MOMENTUM CONSERVATION LAW

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Giles Gibbs
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1 1 AAST/AEDT AP PHYSICS B: Impulse and Momenum Le us run an experimen: The ball is moving wih a velociy of V o and a force of F is applied on i for he ime inerval of. As he resul he ball s velociy changes from V o o V. From he second Newon s law we know F ma We also know ha a V V 0 If we subsiue expression for a ino he equaion for F, we obain F m V V 0 The nex sep is o ransfer from he lef side of he equaion o he righ one. F m V V o mv mv o mv 1 where mv means he change of his produc. A he lef side of he equaion above he expression is F. The produc of he force and he ime inerval when ha force is applied on he objec is defined as an impulse. IMPULSE = F A he righ side of he equaion we have mv The produc of an objec s mass and velociy is defined as momenum MOMENTUM = mv Usually momenum is represened by he symbol P Using ha erminology, we can read he equaion (1) as APPLIED IMPULSE IS EQUAL TO THE MOMENTUM CHANGE MOMENTUM CONSERVATION LAW Le us run anoher experimen. Two balls wih he masses of and m are moving oward one anoher wih he velociies of V o1 and V o. Double subscrip (o1) means -

2 iniial velociy of he firs objec. The balls collide. Afer he collision heir velociies become V 1 and V. According o he 3 Newon s Law F 1 =-F (1) We can express he forces as ( Newon s law) F 1 = ma 1 and F = ma, where a 1 and a are he acceleraions of he balls, If we plug in expressions for acceleraions ino (1) he resul is ma 1 = - ma () We also know, he acceleraion is he raio of he velociy change over he ime or, a 1 V 1 V o1 ; a V V o Afer subsiuion of hose expressions ino () we have V V o1 V 1 m V o We can cancel and muliply mass imes velociies. The resul is V 1 V o1 m V m V o Expressions wih he subscrip o (ha means ha he expression relaes o he evens before collision) we move o he righ side of he equaion, and all ohers we move o he lef. V 1 m V V o1 m V o Now on he lef side we have oal momenum afer and on he righ side - oal momenum before he collision. We have derived he momenum conservaion law. I saes.

3 3 The oal momenum of an isolaed sysem of bodies remains consan, or oal momenum before any ineracions equals o he oal momenum afer hem. Problems solving sraegy 1. Wrie he daa.. Define which objecs had momenum before he ineracion and wrie an expression for he oal momenum of all he objecs before he ineracion. 3. Define which objecs have momenum afer he ineracion and wrie an expression for he oal momenum of all he objecs afer he ineracion. 4. Wrie he momenum conservaion law - oal momenum before he ineracion has o be equal o he oal momenum afer he ineracion 5. From he obained equaion express he unknown variable and subsiue numbers. Example: A kg hockey puck moving a 48 m/s is caugh by a 75-kg goalie a res. Wih wha speed does he goalie slide on ice? = kg In his problem only he puck had a momenum before he V 01 = 48 m/s collision (ineracion). Thus, he oal momenum before is m =75 kg m1v01 V o =0 Afer he collision boh objecs are moving ogeher. The oal momenum afer collision is ( +m )V V =? According o he momenum conservaion law V 01 = ( +m )V In his equaion he only unknown variable is V. We can isolae i and plug in he numbers. V = V 01 /( +m )= kg *48 m/s/(75 kg kg) = m/s JET PROPULSION Le us run an experimen. A es-ube parially filled wih waer and a sopper is insalled on a car. Iniially he sopper and he car are a res. So, he oal iniial momenum m car V car +m sopper V sopper =0 If we hea he waer, he sopper shos and he car sars o move backward.

4 4 According o he momenum conservaion law he oal momenum does no change. Thus, afer he sho he oal momenum also remains zero. If we express he Vcar from he equaion above we have: V car m sopperv sopper m car Sign - shows us, ha he direcion of he sopper s moion is opposie o he direcion of he car s moion. Tha moion wih he oal momenum ha is equal o zero is called je propulsion. The same principle is used for he rocke operaion. Iniially he velociy of he fuel gases and he one of he rocke is zero, so he oal iniial momenum. m rocke V rocke +m fuel V fuel = 0 When he fuel sars o burn, he gases sar o ejec, bu he oal momenum has o remain zero. Thus, The formula provides us wih wo possible ways of increasing he rocke s velociy. One way is o increase he velociy of he ejeced gases. The second is o decrease he rocke s mass Problem: A man wih a mass of 70 kg is sanding a he rear of a boa of mass 80 kg in he middle of he placid lake. The boa is 5 m long. The man walks o he fron of he boa How far does he boa move relaive o he lake boom? Neglec waer resisance. Soluion. As you can observe from he diagram below, when he man moves in one direcion according o he je propulsion principle he boa sars o move in an opposie one. Because he ne momenum should me zero, he equaion for he momenum conservaion law afer projecion is M man V man - M boa V boa =0, V rocke m fuelv fuel m rocke

5 5 or M man V man = M boa V boa (1) Assuming ha he moion is uniform we can express he velociies as V man = D/ = (L-x)/, and V boa = x/, where L is he lengh of he boa and x is he disance ha boa has raveled. When we subsiue hose expressions ino (1), we can cancel ime and isolae x. M man L x M boa x or M man L M man x M boa x x M man L M man M boa Collisions There are wo ypes of collisions - elasic and inelasic. Elasic collision is he collision when boh mechanical energy and momenum are conserved. A good example of an elasic collision is he collision of he wo seel balls. When we have elasic collision only conservaive forces are acing inside he physical sysem Le us assume, ha he ball wih a mass of and wih a velociy of V o1 collides wih he ball m a res. Afer he collision he ball s velociies are V 1 and V especively. Our goal is o evaluae hose velociies. For he observed collision he mechanical energy conservaion law is KE o1 KE 1 KE or mv o1 mv 1 mv

6 6 Afer cancellaion of in he denominaor we have Momenum conservaion law for he same collision is V o1 V 1 m V V o1 V 1 m V We obained a se of wo equaions wih wo unknown variables V 1 and V. The se can be solved The final resul is V 1 m m V 01 and V m V 01 These formulas allows us o make several very ineresing conclusions abou he elasic collision

7 7 1. If >m hen V 1 is posiive. Tha means ha boh balls afer he collision are moving in one and he same direcion.. If <m hen V 1 is negaive. Tha means ha boh balls afer he collision are moving in differen direcions. 3. If =m hen V 1 is 0 and V =V o1. Tha means ha he firs ball sops afer he collision and he second one is moving forward wih he same speed. A good example of his is he collision of he wo billiard balls. 4. If m = (infiniive mass). A good example of his is he collision wih he wall, hen V 1 = -V o1 and V =0. Tha means ha he firs ball renounced wih he same speed. Inelasic collision Only momenum conservaion law is valid for he collision. Mechanical energy conservaion law does no ake place, because parially mechanical energy ransfers ino he hea form of energy. A good example of his collision is he collision of he wo clay balls. The momenum conservaion law for he collision is V o1 m V or V m V 01 We can evaluae he amoun of mechanical energy dissipaed ino hea. H = KE 01 - KE 1+, where KE 01 - is he iniial kineic energy of he firs ball before he collision and he KE 1+ - is he oal energy of he boh balls afer he collision. In a real life we do no have perfecly elasic or oally inelasic collisions. To observe he qualiy of collision physiciss use an elasiciy coefficien. If i is 1 - he collision is elasic. If i is 0 i is compleely inelasic.

8 8 Problem: A bulle of mass m flying horizonally a a speed Vo srikes a block of mass M suspended on a sring wih a lengh of L and ges suck in i. To wha angle o he verical does he block deviae? The problem should be divided ino hree separae problems 1. Wha velociy block obained as he resul of inelasic collision wih he bulle. We apply he momenum conservaion law and obain mv o = (m+m)v block, or V block mv o m M. Wha was he final heigh of he deviaed block? As he resul of he collision block obained KE and i was finally ransformed ino he poenial one. We apply he energy conservaion law and obain KE=PE (m M)V block (m M)gh or h V block g or 3. Wha is he angle of deviaion? From riangle ABC i is obvious ha cos L h l, so cos 1 L h h Home Assignmen: Beyser: pages #4, 7,1, 17, 3, 6, 34, 37, 39,46, 47, 48 Cunell: Concepual Quesions Page 05: # 1,, 3, 4, 5, 7,11, 13, 16, Problems: Page 06 #3, 5, 8, 9, 13, 17, 18, 19,, 5, 30, 31, 35, 39, 51

1. VELOCITY AND ACCELERATION

1. VELOCITY AND ACCELERATION 1.1 Kinemaics Equaions s = u + 1 a and s = v 1 a s = 1 (u + v) v = u + as 1. Displacemen-Time Graph Gradien = speed 1.3 Velociy-Time Graph Gradien = acceleraion Area under