Newon s Laws of Moion MS4414 Theoreical Mechanics Firs Law velociy. In he absence of exernal forces, a body moves in a sraigh line wih consan F = 0 = v = cons. Khan Academy Newon I. Second Law body. The rae of change of momenum of a body is equal o he ne force on he F = dp d Khan Academy Newon II. Third Law force on body A. If body A exers a force on body B hen body B exers an equal and opposie F AB = F BA F BA B Khan Academy Newon III. A F AB Noe ha Newon s hird law does no require he equal and opposie forces o ac along he same line of acion (e.g. wo charged paricles ineracing hrough he magneic field). 1
MS4414, Theoreical Mechanics 2 F BA B A F AB 1 Forces A force is a push or a pull on a body. The unis of force are newons, N. In his course we will encouner he following forces: Conac forces These are forces which preven wo bodies occupying he same space a he same ime. (Ulimaely hey resul from he Pauli exclusion principle of quanum mechanics which saes ha wo paricles canno possess he same se of quanum numbers.) Fricion forces These forces preven or oppose he moion of a body. Graviaional forces Aracive forces beween wo bodies by virue of heir mass. Elasic forces These forces arise as deformed bodies aemp o recover heir original shape. 2 The firs and second laws Newon s firs and second laws can be wrien ogeher as a single equaion F = dp d = d (mv) (1) d where F is he imposed force, m is he mass of he body and v is is velociy. The vecor mv = p is called he momenum. If and only if he mass of he body is consan (no he case Khan Academy Momenum.
MS4414, Theoreical Mechanics 3 for a rocke for example) his equaion can be rewrien as F = m dv d (2) F = ma (3) Worked Example F = ma. Show ha Newon s firs law of moion can be derived from he equaion Worked Example force F? Wha is he rajecoryx() of a paricle of mass,m, subjec o a consan 3 Newon s Third Law and Conservaion of Momenum N.B. proving he law of conservaion of momenum (or angular momenum) ofen comes up in exams. Newon s hird law saes ha if body A exers a force F BA on body B, hen he force exered by body B on body A,F AB, is equal in magniude and opposie in direcion i.e. F BA = F AB.
MS4414, Theoreical Mechanics 4 F BA B A F AB Theorem Conservaion of momenum. If a collecion of paricles inerac wih each oher bu no wih any exernal eniies, he oal momenum pi = of he collecion is independen of ime. i Proof 1 (wih words) Newon s second law of moion saes ha he rae of change of he momenum of a paricle is equal o he oal acing on i. Therefore he rae of change of he oal momenum of all he paricles in he collecion is he oal exered on all he paricles. Newon s hird law saes ha forces come in and pairs. Therefore he oal force acing on he collecion of paricles is. Therefore he rae of change of he momenum of he collecion of paricles is. Therefore he momenum is a (vecor) consan. Proof 2 If here is a collecion of pariclesi, (i = 1...n) ineracing wih each oher bu no wih any exernal paricles. If F ij is he force exered on paricle i by paricle j. The rae of change of he oal momenum of paricle i is given by = d d (m iv i )
MS4414, Theoreical Mechanics 5 The rae of change of he oal momenum of all he paricles = i=1 d d (m iv i ) I can rewrie he sum as 1 2 i=1 (F ij +F ji ) = j=1 i=1 d d (m iv i ) Now using Newon s hird lawf ij = F ji = d d (m i v i ) i=1 Inegraing once where consan is a vecor consan. consan = (m i v i ) i=1 In paricular he oal momenum before and afer a collision is he same. Noe ha he proof required forces o be equal and opposie bu no acing along he same line. (This will be required for conservaion of angular momenum.) 4 Inelasic Collisions Collisions can be elasic or inelasic. Elasic collisions In an elasic collision he oal energy is conserved as well as he oal momenum. The collisions in Newon s cradle are approximaely elasic. Inelasic collisions paricle. In an inelasic collision he colliding paricles coalesce ino a single
MS4414, Theoreical Mechanics 6 4.1 Two paricles in one dimension Two paricles wih masses m 1,2 and velociies v 1,2 collide and coalesce. Wha is he velociy, V, of he final paricle? Iniially he oal momenum of he sysem is p iniial = Afer he collision and coalescence he sysem consiss of a single paricle of mass m 1 +m 2 ravelling wih velociy V. The final momenum of he sysem is given by p final = By conservaion of momenum p final = So m 1 v 1 +m 2 v 2 = (m 1 +m 2 )V and herefore V = To saisicians he final velociy is he mass weighed average of he iniial velociies of he paricles. Worked Example A paricle of mass 1 kg ravelling a 2 m s 1 collides wih a paricle of unknown mass ravelling in he opposie direcion a 4 m s 1. The paricles coalesce and he final assembly moves a a velociy of 1 m s 1 in he direcion he firs paricle was ravelling in. Wha is he mass of he second paricle? Khan Academy Conservaion of Momenum.
MS4414, Theoreical Mechanics 7 Worked Example speeds. Is his reasonable? In films people sho by handguns are ofen hurled backwards a high 4.2 Two paricles in higher dimensions Two paricles wih masses m 1,2 and velociies v 1,2 collide and coalesce. Wha is he (vecor) velociy of he coalesced paricle? Iniial momenum p iniial = Final momenum p final = Conservaion of momenum p iniial = And so V =. Exam Quesion 2007 Three paricles of masses 1 kg, 3 kg and 2 kg, simulaneously collide: Before he collision, he middle paricle was moionless, whereas he velociies of he oher wo were 1 m s 1 and 3 m s 1 (see he diagram). Assuming ha he paricles collide non
MS4414, Theoreical Mechanics 8 elasically and coalesce, find heir velociy afer he collision. Which way will hey be moving? 5 Forces during a collision The law of conservaion of momenum allows us o calculae he velociies of paricles wihou worrying abou he deails of he forces. Someimes i is useful o undersand he deails of hose forces. Consider wo paricles boh of massm, one is moving wih velociyv, he oher is saionary. A F v F B When hey collide, he paricles coalesce. By conservaion of momenum, he velociy of he compound paricle is v/2. A B v 2 As he paricles collide hey exer forces on each oher. Paricle B exers a forcef on paricle A and, by Newon s hird law, paricle A exers an equal and opposie force, F, on paricle B. The force is ime dependenf = F(). The force F() = m dv A d F() = m dv B d
MS4414, Theoreical Mechanics 9 Inegrae wih respec o ime unil boh paricles have he same velociy u and show ha u mus be he velociy prediced by conservaion of momenum u = v/2 F() d = m u 0 v dv A 0 F() d = m u 0 dv B Evaluae he righ hand side inegrals 0 F() d = = Solving he equaion forugivesu = v/2. This resul is independen of he form off(); only he area under he graph of F() is imporan. 6 The Sledgehammer Trick In Jearl Walker s sledgehammer rick, really don ry his a home, here is an inelasic collision in which he sledge hammer, iniially ravelling downwards wih speed v 0, is brough o res. The concree block (while fracuring) exers a force F() on he sledgehammer. Newon s hird law ells us ha his force is exchanged beween: The sledgehammer and he concree block, The concree block and he upper bed of nails, The upper bed of nails and Jearl Walker, Jearl Walker and he lower bed of nails, The lower bed of nails and he ground. Ideally ha force should remain below he value needed o drive he nails ino any bones or imporan organs. The value of he inegral of F() wih respec o ime is fixed. If he collision sars a = 0 and ends a = T T 0 F() d = mv 0.
MS4414, Theoreical Mechanics 10 Bu, by increasing he ime of he collision, he average value of he force. F =, can be reduced FT = mv 0 = F =. v 0 v No block v 0 v Block F No block F Block Area = mv 0 Area = mv 0 The same idea is used in cars wih crumple zones, collapsible seering columns and airbags. v F v 0 Sledgehammer rebounds A imid sledgehammer wielder who does no hi he concree block hard enough o fracure i can be very dangerous