Module - 55.38/5.7 Collision of elasticbodiesCollision of twobodieswhichoccursin avery small intervalof time andduring which thetwobodies exertvery largeforce on eachother is called an impact.whentwo bodies collidewith eachother, the common nofinal to thesurfaces oftwobodies in contact is called lineofimpact. when this lineof impact passesthrough the masscenters ofthecolliding bodies, theimpact is called central impact.If the velocities ofthetwo bodies before collision arealongthelineof impact, the impactis called directimpact, otherwise it is calledindirect orobliqueimpact.(a) DirectimpactG) lndirectimpac!Fig.5-33Directimpact between elasticbodies'Considertwo spheres of masses m, and mrmoving in thesamedirection alongthe samestraight line with velocities u, and u,' If u' ) u" thetwospheres will collide' According tothe lawof conservation of momentum, mrur * ffi, u, = fl,V, * mrV, where V, and Y 'arethevelocities of the spheres afterimpact.Newton's law of collision of elastic bodies statesthat whentwobodies collide witheachother, their velocityof separation bears a constantratioto theirvelocity of approach. Thevelocity of separation is proportional to thevelocityof approach. This constant ofproportionality is calledcoeffrcient ofrestitution andis denotedby e.V"_V,LL -^ut - uzVr-V,:e(u,-ur).Thevalueof eliesbetween0 andl.Forperfectlyelasticbodiese = l andforinglasticbodies, e :0.'Loss of kinetic energy during impact'It is thedifference between the sumof kinetic energJ of themasses before and after theimpact.(a)DirectimPact5.39Module- 5Sumof kineticenerrybeforeimpact : ) *,ur, + ) mrur,Sumofkineticener- .. I --^ I:gy atter rmpact = i, ^ry ,, + i frry r,Therefore,Iossofkinetic energyduring impact_1 ^ I Ir: [, ^, u,, + i mrur2) _ [, ^, V ,, * i mry rr)Example 5.33A car of weight20 kN movingat a speedof 0.5 m/s to the right collideswitha car ofweight35 kNwhich is at rest. If after the collisionthesecond car is observedtomove to therightata speed of 0.3m/s. Determine the coefficient of restitution between thetwocars.Solution.20 x 103ml = -; kgur = 0.5rnlstn, = 3l+kgiu2 = o, v, :0.3 m/s.oApplying momentumequationD,u,* m2U2= m, V, * mrV,2038.74 x 0.5+ 3567.79 x 0=2038.74 V,+ 3567.7gx0.3V,: -0.025mls= 0.025 m/s towards lefte _ V, - V, _ 0.3 - (- 0.025)ur-uz 0.5-0Example5.34 e: o'65A sphere of mass4 kg moving with a velocity of 5 m/s approaches a sphereof mass 3kgmovingwitha velocity of 4 m/s.If a directimpact takes place, find theirvelocityafter theimpact. Take e :0.5.Module - 5 5'40Solution.m, = 4kg u,= 5m'/s e:0'5fr, :3 kg u, :4 m/sFrom momentum equationffiru, * ffiru, = mrv' * *' v'4"5 + 3x4 :4xV,* 3xV'4 V' + 3 V2 : 32 ---------------(i)From Newton's law of collision of elastic bodies'vr-V,U-ut - uzVr-V,:0.5(5-4)4Y'-4V'=2----(11)Adding eqns (i) and (ii),7 Y, :34Vz = 4'86 m/sVr-V,=6.54.56 -V, : 0.5Vr :4'36 m/sExamPte 5'35 dl ^ ^ i-- tt.-intheThree perfectly efastlc ballsA, B and c of masses 2kg,6 kg and lzkgate movmg Isamedireiiionwithvilocities tZ mls,4rnlsand2m/s'respectively' IftheballAstrikeswithball B which in turnstrikes with ball c, prove that the balls A and B will be brought to rest bythe imPact.Solution.fro:Zkg;uo :l?mls e:lffi" =6kg;us =4mlsm":l}kg; uc = 2m/s. Firstconsiderthe collision of ballsAandBffioun+mBuB : froVo*m"V"2xl2+ 6x4= 2 *Vo* 6xVBModule - 5VA+3Yr:24-_-(i)^-V"-Vo'-"o-",1 x (12-4) : V, -VoV" -Vo :8 V ---(ii)Adding eqns(i) and (ii)4Yr:32V":8 m/sFromeqn(i)Vo+3x8:24Vo:0Theball A is broughttorest andthe velocityof ballB is 8 m/s.TheballB strikestheballc with this velocityof 8 m/s. considerthecollisionbetween ball B and ball c.u, : 8 m,/sm, :6 kgu. :2 m,/sm" : 12kgm"\ *mc uc : ffi"V, * -. V.6x8+12x2 :6V"+l2xYcYB+2Yr:12 ----(i)^-v.-vu--us-uc1x (8-2):V.-V"V.-Vr:6 ---(ii)Adding equations(i) and (ii)3V.:18V.:6m/sFrom eqn (ii)6-Vr:6vr:0The ball B is broughtto rest due to the secondimpact.5.41Module - 5 5'42Example 5.36A carof mass15000 kg moving with velocity 1.5m/s collides with another carof mass12000 kg moving with velocity 0.75 mlsin thesame direction' Aftertheimpact the twocarsmove together. Determine the speed of the coupled cars just aftertheimpact' Also calcu-late the lossof kineticenergy due to the impact'Solution.m, : 15000 kg u, = 1.5 m./sm, : 12000kg u, : 0'75m/svl=v2 :vApplying momentum equation.ffi,U, * mittz : m' V' * m' V'15000x1.5 + 12000x0.75 : ( 15000 + 12000 ) Vv : 1.17m/s.Lossof kineticenergy= InitialK.E' - Final K'E'= l:X rrl u,2 +) ^,u,,J - i ,rn, + m,) v'1 I --= , "15000x 1.52+ ; x I2000x0'752 -:1769.85 J.Direct impact of a bodywith a fixedplane'Consider a bodyof massm moving with a velocityu m/s strikesa fixed plane' In directimpact, the line of impactwill be perpendicular to thefixed plane. Sincethe massofthefixedplane is infinity ( very largecompared to themassm ), the law of conservation of momen-tumcannotbeaPPlied. |1.,-Coeffi cientofrestitution,12x27000x1.172ul:uml:mm2: *u^:Vr=0/LV, -V'o--w- lrur-uz i0-v=-u-0V=--ua,lI.)iri| 1/)Fig.5.34Module - 5 5.43The negativesign showsthechangein thedirectionof velocity.ReboundingvelocityVwillbe oppositeto thestrikingvelocityu.Va--uV:euReboundingvelocity = e x strikingvelocity-Whena bodyis droppedfrom a height ho, thevelocitywith whichit strikes thegrqundis,uo = J, glr, .Whena bodyis projected verticallyupwardswith a velocity u, the maximumheightattained by the body will be,- ,)'h = zgConsidera bodywhichis dropped from a heightho upona horizontal fixed plane.Let h,,h, h, etcbetheheightattainedby thebody in the first, second,thirdrebounce etc.5'44Module - 5Thestriking velocity "o = \E; h;,The rebounding velocitY ur = e uo= "JiiG,r,' tJE -erhoh, = fr zgAfterthefirstbounce,thebodystrikesthefixedplanewithvelocityu'andrebounceswith velocitY ur.u2 =eut=QXO2Urh.=Lt )ou3 : avz=eX "',fZ1,5o =.'.,fil,h;L _ u32 _ eu 2gho _ eu*ho,3- 29 2gSimilarlY h, : otho andhs = e'oho etc'ExamPIe 5.37Aball is dropped from a height of 1m on a smooth floor'firstbounce is 81cm, determine'(i) thecoefficient ofrestitution and(ii) the expected height ofthe second bounce'Solution.ho:1mh, = 0.81 m(D h,=e2ho0.81= e2x 1e:0.9,..8 sh"=eo 2Eho)a= "'^[ zEG- eoho{I'iKnowing that the height ofModule- 55.45(ii) h, : oo ho: 0.94x1: 0.656 mExample 5.38Fromwhatheightmust a heavy elastic ballbe droppedon a horizontal floor so thatrebouncingthriceit will reacha heightof 16 m. Take e = (0.5)'/'Solution.hr:16me: ( 0.5 )1/3hr:e6ho16: (0.5)'ox6xho:0.25 hoho:64 mThe ball shouldbe droppedfrom a heightof 64 m.Example 5.39A glass ballis dropped onto a smoothhorizontalfloorfromwhichit bouncesto a heightof 9 m. Onthe secondbounceit rises to a height of 6 m. Fromwhatheighttheballwasdroppedand what is thevalueof e.Solution.h, :9mhr:6 mh' : 'ho:9mhr.:Eoho=6me : 0.816h,= e2ho: 9m0.816'z ho:9ho : 13.52mThe ball wasdropped froma height13.52m abovethefloor.hrr6:-6--_hr9Module - 5 5'46Example 5.40A body of mass m fallsfroma height houpon a fixed horizontal plane' Ifthecoefficient ofrestitution is e, show that the total vertical distance described by the body before it hasfinished rebounding i, ( |l+ ) ho dnd thatthe total timeof travel is: r l *j '' l?!'-il-e'tl ;Solution.Thedistance travelled just before first impact= hoMaximum height reached after first impact h, : e2hoTotal distancetravelled between firstandsecond impact : 2hr:2 ezhoMaximum height attained afterthe second impact= h, = ea ho.Totaldistance traveled between second and third impact : Zhr:2 ea hoSimilarly the totaldistance travelled between third and fourthimpact :2 e6ho and so on'Thetotal distance travelled by thebody,H : ho+ 2 e2ho+2 eaho+ 2 e6 ho+ho+ 2 e2 ho ( l+e2 +e4 + e6 + ---)I:ho* 2ezhot,_ yl.t ^2= ho (1 * ;:J)It-"'+2e21'n" L-r= -.1: 1!td; nol-e-f2t "Timeto fallfreely from aheightho, to= { ?Firstrebouncingvelocity, u, :ouo: {'?gt1Module - 5Heightattainedin the first rebounds, h, : e2ho"(Eh.Timeto attainh,, ,, : {, =Timeto attainedheighth,tr:Ei:t-!eE"'r\Ve_^,Ei-'1/ sTotaltimeof traveT : to* 2 tt + 2 tr* 2 t, + .....EG-I sEG={,t=(=)(1 + 2e )' l-e'-l-e+2e(-.-)l-eEC1/ t2e2 hoob2hoob5.48Module - 5ExamPle5.41fJJu :';:** rrom a height 5 m upon : ''';l ::'f:*:i:::i:rii*:,T$Hn:::,",fJ,11',T$;1:f,ilfl l"Ti:il#"iH'#[l*:i*:roreithasnnishedrebound';n}ffi;;;" ;; total time taken toffavel thetotal distanceSolution.ho:5me = 0.'15The total distance travelled'-2ltcg = 1l-i) ho'1 -e"_,t + 0.75'_y* s'1 - 0.75'= 17.86 m'Total time oftravelI + 0.75.-t I- t1 - 0.75'= 7.07 sOblique impact between elastic bodies'In centrar impact, the line of impactf:'":,Y:1;1:li:ffi;;fT;r"t};|la;;:,r"'J"[Hi:1-1l1lJ]:':::;i,llJ"Hl'ffi;;sdi'rineorimpactrhecoilisionisto be obliqu" wt "nl,rst i"for" tt " i.pact the direion of velocities areinclined to the lil'*pll'n.ia", two bodies of masses m, and mrmovingwith verocities u, and u,Let *, andbe the inclination or veloc itie s u, ld ", -'. 11:t:, f:-".t# |i: ny'*Xi Il ?11+ e, EGr =(;)! ,be the inclination of velocities u' ano u2 *:':'^:::.:;; ri" ;*ou"tas shown in frg' 536'and 0, with the fi" "fitpuct bethe veiocities just after the impacModule- 55.49uziq, 7i.,}\ x\ >t,vt_ velocityofseperationalongthelineofimpact,r\": --(I)Y2Fig.5.36Since theforceof impactis alongthelineof impactandtheforce of impacton eachbodyis equalandopposite,the momentumis conservedalongthelineof impact. Equatingtheinitialandfinal momentum of bodiesalongthe lineof impact,frl,xU,COSC[l * mrxU2 COSGz:flrV,COS 0, * 4,VrCOS 0,.....(i)Coefficientof restitution,V, cos 0, - V, cos 0,---(2)ui cos 0t - uz cos 0. 2Sincethere is no force perpendicularto the line of impact,the componentof velocityalong the perpendiculartothelineof impactremains constant.u,sin cr, : V, siner ----- (3)andursinur: Y, sin0, --- (4)Fromthe above fourequationswe can solve for V, V2, 0, and 0r, themagnitudeanddirectionofvelocities just after impact.Example5.42The magnitudeanddirectionof thevelocities of two identical smooth ballsbeforetheystrikeeach otherare showninfig5.37. Takinge : 0.9,determinethemagnitudeand direc-tion of thevelocityof each ball afterthe impact.v, v2d 2= 600" : l5m/sq1Fi5.5.37Module - 5 5'50Solution.The momentum alongthe line of impact is conserved'mul cos cli - mu2 cos cL2- - mV, cos 0, + mV' cos0'10 cos 30 - 15 cos60 - - Vr cos 0' * V' cos 0'- V, cos er + V2 cos 0 = 1'16 -----( i )V. cos 0, - (-V,cos 0,)Coefficientofrestitution' e = @V, cos 0, *V, cos er: 0'9 [10 cos30 + 15 cos 60] = 14'tO -------(ii)Adding equations (i) and (ii)'2 V, cos 0r: 15'7V, cos 0, :7 -85 ---(iiDSincethecomponentofvelocityperpendiculartothelineofimpactremainsconstant,u,sin o,: V, sin 0, -----(iv)ursin crr: V, sin0, ----------(v)From eqn (v) V, sin 0, : 15 sin 60= l2.ggFrom eqn ( iii) V, cos 0, : 7 '85tzn 0, :1.6550z : 58'860V, cos 58.86 = 7'85V2 = 15'18 m/s'From eqn (ii) 15'18 cos 58'86 + V'cos 0t :1'4'54V, cos0r:6.69From eqn (iv) V, sin 0, : ulsin Gr = 10 sin 30 : 55tan0r = ;;,0 '= 36'77oV, cos 36-77 = 6'69V': 8'35 m/s'Module- 5Obliqueimpactof a bodywith a fixedplane.Considera body of massm strikingona fixed horiz.ontalplanewitha velocityu, inclinedcr with thelineof impactas shown in fig 5 .3 8.LetV be the velocityof bodyaftertheimpactand0 be the directionof velocityafter impact.Sincethecomponentof velocityperpendicularto the lineof impactremains constan!usino: Vsin0 ---(i)Vcos0Coefficientofrestitution, =ucos0eucoscr= Vcos0-:-(ii)Fromthe above equationswe can solveforV and0themagnitudeand directionof bodyafterimpact.Fig.5.38Example 5.43A ballmovingwith avelocityof 6 m/ssfrikesonafixedhorbantalplaneat anangleof30withhorizontal.Ifthe coefficient of restitutionis 0.8, determinethe magnitudeanddirectimofvelocityofthe ballasit rebounds from theplane.Solutioo:90 -30:600u : 6 m,/s.e:0.8Sincecomponentof velocityperpendiculartotheline of impactremains constant,usincr : vsin0V sin0 :6 sin 60 : 5.196.Coefficientof restitution " - v cos eFig.5.36eucosu, - Y uuseV cos 0 : 0.8x6 cos605.5ttaneeV cos 65.21v2.42.16565.2102.45.72 mlsFig.5.39