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1 . Two points A and B ie on a smooth horizonta tabe with AB = a. One end of a ight eastic spring, of natura ength a and moduus of easticity mg, is attached to A. The other end of the spring is attached to a partice P of mass m. Another ight eastic spring, of natura ength a and moduus of easticity mg, has one end attached to B and the other end attached to P. The partice P is on the tabe at rest and in equiibrium. 5a (a) Show that AP =. 3 () The partice P is now moved aong the tabe from its equiibrium position through a distance 0.5a towards B and reeased from rest at time t = 0. At time t, P is moving with speed v and has dispacement from its equiibrium position. There is a resistance to motion of magnitude g mω v whereω =. a d d (b) Show that + ω + 3ω = 0. (5) d (c) Find the veocity,, of P in terms of a, ω and t. (8) (Tota 7 marks). A ight eastic spring AB has natura ength a and moduus of easticity mn a, where n is a constant. A partice P of mass m is attached to the end A of the spring. At time t = 0, the spring, with P attached, ies at rest and unstretched on a smooth horizonta pane. The other end B of the spring is then pued aong the pane in the direction AB with constant acceeration f. At time t the etension of the spring is. d (a) Show that + n = f. (6) (b) Find in terms of n, f and t. (8) Edece Interna Review

2 Hence find (c) the maimum etension of the spring, (3) (d) the speed of P when the spring first reaches its maimum etension. () (Tota 9 marks) 3. A ight eastic spring has natura ength and moduus of easticity mg. One end of the spring is fied to a point O on a rough horizonta tabe. The other end is attached to a partice P of mass m which is at rest on the tabe with OP =. At time t = 0 the partice is projected with speed (g) aong the tabe in the direction OP. At time t the dispacement of P from its initia position is and its speed is v. The motion of P is subject to air resistance of magnitude mvω, where g ω =. The coefficient of friction between P and the tabe is 0.5. (a) Show that, unti P first comes to rest, d d + ω + ω = 0.5g. (6) (b) Find in terms of t, and ω. (6) (c) Hence find, in terms of ω, the time taken for P to first come to instantaneous rest. (3) (Tota 5 marks) Edece Interna Review

3 . A sma ba is attached to one end of a spring. The ba is modeed as a partice of mass 0. kg and the spring is modeed as a ight eastic spring AB, of natura ength 0.5 m and moduus of easticity.5 N. The partice is attached to the end B of the spring. Initiay, at time t = 0, the end A is hed at rest and the partice hangs at rest in equiibrium beow A at the point E. The end A then begins to move aong the ine of the spring in such a way that, at time t seconds, t, the downward dispacement of A from its initia position is sin t metres. At time t seconds, the etension of the spring is metres and the dispacement of the partice beow E is y metres. (a) Show, by referring to a simpe diagram, that y + 0. = + sin t. (3) d y (b) Hence show that + 9y = 98sin t. (5) 98 Given that y = sin t is a particuar integra of this differentia equation, 5 (c) find y in terms of t. (5) (d) Find the time at which the partice first comes to instantaneous rest. () (Tota 7 marks) Edece Interna Review 3

4 5. A P α B A ight eastic spring has natura ength and moduus of easticity mg. One end of the spring is attached to a point A on a pane that is incined to the horizonta at an ange α, where tan α = 3. The other end of the spring is attached to a partice P of mass m. The pane is rough and the coefficient of friction between P and the pane is. The partice P is hed at a point B on the pane where B is beow A and AB =, with the spring ying aong a ine of the pane, as shown in the figure above. At time t = 0, the partice is projected up the pane towards A with speed (g). At time t, the compression of the spring is. (a) Show that d d g. + ω = g, where ω = t (6) (b) Find in terms of, ω and t. (7) (c) Find the distance that P traves up the pane before first coming to rest. () (Tota 7 marks) 6. A partice P of mass m is fied to one end of a ight eastic string, of natura ength a and moduus of easticity man. The other end of the string is attached to a fied point O. The partice P is reeased from rest at a point which is a distance a verticay beow O. The air resistance is modeed as having magnitude mnv, where v is the speed of P. Edece Interna Review

5 (a) Show that, when the etension of the string is, d d + n + n = g (b) Find in terms of t. (5) (7) (Tota marks) Edece Interna Review 5

6 . (a) mge mg( a e) T = ; T = a a (either) T = T e= ( a e) a e = 3 a 5a AP = a + = * * 3 3 (b) T T mω = m mg a mg a + mω = m a 3 a 3 A3 3g + ω + = 0 a + ω + 3ω = 0 * * 5 (c) λ + ωλ + 3ω = 0 ( λ+ 3 ω)( λ+ ω) = 0 λ = 3ω or λ = ω ωt 3ωt = Ae + Be ωt 3ωt = ωae 3ωBe t = 0, = a, = 0 a = A + B 0 = ωa 3ωB 3 A = a, B = a 3 3ωt = v = aω (e ωt e ) 8 [7]. (a) ( ), T = m y Hooke s Law: mn a T = = mn a + y = ft + y = ft + y = f so, ( ), mn = m y m( f ) + n = f B = D * * 6 (b) C.F. : = Acos nt + B sin nt Edece Interna Review 6

7 P.I. : = n f Gen soution: = Acos nt + Bsin nt+ n f = Ansin nt + Bncos nt foow their PI ft f t = 0, = 0 A = n t = 0, = 0 B = 0 f = ( cos nt) 8 n (c) = 0 nt = π f f ma = ( ) = 3 n n (d) y = ft π fπ = f 0 n n [9] g 3. (a) T mg mv = m A3,,,0 mg g mg m = m d d + ω + ω = 0.5g (AG) 6 (b) u + ωu + ω = 0 u = ω (twice) CF is = e ωt (At + B) g PI is = ω t GS is = e ( At + B) g t = 0, = 0 B = ω d ωt = ωe (At + B) + Ae ωt d t = 0, 3 3 g 0. g = g = ω A = ω = = g + ω Edece Interna Review 7

8 so = e ωt 3 ωt + = e ωt (3ωt + ) 6 (c) d = 0 ωe ωt (At + B) + Ae ωt = 0 t = 3 3ω [5]. (a) A sint E y e T v 0.g accn..5e At E, = 0.g 0.5 e = (or ) y = sin t (or ) y = sin t + * 3 Hooke s aw to find etension at equiibrium cao Q specifies reference to a diagram. Correct reasoning eading to given answer Edece Interna Review 8

9 (b) R( ) 0.g T = 0. y M*.5 0.g = 0. y (0. + y sin t) = 0. y DM* (.9y sin t = 0. y ) d y + 9y = 98 sin t * cso 5 Use of F = ma. Weight, tension and acceeration. Condone sign errors. Substitute for tension in terms in terms of Use given resut to substitute for in terms of y Correct unsimpified equation Rearrange to given form cso. (c) CF is y = A cos 7t + B sin 7t 98 Hence GS is y = A cos 7t + B sin 7t + sin t 5 98 t = 0, y = 0: 0 = A so, y = B sin 7t + sin t 5 96 y = 7 B cos 7t + cos t t = 0, y = 0 : 0 = 7B + B = y = (7 sin t sin 7t) 5 Correct form for CF GS for y correct Deduce coefficient of cos θ = 0 Differentiate their y and substitute t = 0, y = 0 y in terms of t. Any eact equivaent. Edece Interna Review 9

10 (d) y = ( cos t cos 7t) 5 y = 0 cos t = cos 7t 7t = kπ ± t k = 9t = π (or 5t = π) π t =, accept 0.698s, 0.70s. 9 y correct set y = 0 sove for genera soution for t: 7t = kπ ± t 9t 5t 9t 5t or: sin sin = 0 sin = 0 or sin = 0 Seect smaest vaue [7] 5. (a) F= R R = mg cosα mg T = ( ): F mg sin α T = m 3 mg. mg mg = m 5 5 d w + = g 6 Edece Interna Review 0

11 (b) w = s L m + w = 0 m =± wi C.F. = Asin wt + Bcoswt P.I. g = = w G.S. = Asin wt + Bcoswt t = 0, = 0 B = = wacoswt wbsin wt g g t = 0, = g : = A A = = (sin wt + coswt ) 7 (c) = 0 wacos wt wbsin wt = 0 A tanwt = = B π wt = (first vaue) = ( + ) = ( ) [7] Edece Interna Review

12 6. (a) O. mn T. a.. mg mg T mn = m mg + man a mn = m n + n = g (*) 5 (b) AE: u + nu + n = 0 (u + n) = n u = n ± ni CF: = e nt g (A cos nt + B sin nt), PI: = n GS: = e nt g (A cos nt + B sin nt) + n g t = 0, = a, = 0: A = a n = e nt ( An sin nt + Bn cos nt) ne nt (A cos nt + B sin nt) = e nt g a (cos nt + sin nt) + n g n 7 [] Edece Interna Review

13 . Part (a) was answered we, and candidates tended to be successfu in appying Hooke s aw and equating the tensions in order to find the distance AP at equiibrium. In part (b) cear diagrams showing the point from which was measured were an advantage here and were often acking, resuting in tension being in the wrong directions and/or with the wrong magnitudes. With defined in the question, candidates often had difficuty in obtaining correct epressions for the etension in each string in terms of and fudges aimed at obtaining the given differentia equation were very common. Some candidates strugged with the mechanics invoved and did not incude a the reevant terms in setting up their equation of motion. In part (c) a great many candidates were abe to sove the second order differentia equation correcty, demonstrating a sound grasp of the pure mathematics invoved, athough a few were epecting a trigonometric soution and forced their auiiary equation to produce one. Many candidates were abe to use correct boundary conditions to find the unknowns and give a correct fina resut. The use of a or 0 in pace of 0.5a for the initia dispacement of P was surprisingy common. A few candidates overooked the request to give the veocity of P as their fina answer.. In part (a) many candidates caimed to have derived the given equation, but very few actuay did so correcty having considered the equation of motion of the partice P. Those candidates who started with a cear diagram were far more ikey to reaise that they needed to consider both the distance moved by P and the etension in the spring. It was very common for candidates to score ony one mark here, for finding the tension in the spring correcty. Amost a candidates in part (b) were confident in attempting to sove the second order differentia equation, athough severa did not choose a correct form for the compementary function, and a few strugged with the particuar integra. It was common to see the correct soution for. In part (c) it was reassuring to find many candidates knowing that the maimum vaue of ( cos nt) is, athough was a popuar aternative answer. Candidates who did not use the basic properties of the trig function were often abe to find the maimum etension by using cacuus, but here too there were some difficuties in identifying the vaue(s) of t for which sin nt = 0. The response to part (d) confirmed that many candidates had itte or no idea of the correct derivation of the equation in (a). Most candidates beieved that the speed of P and the rate of change of the etension in the spring were equa. Correct responses were seen, but usuay ony from the stronger candidates. Edece Interna Review 3

14 3. Estabishing the equation of motion of the system was usuay quite we done, athough fuer and more convincing epanations shoud have been given in many cases. Soving the differentia equation proved more taing. There were frequent errors in the compementary function, the addition of the particuar integra, and the stage in the soution at which arbitrary constants shoud be found. Few candidates paid attention to the dimensiona consistency of their equations. The reaisation that is a ength and that the particuar integra coud aso be epressed as a ength ony ( /) woud have simpified many soutions and enabed to be found in terms of, w and t ony. Those who did obtain the correct genera soution tended to be successfu in using the initia conditions to find the constants. Some candidates with compicated epressions for were abe to simpify their fina answer to 3ω but many were content to give a fina answer that was dimensionay inconsistent with a time.. (a) Most candidates started by finding the correct vaue for the equiibrium etension. Those who were abe to obtain the given resut convincingy initiay incuded the natura ength in their equation. The question did ask candidates to refer to a simpe diagram in their epanation some of the diagrams suggested that candidates had itte appreciation of the situation described in the question. (b) A significant number of candidates omitted this part of the question and others tried to obtain the given resut by differentiating the reationship obtained in part (a). Those who began with Newton s Second Law, with the correct acceeration, generay scored fu marks for this part athough it was quite common to see confusion between and y which was fudged to obtain the required resut. (c) Candidates were usuay abe to give the correct form of the genera soution and then find vaues for the constants of integration by correct methods. Candidates who used ony the compementary function in their working to find the vaues of the two constants scored itte or nothing for the remainder of the question. Simiary, there were some candidates who did not obtain the correct form for the compementary function who coud not then score many further marks.. Some candidates attempted to use incorrect initia conditions based on t = 0, = 0, = 0, and using the resut from part (a). Severa candidates created additiona work for themseves by working from the genera form of the particuar integra to deduce the given form, others substituted the given term into the differentia equation to show that it works. (d) Most candidates differentiated their soution to part (c) and put it equa to 0. Many candidates obtained the correct trigonometric equation cos t = cos 7t. The majority of candidates did not know how to sove this equation and stopped at this point. Those who were abe to sove it generay used the correct sum/product formua, rather than the genera soution for the cosine function. Edece Interna Review

15 5. (a) Most candidates were abe to make a reasonabe attempt athough there were some sign errors in Newton s second aw. Some weaker candidates missed out the component of the weight. (b) (c) The auiiary equation was usuay soved correcty. Subsequenty a common error was either not to find a particuar integra or to attempt to find it having aready used the initia conditions on the compementary function this was heaviy penaised. Athough the straightforward method of equating the veocity to zero was usuay known, those candidates who had not simpified their answer to part (b) were often unabe to compete this part. There were a very sma number who attempted to use an energy method, occasionay correcty. 6. No Report avaiabe for this question. Edece Interna Review 5

Mechanics 3. Elastic strings and springs

Mechanics 3. Elastic strings and springs Chapter assessment Mechanics 3 Eastic strings and springs. Two identica ight springs have natura ength m and stiffness 4 Nm -. One is suspended verticay with its upper end fixed to a ceiing and a partice