collson (at the short nternal of contact) 8 Impulseomentum: Crash Test Example Collsons Characterstcs The momenta before and after the collson between the car and the wall can be determned (p = m v) Fnd the mpulse: I = Δp = p f p F = Δp / Δt We use the term collson to represent an event durng whch two partcles come close to each other and nteract by means of forces The tme nterval durng whch the velocty changes from ts ntal to fnal values s assumed to be very short 9 0
6 Collsons Example Collsons Example Collsons may be the result of drect contact Its momentum s conserved Another type of collson needs not nclude physcal contact between the objects. E.g., the collson of a proton, and a Alpha partcle (the nucleus of a helum atom). As both are postvely charged, they repel each other due to the strong electrostatc force between them at close separaton and never come nto physcal contact. There are stll forces between the partcles Ths type of collson can be analyzed n the same way as those that nclude physcal contact Types of Collsons Perfectly Inelastc Collsons In an nelastc collson, knetc energy s not conserved although momentum s stll conserved If the objects stck together after the collson, t s a perfectly nelastc collson In an elastc collson, momentum and knetc energy are conserved Perfectly elastc collsons occur on a mcroscopc level (eg, atomc or subatomc level) In macroscopc (eg, what you can see wth your eyes) collsons, only approxmately elastc collsons actually occur. E.g., some energy s lost n sound or n heat (sparks) durng collson 3 Snce the objects stck together, they share the same velocty after the collson m v + m v = (m + m ) v f v f = m v + m v m + m 4
7 Example. A bullet of mass m s fred nto a block of mass ntally at rest at the edge of a frctonless table of heght h. The bullet remans n the block, and after mpact the block lands a dstance d from the bottom of the table. Determne the ntal speed of the bullet. Answer: v m h d Collsons, cont In an perfectly nelastc collson, some knetc energy s stll lost, but the objects do not stck together Elastc and perfectly nelastc collsons are two extreme cases, most actual collsons fall n between these two types omentum s conserved n any type of collsons 5 6 Elastc Collsons Both momentum and knetc energy are conserved mv + mv = mv + m v f f mv + mv = mv + m v f f From: we have m (v v f ) = m (v f v ), factorng both sdes : m (v v f )(v + v f ) = m (v f v )(v f + v ) () From: we have m (v v f ) = m (v f v ) () Dvde () by (), we have v + v f = v f + v, or, v v =  (v f v f ) (3) Equaton (3) shows that the relatve velocty of the two partcles before collson, v v, equals the negatve of ther relatve velocty after the collson (v f v f).. 7 8
8 Fnal veloctes v f,v f. To solve for v f, we rearrange (3) and obtan: v f = v f + v v, and substtute t n the equaton of lnear momentum to get v f. Example. Two blocks are free to slde along the frctonless wooden track ABC shown n Fgure. A block of mass m = 5.00 kg s released from A. Protrudng from ts front end s the north pole of a strong magnet, repellng the north pole of an dentcal magnet embedded n the back end of the block of mass m = 0.0 kg, ntally at rest. The two blocks never touch. Calculate the maxmum heght to whch m rses after the elastc collson. m v f = m v + m v m v f m v f = m v + m v m (v f + v v ) Answer: Let v be the speed of m at B before collson. m v = m gh ( )( ) v = = 9.90 m s (m + m ) v f = (m m ) v + m v (m m ) v m v v f = + (m + m ) (m + m ) Let v f be the speed of m at B after collson. m m = = ( 9.90 ) m s = 3.30 m s 3 f v m+ m v Smlarly: (m m ) v m v v f = + (m + m ) (m + m ) 9 At the hghest pont (after collson) mghm ax = m( 3.30) h ( 3.30 m s) = = 9.80 m s ( ) max m 30 In the drawng on the rght for Executve Stress Relever, The ntal total momentum s mv, and the fnal total momentum s m(v/) + m (v/) = mv. Ths obeys the conservaton of lnear momentum. Explan f the scenaro shown s possble or not f the collson s elastc. Answer: Consderng the energy aspect based on elastc collson. The ntal energy n the system s mv. The fnal energy n the system s m(v/) + m(v/) = mv. Ths has volated 4 the conservaton of energy n elastc collson. So the above scenaro s not possble. 3 TwoBody Collson wth a Sprng Example. A 5.00g bullet movng wth an ntal speed of 400 m/s s fred nto and passes through a.00kg m block, as n Fgure. The block, ntally at rest on a frctonless, horzontal surface, s connected to a sprng of force constant 900 N/m. If the block moves 5.00 cm to the rght after mpact, fnd (a) the speed at whch the bullet emerges from the block and (b) the mechancal energy converted nto nternal energy n the collson. Answer: Assume for an approxmaton that the block quckly reaches ts maxmum velocty V, and the bullet kept gong wth a constant velocty v. The block then compresses the sprng and stops after movng a dstance of 5.00 cm. Next by conservaton of lnear momentum, mv= V+ mv = 00 m/s 3
9 Answer: Fnal energy: 00 m/s Intal energy (Knetc energy n bullet): (b) the mechancal energy converted nto nternal energy n the collson. Knetc energy n bullet Fnal energy  Intal energy: 5 J +.5 J 400 J = J 374 J was lost. The lost energy s manly converted to heat. Potental energy n sprng 33 TwoDmensonal Collsons The momentum s conserved n all drectons Use subscrpts for dentfyng the object ndcatng ntal or fnal values the velocty components If the collson s elastc, use conservaton of knetc energy as a second equaton 34 TwoDmensonal Collson, example TwoDmensonal Collson, example cont Partcle s movng at velocty v and partcle s at rest In the xdrecton, the ntal momentum s m v In the ydrecton, the ntal momentum s 0 After the collson, the momentum n the xdrecton s m v f cos θ + m v f cos φ After the collson, the momentum n the ydrecton c m v f sn θ + m v f sn φ. Let the angle take cares of the sgns
10 ProblemSolvng Strateges TwoDmensonal Collsons If the collson s nelastc, knetc energy of the system s not conserved, and addtonal nformaton s needed If the collson s perfectly nelastc, the fnal veloctes of the two objects are equal (both objects together). Solve the momentum equatons for the unknowns. ProblemSolvng Strateges TwoDmensonal Collsons If the collson s elastc, the knetc energy of the system s conserved Equate the total knetc energy before the collson to the total knetc energy after the collson to obtan more nformaton on the relatonshp between the veloctes TwoDmensonal Collson Example Before the collson, the car has the total momentum n the xdrecton only, and the van has the total momentum n the ydrecton only After the collson, both have x and ycomponents Example. Two automobles of equal mass approach an ntersecton. One vehcle s travelng wth velocty 3.0 m/s toward the east and the other s travelng north wth speed v. Nether drver sees the other. The vehcles collde n the ntersecton and stck together, leavng parallel skd marks at an angle of 55.0 north of east. The speed lmt for both roads s 35 m/h and the drver of the northwardmovng vehcle clams he was wthn the speed lmt when the collson occurred. Is he tellng the truth? Answer: 8.6 x3600/609 = 4.6 m/h 39 40
11 A basket ball (Ball A) of mass 500g and a table tenns ball (Ball B) of mass 0g are placed as shown n fgure. The two balls are released at a certan heght at the same tme wth the table tenns ball on top of the basket ball. What can you observe and why? Ball B Ball A Ball A Ball B What f Ball A s above Ball B? Why? For a bg contaner truck and a small motorbke travelng n the same drecton along a road, and approachng a red traffc lght, and both drvers want to stop at the traffc lght, what wll be the dfference for () the truck to ht the back of the motorbke and () the motorbke to ht the back of the truck? 4 4 The Center of ass The Center of ass There s a specal pont n a system or object, called the center of mass, that moves as f all of the mass of the system s concentrated at that pont The system wll move as f an external force were appled to a sngle partcle of mass located at the center of mass s the total mass of the system 43 44
12 Center of ass, Coordnates The coordnates of the center of mass are mx my mz x y z C = C = C = where s the total mass of the system Center of ass, poston The center of mass can be located by ts poston vector, r C r C mr = r s the poston of the th partcle, defned by r = x ˆ+ y ˆj + z kˆ Center of ass, Example Both masses are on the xaxs The center of mass s on the xaxs The center of mass s closer to the partcle wth the larger mass Center of ass, Extended Object Thnk of the extended object as a system contanng a large number of partcles The partcle separaton s small, so the mass can be consdered a contnuous mass dstrbuton 47 48
13 x mx my C = yc = Center of ass, Extended Object, Coordnates The coordnates of the center of mass of the object are xc = xdm yc ydm = zc = zdm Center of ass, Extended Object, Poston The poston of the center of mass can also be found by: C = dm r where r s a vector r The center of mass of any symmetrcal object les on an axs of symmetry and on any plane of symmetry 5 Center of ass, Example An extended object can be consdered a dstrbuton of small mass elements, Δm The center of mass s located at poston r C 5
14 Example. Fnd the center of mass of a rod of mass and length L. Answer: Let λ (read as lambda ) denote the lnear mass densty, or the mass per unt length, then λ =. L If we dvde the rod nto elements of length dx, then dm λ =, or dm = λ dx. dx L xdm o L X cm = = xλdx o λ x L λl L = [ ] = = 0 53 How to do ths? The dsc s to roll downhll from left to rght, and to roll uphll from rght to left when t s released from rest. 54 Cone Rollng up the slope V Track How to do ths? 55 56
15 Cone Rollng up the slope Explan the physcs behnd ths observaton. 57 The cone s not loaded. No cheatng! cm Cone Rollng up the slope 8. cm Hgher Lower 59 Unstable equlbrum poston!! 60
16 oton of a System of Partcles Assume the total mass,, of the system remans constant We can descrbe the moton of the system n terms of the velocty and acceleraton of the center of mass of the system We can also descrbe the momentum of the system and Newton s Second Law for the system 6 Velocty and omentum of a System of Partcles The velocty of the center of mass of a system of partcles s v dr dt C C = = mv The momentum can be expressed as v = m v = p = p C The total lnear momentum of the system equals the total mass multpled by the velocty of the center of mass tot 6 Acceleraton of the Center of ass Forces In a System of Partcles The acceleraton of the center of mass can be found by dfferentatng the velocty wth respect to tme = dvc ac m dt = a 63 The acceleraton can be related to a force a C = F If we sum over all the nternal forces, they cancel n pars and the net force on the system s caused only by the external forces 64
17 Newton s Second Law for a System of Partcles Snce the only forces are external, the net external force equals the total mass of the system multpled by the acceleraton of the center of mass: ΣF ext = a C The center of mass of a system of partcles of combned mass moves lke an equvalent partcle of mass would move under the nfluence of the omentum of a System of Partcles The total lnear momentum of a system of partcles s conserved f no net external force s actng on the system v C = p tot = constant when ΣF ext = 0 net external force on the system oton of the Center of ass, Example Rocket Propulson A projectle s fred nto the ar and suddenly explodes Wth no exploson, the projectle would follow the dotted lne After the exploson, the center of mass of the fragments stll follows the dotted lne, the same parabolc path the projectle would have The operaton of a rocket depends upon the law of conservaton of lnear momentum as appled to a system of partcles, where the system s the rocket plus ts ejected fuel followed wth no exploson 67 68
18 Rocket Propulson, Rocket Propulson, 3 The ntal mass of the rocket ncludng the major porton of the total fuel, plus, a very very very small amount of fuel to be burned mmedately s + Δm at tme t and velocty v The ntal momentum of the system s At some tme t + Δt, the rocket s mass has been reduced to and an amount of fuel, Δm has been ejected The rocket s speed has then ncreased by p = ( + Δm) v Δv Rocket Propulson, 4 Because the gases are gven some momentum when they are ejected out of the engne, the rocket receves a compensatng momentum n the opposte drecton Therefore, the rocket s accelerated as a result of the push (or thrust) from the exhaust gases In free space, the center of mass of the system (rocket plus expelled gases) moves unformly, ndependent of the propulson process 7 Let v be the ntal speed of the rocket (mass = ) wth ts fuel (mass = Δm). Over a short tome of Δt, the rocket ejects fuel of mass Δm, and the speed s ncreased to v + Δv, where Δv s the change n the speed of rocket. If the fuel s ejected wth a speed v e relatve to rocket, the velocty of the ejected fuel wth respect to the Earth s v v e. By equatng the total ntal momentum of the system to the total fnal momentum, we have ( + Δm)v = (v+ Δv) + Δm(v  v e ) Smplfyng the equaton, we have Δv = v e Δm But the ncrease n exhaust mass m s equal to the decrease mass. So we have Δm =  Δ. Ths gve Δv = v e ( Δ). 7
19 By ntegratng Δv = v e ( Δ), we have vf v = veln f dv = v e d v f dv v =  v e d f v f v =  v e [ ln ] f vf v = veln f 73 The ncrease n rocket speed s proportonal to the speed of the escape gases (v e ) So, the exhaust speed should be very hgh The ncrease n rocket speed s also proportonal to the natural log of the rato / f So, the rato should be as hgh as possble, meanng the mass of the rocket should be as small as possble and t should carry as much fuel as possble What f / f s close to? 74 dv = v e d The thrust on the rocket s the force exerted on t by the ejected exhaust gases dv d Thrust = = ve dt dt The thrust ncreases as the exhaust speed ncreases The thrust ncreases as the rate of change of mass ncreases The rate of change of the mass s called the burn rate 75 Example. A sze C5 model rocket engne has an average thrust of 5.6 N, a fuel mass of.7 grams, and an ntal (total) mass of 5.5 grams. The duraton of ts burn s.90 s. (a) What s the average exhaust speed of the engne? (b) If ths engne s placed n a rocket body of mass 53.5 grams, what s the fnal velocty of the rocket f t s fred n outer space? Assume the fuel burns at a constant rate. Answer: (a) (b)
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