Review of Kinematics Using Vectors Physics 3A (Fall 2001)

Transcription

1 Review f Kinematics Using Vectrs Physics 3A (Fall 2001) This is a shrt summary t review the tw main kinematics equatins and cmpare the use f cmpnents versus vectr diagrams. Equatin 1: x = v t + ½ a t 2. Methd 1: vectr diagram: Here we are applying the graphical rules fr adding vectrs t cnstruct a triangle. The additin we are ding is c = a + b, where c = x, a = v t, and b = ½ a t 2. S, frm additin f vectrs, we knw that the tails f c and a must tuch, the head f a tuches the tail f b, and the heads f c and b must tuch. This tells us hw t cnnect x, v t, and ½ a t 2. S, the main questin is which t draw first? The best prcedure is t apply the fllwing three guidelines: 1) Attempt t draw the psitin vectr first. Generally, yu knw smething abut the initial and final psitin f the bject, this tells yu hw t draw x, as it just cnnects the initial and final psitin. The situatin where yu wuld nt knw this t begin with is when yu are being asked t find the initial r final psitin. 2) Fr prjectile mtin, draw ½ a t 2 next as it is always vertical, and by cnstructin must end at the final psitin. It starts where v t ends. 3) By cnstructin, v t starts frm the initial psitin and ges at an angle that is either given, r yu need t find, and then meets up with ½ a t 2. The types f picture yu will end up with have the fllwing characteristics: v up Final height = initial height final height < initial height final height > initial height v hrizntal v dwn

2 Fr the cases where the vectr additin des nt give a right triangle, a dashed line is drawn t indicate the hrizntal. In these cases, the difference in initial and final height cmbined with the ½ at 2 vectr will give the length f ne side f the right triangle. Hw des this cmpare t cmpnents? Cmpnent equatins: x = v x t + ½ a x t 2 and y = v y t + ½ a t 2. These tw equatins are the nes that yu wuld get frm using sine and csine f the angle between v and the hrizntal. Ntice, by drawing the triangle, yu als have the ptin t immediately use tangent, Pythagrean therem, r cmplete the diagram as a rectangle and equate ppsite sides f the rectangle. If yu g back and lk at the example prblems, extra prblems, and exam prblems, yu will find cases where this was useful. EQUATION 2: v = v + at Fr this equatin, if it is prjectile mtin, the best way t draw it is as fllws: 1) Draw v first. Is it up, dwn, r hrizntal? Generally, even when yu are slving fr the angle, yu at least knw if it was thrwn up r dwn. 2) Draw at cming dwn frm v. 3) Draw in v f. Start v f frm v and draw it up, dwn, r hrizntal based n the infrmatin yu have abut its final directin. If this is unknwn, pick a likely directin, and use the infrmatin t slve fr the unknwn angle. Cnnect at t v f. WARNING: In this case, whether r nt at ges abve r belw the hrizntal line has nthing t d with the initial and final psitin f the bject, nly the initial and final directin f mtin. Tw example diagrams fr an bject launched up, and the relatinship t the actual mtin: The path f the bject is shwn as a dashed line. The crrect diagrams fr the velcity fr the tw cases are: A B FINAL NOTE: There are nly tw angles that are guaranteed t be the same in the psitin and velcity diagrams. These are: (1) the angle in the velcity diagram between v and at is the same as the angle in the psitin diagram between v t and ½ at 2 ; (2) the angle between v and the hrizntal in the velcity

3 diagram is the same as the angle between v t and the hrizntal in the psitin diagram. This is because multiplicatin by a scalar des nt change the directin f a vectr. Here are fur example questins that yu shuld try t wrk ut. All fur questins are based n the diagram and situatin belw: 10 m 2 m 20 m Yu are standing n a 10 m high cliff. A tube is 20 m away in the hrizntal directin. The tube is 2 m high. Answer the fllwing questins, ignring air resistance. 1) Suppse that yu thrw a ball up at an angle f 30 frm the hrizntal, hw lng befre it hits the tp f the tube? 2) Given that thrw, what was the initial speed and its final velcity? 3) What initial speed is required if the ball is launched hrizntally? 4) Hw high is the cliff if the ball is launched at an angle f 5 belw the hrizntal and it lands just in frnt f the tube traveling at an angle f 15 frm the vertical? (The distance frm the cliff t the tube is still 20 m.) A DISCUSSION OF THE POSSIBLE WAYS TO ANSWER TO ALL QUESTIONS ARE ON THE NEXT PAGE. NOTE: THESE DO NOT REPRESENT SOLUTIONS IN THE SENSE THAT THE ARE NOT PRESENTED IN THE SOLUTION FORM. THIS IS DOEN TO PROVIDE A LITTLE EXTRA DISCUSSION OF HOW TO DO EACH PROBLEM.

4 1) The tw diagrams fr this situatin are: Sme things t nte are the fllwing. In the psitin diagram, the vectr x starts at the initial height f 10 m and ges t the tp f the tube at 2 m. Therefre, the dashed line representing its vertical cmpnent is nly 8 m lng. As discussed, the vectr ½ at 2 must start at the head f the vectr v t. Since this vectr start initially in the upward directin, ½ at 2 must be LONGER then the vertical cmpnent f the vectr x. Hwever, by drawing a hrizntal line, we are able t make a right triangle with a vertical side ppsite the 30 angle whse length is given by h, a quantity that we can calculate. The velcity diagram is relatively straightfrward fr this prblem. Fr the fllwing calculatins, all variables are defined in the vectr diagrams. Fr the first prblem, we want t find time. N single equatin will give us time as h, v, and v f are all unknwn. Hwever, we can get tw equatins with tw unknwns using the upper right triangle in the psitin diagram: Dividing the secnd equatin by the first gives: ( vt)cs(30) = 20 m ( vt )sin(30) = h= 2 at 8 m 2 at Slving fr time gives: t 2 = (20 tan(30) + 8) / 5 r t = 2 s. tan(30) = ( 8) / 20. This methd required a minimum f 2 equatins with tw unknwns. We had the pssibility f many ways f getting these equatins and unknwns. If yu used cmpnents fr a prblem like this, yu still wuld have tw equatins. Hwever, nw yu wuld have t make sign chices. Chsing up and t the right as psitive, the tw equatins becme: x= 20 m = [ v cs(30)] t y = 8 m = [ v sin(30)] t 2 at Ntice, this assumes that yu use a = 10 m/s 2 and put the minus sign explicitly in the equatin. If yu use a = -10 m/s 2, then the secnd equatin wuld have +1/2 at 2. Therefre, in the cmpnent methd yu have t think carefully abut minus signs. In the graphical methd, the directin is accunted fr when yu draw the vectrs. Minus signs nly ccur due t gemetry, as in the h = 1/2at 2 8 m equatin. All variables becme magnitudes because they represent lengths f the vectrs.

5 2) Fr the first part, the equatin, v = 20/[2 cs(30)] = 11.5 m/s, is easily recgnized by either methd. Yu just use the equatins yu already had in part 1. T find v f by the graphical methd, we cnsider the triangle fr the velcity vectrs: Nw we just apply the law f csines t find the magnitude f v f and the law f sines t find θ v = ( v ) + ( at) 2( v )( at)cs(60) v f f = + = (11.5)(20) = 17.4 m/s sin( θ ) sin(60) = 20 m/s 17.4 m/s r sin( θ ) = 20sin(60) /17.4 θ = 84.5 S, the final answer is a velcity f 17.4 m/s t the right and dwn 54.5 frm the hrizntal (ntice the directin f v f is nt θ but θ - 30 ). Other ptins included slving fr the angle in the lwer right f the triangle and giving the directin f v f relative t the vertical. This wuld invlve the law f sines applied t the v and v f sides f the triangle. Again, by cmpnents ne wuld have slightly additinal wrk: vfx = v cs(30) v = v sin(30) at fy f fx fy v fy tan( θ ) = v fx Here ne has t make a sign chice again. The nly advantage is that in this case θ is actually the directin f v f. One des nt have t subtract frm 30. v = v + v

6 3) This ne is relatively straightfrward by either cmpnents r a diagram. One nly needs the psitin diagram: By either the diagram r cmpnents: 8 m = ½ at 2 (ntice, here the sign chice is straightfrward) t = 1.3 s v t = 20 m gives v = 15.4 m/s 4) This is the prblem that shws the full pwer f the graphical methd. This methd will always be better when infrmatin abut angles is the main infrmatin in the prblem. In this case, we are given the initial and final directin f travel s we knw the angles in the velcity diagram (as indicated). Please nte that a cmmn errr here is t use the final angle f travel in the psitin diagram. THIS IS INCORRECT. The angle in the psitin diagram fr x has nthing t d with the directin f travel, nly the relatin between its hrizntal and vertical cmpnents. See the abve diagrams f a trajectry t cnvince yurself f this. Sme cmments n the velcity diagram, because this is where all the wrk fr this prblem really is. First, it is nt t scale. Secnd, the angles f 5 and 15 are what are given in the prblem. Since at is always vertical, we knw where the 15 angle ges. Secnd, using that at is perpendicular t the hrizntal, we can put in the 75 angle. Finally, using the fact that alternate interir angles fr a line

7 crssing tw parallel lines are equal, we can cnclude that the angle in the triangle is 70. As yu will see, this methd eliminates are large amunt f algebra in exchange fr remembering sme gemetry. S, hw t d this prblem? The gal is t find h. If we knw time and the initial speed, we have h frm the psitin diagram. S we have 3 unknwns. Tw mre equatins invlving nly h, v and t are required. One can cme frm the psitin diagram. The ther cmes frm the velcity diagram using the Law f Sines. Therefre, we have 3 equatins and 3 unknwns. Slightly harder than ur ther prblems, but nt t bad as the last equatin we simply have t plug int: Frm the psitin diagram: vt sin(5) = h 2 at vt cs(5) = 20 m r v = 20 /[ tcs(5)] Frm the velcity diagram: v at = sin(15) sin(70) sin(70) v t = using equatin 2 frm the psitin diagram asin(15) 2 sin(70)20 2 t = = 7.3 s r t = 2.7 s asin(15) cs(5) Plugging back int equatin 2 gives: v = 7.4 m/s Plugging int equatin 1 gives: h= vtsin(5) + 2 at = 38.2 m If yu used cmpnents, yur nly real chice is t use the fllwing fur equatins and fur unknwns: hrizntal pstin: v cs(5) t = 20 m vertical pstin: v sin(5) t + 2 at = h hrizntal velcity: vf cs(75) = vcs(5) vertical velcity: v sin(75) = v sin(5) + at f Again, yu have t make the crrect chices f sign. The bttm line is the fllwing: I wuld always draw the vectr additin diagrams. This will help yu make sure yu knw what is happening. Then, decide which methd yu are mre cmfrtable with. If yu are cmfrtable with trig and gemetry, read the mst useful equatins ff f the diagrams, and yu will nt have t wrry abut signs and much algebra. If yu are much mre cmfrtable with algebra, then prceed t write dwn the equatins fr the cmpnents. Either way, what the vectr diagram help with is crrectly lcating the relevant angles in the prblem, s that fr either methd, yu d nt use an incrrect angle. FINAL NOTE: Only use the third equatin: v 2 2 f = v + 2ax when the vectrs a and x are in the SAME DIRECTION. Again, the vectr diagram will make clear when this ccurs!