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A-Level · Further Mechanics 4 (Elastic Strings, Springs and Elastic Energy)

FP.FM4 Further Mechanics: Elastic Strings, Springs and Elastic Energy

AQA 7367 · Calculator allowed · about 140 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Formulae and modelling assumptions used in this worksheet

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Unless a question states otherwise, take g = 9.8 m s^-2 and model each object as a particle, and each string or spring as light. Hooke's law states that the tension (or thrust) in an elastic string or spring of natural length l and modulus of elasticity lambda, when it is extended (or compressed) by x, has magnitude T = lambda*x/l; an elastic string can only pull (it exerts no force, and is described as slack, whenever its length is less than l), while a spring can both pull and push. The elastic potential energy (EPE) stored in a stretched or compressed string or spring is EPE = lambda*x^2/(2*l). The principle of conservation of mechanical energy states that, provided no energy is lost to resistances such as friction, the sum of the kinetic energy, gravitational potential energy and elastic potential energy of a system remains constant; more generally, the work-energy principle states that the total work done by all the forces acting on a particle (including any work done against resistances) equals its change in kinetic energy.

One end of a light elastic string of natural length 1.4 m is attached to a fixed point A. A particle P is attached to the other end of the string and hangs in equilibrium vertically below A, with the string stretched to a length of 1.75 m. The modulus of elasticity of the string is 98 N.

Figure (to be drawn): Diagram: fixed point A at the top, a light elastic string hanging vertically below it and extended to a total length of 1.75 m, with particle P attached at the lower end and weight mg acting vertically downward on P.

(a)Calculate the tension in the string.(2)

(b)Given that P is in equilibrium, find the mass of P.(2)

(Total for Question 1 is 4 marks)

A light spring has natural length 0.8 m and modulus of elasticity 60 N. It is compressed, first to a length of 0.65 m, and then further to a length of 0.50 m.

Figure (to be drawn): Diagram: spring of natural length 0.8 m shown compressed to a length of 0.65 m, and then compressed further to a length of 0.50 m, with the compression x marked in each case.

(a)Find the elastic potential energy (EPE) stored in the spring when it is compressed to a length of 0.65 m.(3)

(b)Find the EPE stored in the spring when it is compressed further, to a length of 0.50 m.(2)

(c)Hence find the additional work done in compressing the spring from a length of 0.65 m to a length of 0.50 m.(3)

(Total for Question 2 is 8 marks)

The tension in a light elastic string or spring of natural length l and modulus of elasticity λ, when extended (or compressed) by an amount s, is T = λ*s/l. Starting from the fact that the work done in stretching the string by a further small amount ds against this tension is T ds, use integration to show that the elastic potential energy stored when the total extension is x is EPE = λ*x²/(2*l).

(Total for Question 3 is 4 marks)

A particle of mass 2 kg rests in equilibrium on top of a light vertical spring of natural length 0.6 m and modulus of elasticity 147 N, which stands on horizontal ground.

Figure (to be drawn): Diagram: a vertical spring of natural length 0.6 m standing on the ground, compressed under the weight of a particle resting on its top end; weight mg acts vertically downward on the particle, and thrust T acts vertically upward from the spring.

(a)Find the compression of the spring.(3)

(b)Find the length of the spring in this position.(1)

(c)State one assumption made about the spring in this model, and briefly explain why it is needed.(1)

(Total for Question 4 is 5 marks)

A particle P of mass 4 kg rests in equilibrium on a smooth plane inclined at 30 degrees to the horizontal. P is attached to one end of a light elastic string of natural length 1.0 m, lying along a line of greatest slope, whose other end is fixed to a point A at the top of the plane. The string is stretched to a length of 1.2 m.

Figure (to be drawn): Diagram: particle P on a smooth plane inclined at 30 degrees to the horizontal, connected by a light elastic string up the line of greatest slope to a fixed point A at the top of the plane; weight mg acts vertically downward, normal reaction N acts perpendicular to the plane, and tension T acts up the slope along the string.

(a)Show that the modulus of elasticity of the string is 98 N.(4)

(b)The particle P is now replaced by a different particle, of mass 6 kg, attached to the same string. Find the new extension of the string when the system is again in equilibrium.(3)

(Total for Question 5 is 7 marks)

A fitness website advertises a resistance band, modelled as a light elastic string, with natural length 0.5 m and modulus of elasticity 80 N. The advert claims that when the band is stretched to a length of 0.75 m, the elastic potential energy stored in it is 5 J. Verify, showing your working clearly, whether this claim is correct.

(Total for Question 6 is 4 marks)

A small block of mass 0.5 kg rests against one end of a light spring of natural length 0.5 m and modulus of elasticity 100 N, on a smooth horizontal table. The other end of the spring is fixed to a wall. The spring is compressed so that its length is 0.3 m, and the block is released from rest.

Figure (to be drawn): Diagram: horizontal table with a spring fixed to a wall at one end and a block at the other end; the spring is shown compressed to a length of 0.3 m (natural length 0.5 m); beyond the point where the spring reaches its natural length, a rough section of the table of length 1.2 m is marked, followed by a further smooth section.

(a)Find the elastic potential energy stored in the spring at the moment of release.(2)

(b)Find the speed of the block at the instant it leaves the spring (i.e. when the spring first returns to its natural length).(3)

(c)Beyond the point where the spring reaches its natural length, the table is rough for a further distance of 1.2 m, with coefficient of friction 0.4, after which it is smooth again. Find the speed of the block after it has crossed this rough section.(5)

(Total for Question 7 is 10 marks)

Aisha claims that, for a given light elastic string, doubling the extension (while keeping the modulus of elasticity and natural length unchanged) will double the elastic potential energy stored in the string.

(a)By considering the formula for elastic potential energy, explain algebraically whether Aisha's claim is correct.(3)

(b)Illustrate this result using a string of natural length 1.2 m and modulus of elasticity 60 N, by calculating the EPE stored for extensions of 0.2 m and 0.4 m.(3)

(Total for Question 8 is 6 marks)

A particle P of mass 1.2 kg rests in equilibrium on a smooth horizontal table. P is connected to two fixed points, A and B, on the table, with A, P and B lying in a straight line and AB = 3 m. P is attached to A by a light elastic string of natural length 1 m and modulus of elasticity 50 N, and to B by a second light elastic string of natural length 1 m and modulus of elasticity 30 N. Both strings are taut.

Figure (to be drawn): Diagram: horizontal table viewed from above; points A, P and B lie on a straight line with AB = 3 m; a light elastic string of natural length 1 m and modulus 50 N connects A to P, and a second light elastic string of natural length 1 m and modulus 30 N connects P to B, both strings shown taut and pulling P in opposite directions along the line AB.

(a)Using the fact that AB = 3 m, write down an equation connecting the extension x_A of string AP and the extension x_B of string PB.(2)

(b)By equating the tensions in the two strings, find the values of x_A and x_B.(5)

(c)Find the tension in each string.(1)

(d)Explain why the mass of P does not appear anywhere in the working used to find x_A and x_B.(2)

(Total for Question 9 is 10 marks)

A particle P of mass 5 kg is attached to one end of a light elastic string of natural length 2 m and modulus of elasticity 588 N. The other end of the string is attached to a fixed point A. Initially P is held at the same level as A and is released from rest, falling vertically downwards in a straight line.

Figure (to be drawn): Diagram: fixed point A with P held initially at the same level as A (string slack); P then falls vertically, the string becoming taut once P has fallen 2 m, and stretching further as P continues to fall to its lowest point.

(a)Explain why the string does not begin to exert any force on P until P has fallen a distance of 2 m.(1)

(b)Find the speed of P at the instant the string becomes taut.(3)

(c)Show that the maximum extension of the string is 1 m.(6)

(d)Find the maximum tension in the string during the motion.(2)

(Total for Question 10 is 12 marks)

A block of mass 0.8 kg is moving at 5 m/s on a rough horizontal floor when it strikes one end of a light spring of natural length 1 m and modulus of elasticity 200 N, whose other end is fixed to a wall. The coefficient of friction between the block and the floor is 0.2, and this friction acts throughout the block's contact with the spring.

Figure (to be drawn): Diagram: block moving at 5 m/s towards a spring fixed to a wall on a rough horizontal floor; the spring is shown compressed by an amount x as the block decelerates, with the friction force and the spring's thrust both acting to oppose the block's motion.

(a)Show that the maximum compression x of the spring satisfies the equation 100*x² + 1.568*x - 10 = 0.(4)

(b)Solve this equation to find the maximum compression of the spring, giving your answer to 3 significant figures.(2)

(c)Given that the same frictional force acts over the same distance as the block moves back from the point of maximum compression to the point where the spring returns to its natural length, find the speed of the block at that point.(4)

(d)State one modelling assumption, other than the coefficient of friction being constant, that has been made in this problem, and briefly explain how it might affect the accuracy of the calculated speeds.(2)

(Total for Question 11 is 12 marks)

A particle P of mass 2 kg lies on a smooth plane inclined at 30 degrees to the horizontal. P is attached to one end of a light elastic string of natural length 1.6 m and modulus of elasticity 120 N, lying along a line of greatest slope; the other end of the string is fixed to a point A at the top of the plane. Initially P is held at A (so the string is not stretched) and is released from rest, sliding down the line of greatest slope.

Figure (to be drawn): Diagram: particle P held initially at point A at the top of a smooth incline of angle 30 degrees, connected by a light elastic string (natural length 1.6 m) down the line of greatest slope; P released from rest and sliding down, with the string becoming taut after 1.6 m and then stretching further as P continues down the slope.

(a)Find the speed of P at the instant the string becomes taut.(3)

(b)Show that the maximum extension x of the string satisfies the equation 37.5*x² - 9.8*x - 15.68 = 0, and hence find x, giving your answer to 3 significant figures.(6)

(c)Find the maximum tension in the string during the motion.(3)

(Total for Question 12 is 12 marks)

A bungee-style fairground ride uses a ball of mass 4 kg attached to one end of a long light elastic string of natural length 8 m. The other end of the string is fixed to a point O at the top of a tower. The ball is released from rest at O and falls vertically. The ball comes to instantaneous rest at its lowest point, having fallen a total distance of 15 m below O.

Figure (to be drawn): Diagram: fixed point O at the top of a tower; ball released from rest at O and falling vertically; the string remains slack for the first 8 m (its natural length), becomes taut, and stretches further until the ball reaches its lowest point, 15 m below O in total.

(a)Find the speed of the ball at the instant the string becomes taut.(3)

(b)Given that the ball falls a total distance of 15 m before coming to instantaneous rest, find the modulus of elasticity of the string.(5)

(c)Find the maximum tension in the string during the motion, and state at what point of the motion this maximum occurs.(3)

(d)The ball subsequently moves back upwards. State, with a reason, the condition needed for this subsequent motion to be modelled using simple harmonic motion, and explain why this model would not apply for the whole of the return journey.(2)

(Total for Question 13 is 13 marks)

Mark scheme · FP.FM4 Further Mechanics: Elastic Strings, Springs and Elastic Energy

Question 1

(a) M1 uses Hooke's law T = λ*x/l with x = 1.75 - 1.4 = 0.35 (m)
(a) A1 T = 24.5 N, cao
(a) Answer: T = 24.5 N
(b) M1 equates the tension to the weight: mg = T, ft (a)
(b) A1 m = 2.5 kg, cao
(b) Answer: m = 2.5 kg

Question 2

(a) B1 uses EPE = λ*x²/(2*l) with x = 0.8 - 0.65 = 0.15 (m)
(a) M1 substitutes correctly: 60*0.15²/(2*0.8)
(a) A1 EPE = awrt 0.844 J (exact value 0.84375 J), cao
(a) Answer: EPE = awrt 0.844 J (exact 0.84375 J)
(b) M1 substitutes x = 0.8 - 0.5 = 0.3 into EPE = λ*x²/(2*l)
(b) A1 EPE = 3.375 J, cao
(b) Answer: EPE = 3.375 J
(c) M1 recognises the additional work done equals the difference in EPE values, ft (a) and (b)
(c) A1 3.375 - 0.84375, oe, ft
(c) A1 awrt 2.53 J (exact 2.53125 J), cao
(c) Answer: Additional work done = awrt 2.53 J (exact 2.53125 J)

Question 3

M1 sets up EPE = integral from 0 to x of T ds = integral from 0 to x of (λ*s/l) ds
M1 integrates correctly: (λ/l)*[s²/2] (+ c)
A1 substitutes limits: (λ/(2*l))*x² - (λ/(2*l))*0²
A1 EPE = λ*x²/(2*l), cso
Answer: EPE = λ*x²/(2*l) (shown)

Question 4

(a) M1 at equilibrium, thrust in spring equals weight of particle: T = mg = 2*9.8
(a) M1 uses Hooke's law T = λ*x/l, rearranged for x
(a) A1 x = 0.08 m, cao
(a) Answer: x = 0.08 m
(b) B1 length = 0.6 - 0.08 = 0.52 m, ft (a)
(b) Answer: 0.52 m
(c) B1 valid assumption with reason, e.g. the spring is modelled as light (of negligible mass), so its own weight does not need to be included in the equilibrium equation; or the spring has not exceeded its elastic limit, so T = λ*x/l remains valid, oe
(c) Answer: The spring is modelled as light, so its own weight does not contribute to the equilibrium equation.

Question 5

(a) M1 resolves forces along the plane at equilibrium: T = mg*sin(30 degrees)
(a) A1 T = 19.6 N, cao
(a) M1 uses Hooke's law rearranged for λ: λ = T*l/x, with x = 1.2 - 1.0 = 0.2
(a) A1 λ = 98 N, cso
(a) Answer: λ = 98 N (shown)
(b) M1 new tension T = 6*9.8*sin(30 degrees)
(b) M1 uses Hooke's law rearranged for x: x = T*l/λ, ft (a)
(b) A1 x = 0.3 m, cao
(b) Answer: x = 0.3 m

Question 6

M1 finds the extension x = 0.75 - 0.5 = 0.25 (m)
M1 substitutes into EPE = λ*x²/(2*l): 80*0.25²/(2*0.5)
A1 EPE = 5 J, cao
A1 concludes the claim is correct, since the calculated value equals the advertised 5 J, cso
Answer: EPE = 5 J; the claim is correct.

Question 7

(a) M1 uses EPE = λ*x²/(2*l) with x = 0.5 - 0.3 = 0.2
(a) A1 EPE = 4 J, cao
(a) Answer: EPE = 4 J
(b) M1 uses conservation of energy: EPE = (1/2)*m*v², ft (a)
(b) M1 rearranges for v²: v² = 2*4/0.5
(b) A1 v = 4 m/s, cao
(b) Answer: v = 4 m/s
(c) M1 work done against friction = μ*m*g*d
(c) A1 = 0.4*0.5*9.8*1.2 = 2.352 J, awrt
(c) M1 applies the work-energy principle: (1/2)*m*v² = (their KE leaving spring) - (their friction work), ft (b)
(c) M1 rearranges for v²
(c) A1 v = 2.57 m/s, awrt, cao
(c) Answer: v = awrt 2.57 m/s

Question 8

(a) M1 states EPE = λ*x²/(2*l), so EPE is proportional to x² for fixed λ and l
(a) A1 if x is doubled, EPE increases by a factor of 2² = 4, not 2
(a) A1 concludes Aisha's claim is incorrect, oe
(a) Answer: Aisha's claim is incorrect: since EPE is proportional to x², doubling x quadruples the EPE, rather than doubling it.
(b) M1 substitutes x = 0.2: EPE = 60*0.2²/(2*1.2)
(b) A1 EPE = 1 J, cao
(b) A1 substitutes x = 0.4 to find EPE = 4 J, confirming the factor of 4 (not 2), cao
(b) Answer: EPE(0.2 m) = 1 J; EPE(0.4 m) = 4 J (a factor of 4, confirming part (a))

Question 9

(a) B1 recognises that the sum of the natural lengths is 1 + 1 = 2 m
(a) B1 x_A + x_B = 3 - 2 = 1 (m), oe
(a) Answer: x_A + x_B = 1
(b) M1 writes tension in each string using Hooke's law: T_A = 50*x_A/1, T_B = 30*x_B/1
(b) M1 equates the tensions (P is in equilibrium with no other horizontal forces): 50*x_A = 30*x_B
(b) M1 solves simultaneously with x_A + x_B = 1 from (a), ft
(b) A1 x_A = 0.375 m, cao
(b) A1 x_B = 0.625 m, cao
(b) Answer: x_A = 0.375 m, x_B = 0.625 m
(c) B1 T = 18.75 N (both strings), ft (b)
(c) Answer: T = 18.75 N
(d) B1 the table is smooth and horizontal, so P's weight is balanced vertically by the normal reaction from the table, and does not act along the line AB
(d) B1 the only horizontal forces on P are the two tensions, so the equilibrium equation 50*x_A = 30*x_B involves only the string properties, not the mass, oe
(d) Answer: The weight of P acts vertically and is balanced by the normal reaction from the table; the only forces acting along AB are the two tensions, so the horizontal equilibrium equation does not involve P's mass.

Question 10

(a) B1 an elastic string can only pull (exert tension) and cannot push, so it is slack and exerts no force while its length is less than its natural length of 2 m, oe
(a) Answer: The string is slack (and so exerts no tension) until P has fallen its natural length of 2 m, since a string cannot push.
(b) M1 uses v² = u² + 2*a*s with u = 0, a = g = 9.8, s = 2
(b) A1 v² = 39.2
(b) A1 v = 6.26 m/s, awrt (or exact √39.2)
(b) Answer: v = awrt 6.26 m/s
(c) M1 uses conservation of energy from release to the lowest point (speed = 0 at both ends): loss in GPE = EPE gained
(c) A1 correct equation: mg*(2+x) = 588*x²/(2*2), i.e. 49*(2+x) = 147*x²
(c) M1 rearranges to a 3-term quadratic: 147*x² - 49*x - 98 = 0 (or a simplified equivalent, e.g. 3*x² - x - 2 = 0)
(c) M1 solves the quadratic (by factorising or the quadratic formula)
(c) A1 rejects the negative root x = -2/3 as not physically valid
(c) A1 x = 1 m, cso
(c) Answer: Maximum extension = 1 m (shown)
(d) M1 uses T = λ*x/l at maximum extension, ft (c)
(d) A1 T = 294 N, cao
(d) Answer: T = 294 N

Question 11

(a) M1 initial KE = (1/2)*0.8*5² = 10 (J)
(a) M1 EPE stored at compression x = 200*x²/(2*1) = 100*x²
(a) M1 work done against friction over compression x = μ*m*g*x = 0.2*0.8*9.8*x = 1.568*x
(a) A1 applies the work-energy principle (KE = EPE + friction work) and rearranges to the given equation, cso
(a) Answer: 100*x² + 1.568*x - 10 = 0 (shown)
(b) M1 uses the quadratic formula (or equivalent method) to solve, ft (a)
(b) A1 x = 0.308 m, awrt, cao
(b) Answer: x = awrt 0.308 m
(c) M1 EPE released = 100*x², ft (b)
(c) M1 KE on leaving spring = (their EPE) - (friction work over x = 1.568*x), ft
(c) M1 rearranges (1/2)*0.8*v² = (their KE) for v
(c) A1 v = 4.75 m/s, awrt, cao
(c) Answer: v = awrt 4.75 m/s
(d) B1 valid assumption, e.g. the spring is modelled as light (massless), so its own kinetic energy during compression/extension is ignored
(d) B1 explains the likely effect, e.g. if the spring has mass, some of the initial kinetic energy would be transferred to the spring itself, so the calculated speeds would be slightly too high, oe
(d) Answer: The spring is modelled as light; if it actually has mass, some energy would go into moving the spring itself, so the true final speed would be slightly lower than calculated.

Question 12

(a) M1 uses v² = 2*(g*sin(30 degrees))*1.6, or an equivalent energy method
(a) A1 v² = 15.68
(a) A1 v = 3.96 m/s, awrt
(a) Answer: v = awrt 3.96 m/s
(b) M1 uses conservation of energy from release to the point of maximum extension (speed = 0 at both ends): loss in GPE = EPE gained
(b) A1 correct equation: 2*9.8*sin(30 degrees)*(1.6+x) = 120*x²/(2*1.6), i.e. 9.8*(1.6+x) = 37.5*x²
(b) M1 rearranges to the given quadratic form
(b) M1 solves using the quadratic formula
(b) B1 rejects the negative root as not physically valid
(b) A1 x = 0.790 m, awrt, cao
(b) Answer: x = awrt 0.790 m
(c) M1 uses T = λ*x/l, ft (b)
(c) M1 substitutes λ = 120, l = 1.6, x from (b)
(c) A1 T = 59.3 N, awrt, cao
(c) Answer: T = awrt 59.3 N

Question 13

(a) M1 uses v² = 2*g*8 (free fall over the natural length before the string becomes taut)
(a) A1 v² = 156.8
(a) A1 v = 12.5 m/s, awrt
(a) Answer: v = awrt 12.5 m/s
(b) M1 identifies the extension of the string at the lowest point: x = 15 - 8 = 7 (m)
(b) M1 uses conservation of energy over the whole fall (speed = 0 at O and at the lowest point): loss in GPE = EPE gained
(b) A1 correct equation: 4*9.8*15 = λ*7²/(2*8)
(b) M1 rearranges for λ
(b) A1 λ = 192 N, cao
(b) Answer: λ = 192 N
(c) M1 uses T = λ*x/l with x = 7, ft (b)
(c) A1 T = 168 N, cao
(c) B1 occurs at the lowest point of the motion (i.e. at the point of maximum extension), oe
(c) Answer: T_max = 168 N, occurring at the lowest point (maximum extension).
(d) B1 SHM requires the string to remain taut throughout, i.e. its extension x ≥ 0, so that the restoring force is proportional to the displacement from the equilibrium position at all times considered
(d) B1 once the ball rises above the point where the string is at its natural length, the string would need to push the ball to continue the SHM model, but a string cannot exert thrust, so it goes slack and the SHM model breaks down beyond this point, oe
(d) Answer: SHM requires the string to stay taut (extension x ≥ 0); once the ball rises past the natural-length point the string would need to push, which it cannot do, so it goes slack and the SHM model no longer applies.

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