Mechanics 3. Elastic strings and springs

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Dominick Weaver
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1 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 of mass kg hanging in equiibrium from its ower end. (i) Cacuate the extension of the spring. [] The second spring is then attached to the partice and its other end is attached to the foor. The system is in equiibrium with the two springs in the same vertica ine. The distance between the foor and ceiing is.5 m and the extension of the upper spring is e metres. The situation is shown in the diagram beow. m.5 m e m (ii) Write down the extension of the ower spring in terms of e. [] (iii) Write down the equiibrium equation for the partice and hence cacuate e. [4] (iv) Cacuate the tensions in the springs. []. Each of two ight eastic strings, AB and BC, has moduus N. AB has natura ength.5 m and BC has natura ength.8 m. The strings are both attached at B to a partice of mass.75 kg. The ends A and C are fixed to points on a smooth horizonta tabe such that AC = m, as shown in the diagram. A m B C Initiay the partice is hed at the mid-point of AC and reeased from rest. (i) Find the tension in each string before reease and cacuate the acceeration of the partice immediatey after it is reeased. [5] MEI, 9/5/8

The partice is now moved to the position where it is in equiibrium. The extension in AB is e m. (ii) Cacuate e. [4] The partice is now hed at A and reeased from rest.

2 The partice is now moved to the position where it is in equiibrium. The extension in AB is e m. (ii) Cacuate e. [4] The partice is now hed at A and reeased from rest. (iii) Show that in the subsequent motion BC becomes sack. Cacuate the furthest distance of the partice from A. [6] 3. A ight eastic string AB, of natura ength.8 m and moduus 5 N, is attached at A to a ceiing which is.4 m above the foor. A sma ba of mass. kg is attached to the other end B and hangs in equiibrium. (i) Cacuate the ength of the string. [3] (ii) The ba is pued down unti it touches the foor with AB vertica and it is then reeased from rest. Cacuate the speed at which it hits the ceiing. [4] A second ight eastic string of moduus 5 N and natura ength m, where <.4, is attached to the ba at B and to the foor verticay beow A. The ba is hed at rest on the foor with AB vertica and it is then reeased. (iii) Find the range of vaues of for which the ba wi sti hit the ceiing. [8] 4. A mechanism for firing a ba in a tabe-top game is shown in the diagrams. A moving piston of mass. kg sides freey in a barre, which is fixed. A spring of natura ength.5 m and negigibe mass is fitted inside the barre so that before being primed for use the configuration is that shown in the diagram beow. You may assume that the force due to the compression of the spring may be modeed by Hooke's Law with a stiffness of Nm -.. m barre spring piston (i) Show that the energy stored in the spring is.5 J when it is in the configuration above. [] The mechanism is primed for firing by puing the piston back a distance d, in metres, where d <., as shown in the diagram beow. (ii) Show that the spring has been compressed by a distance (d +.5) m and that the work done in priming the mechanism is d( + d). [3] MEI, 9/5/8

3 The mechanism is primed with d =.8m and a ba of mass. kg is paced in contact with the piston. (iii) Cacuate the aunching speed of the ba (i.e. the speed at which the piston hits the case), if the mechanism is reeased horizontay. [3] (iv) Cacuate the aunching speed of the ba if the mechanism is fired verticay upwards. [3] Tota 5 marks Soutions to Chapter assessment. (i) Verticay: T = g Hooke s Law: T = kx g = 4x 9.8 x = =.49 4 The extension is.49 m. T g (ii) Extension = (.5 e) =.5 e (iii) Verticay: T = g + T Hooke s Law for upper spring: T = 4e Hooke s Law for ower spring: T = 4(.5 e) = 4e 4e = e 8e = 9.6 e =.745 T g + T (iv) T = 4e = = 9.8 The tension in the upper spring is 9.8 N. T = 4e = = 9. The tension in the ower spring is 9. N. MEI, 9/5/8

4 . (i) Initiay AB = m so extension =.5 λ x.5 For AB: T = = =.5 The tension in AB is N. Intiay BC = m so extension =. λ x. For BC: T = = = 5.8 The tension in BC is 5 N. Newton s nd aw: 5 =.75a a = The acceeration of the partice is ms -. (ii) In equiibrium, extension of AB is e m and extension of BC is (.7 e) m. λ x e For AB: T = = = 4e.5 λ x (.7 e) For BC: T = = = 7.5 5e.8 In equiibrium the tensions are equa, so 4e = 7.5 5e 65e = 7.5 e =.69 (3 s.f.) (iii) Initiay extension of BC is. m λ x. E.P.E. in BC = = = 8 J.8 If BC becomes sack, extension of AB wi be.7 m. λ x.7 In this case, E.P.E. in AB = = = 9.8 J..5 Since this is ess than the origina energy in the system, the partice must have kinetic energy of 8. J at this point, so BC does become sack. At furthest distance from A, K.E. =, BC is sack and so E.P.E. in AB must be equa to the origina energy in the system. λ x x E.P.E. in AB = = = x.5 x = 8 x =.949 (3 s.f.) The ength of AB is therefore.45 m (3 s.f.) MEI, 9/5/8

5 3. (i) Verticay: T =.g λ x Hooke s Law: T = 5 x.g = x = = Length of string =.36 m. T.g (ii) When ba is on foor, extension =.6 m λ x 5.6 E.P.E. stored in string = = = 8 J.8 When ba hits ceiing, G.P.E. gained = mgh = = 4.74 J and K.E. gained = mv =.v =.v By conservation of energy: E.P.E. ost = G.P.E. gained + K.E. gained 8 = v v = 5.74 (3 s.f.) Speed = 5.74 ms - (3 s.f.) MEI, 9/5/8

6 (iii) When ba hits ceiing, extension of second string =.4 λ x 5(.4 ) E.P.E. stored in second string when ba hits ceiing = = By conservation of energy: E.P.E. ost by first string = E.P.E. gained by second string + G.P.E. gained + K.E. gained If the ba hits the ceiing, the K.E. gained is greater than or equa to zero E.P.E. ost by st string E.P.E. gained by nd string G.P.E. gained 5(.4 ) (.4 ) ( ) Using the quadratic formua, roots of equation = 3.59 ± are =.6 or From the graph, the soution set of the inequaity is Since <.4, the possibe vaues of are.6 < (i) Compression of spring =.5 m E.P.E. stored = kx =.5 =.5 J. (ii) Origina compression =.5 m Further compression = d m so tota compression of spring = (.5 + d) m. Work done = increase in E.P.E. = ( d +.5).5 = d + d +.5,5 = d(d + ) MEI, 9/5/8

7 (iii) For d =.8, work done =.8(.8 + ) =.44 Work done = K.E. gained.44 =.v.44 =.v v = 3.79 (3 s.f.) Incude mass of piston as we as mass of ba (iv) Work done = K.E. gained + G.P.E. gained.44 =.v =.v v =.83 v = 3.58 (3 s.f.) MEI, 9/5/8

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PhysicsAndMathsTutor.com . 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