Energy, work and power — Paper 5 · A Level Mathematics 9709

4 A particle P of mass 0.4 kg is attached to one end of a light elastic string of natural length 1.5 m and modulus of elasticity 6 N. The other end of the string is attached to a fixed point O on a rough horizontal table. P is released from rest at a point on the table 3.5 m from O. The speed of P at the instant the string becomes slack is 6 m s−1. Find (i) the work done against friction during the period from the release of P until the string becomes slack, [5] (ii) the coefficient of friction between P and the table. [2]

7 marks

Mark scheme: 4 (i) Initial EE = 6 x 22 ÷ (2 x 1.5) B1 Final KE = ½ 0.4 x 62 B1 M1 For using WD against friction = initial EPE – final KE WD = 6 x 4 ÷ (2 x 1.5) – ½ 0.4 x 62 A1 ft Any correct form WD against friction is 0.8 J A1 5 (ii) (0.8 = µ 0.4g x 2) M1 For using WD = F x d and F = µ R Coefficient is 0.1 A1 ft 2 ft µ = WD ÷ 8

This question in 9709/51 May/June 2005

5 A particle P of mass 0.2 kg is attached to the mid-point of a light elastic string of natural length 5.5 m and modulus of elasticity λ N. The ends of the string are attached to fixed points A and B which are at the same horizontal level and 6 m apart. P is held at rest at a point 1.25 m vertically above the mid-point of AB and then released. P travels a distance 5.25 m downwards before coming to instantaneous rest (see diagram). By considering the changes in gravitational potential energy and elastic potential energy as P travels downwards, find the value of λ. [8]

8 marks

Mark scheme: 5 Loss in GPE = 0.2g×5.25 (10.5 J) B1 AP is 3.25 initially and 5 finally B1 M1 For using EE = λ x2 ÷ (2L) L must be correct (2.75 or 5.5) For any correct expression for Initial EPE or for Final EPE A1 ft ft incorrect AP [2×0.52λ ÷ (2×2.75) for initial or 2×2.252λ ÷ (2×2.75) for final] Gain in EPE = (81-4)λ /44 =1.75λ A1 Any correct expression M1 For applying the principle of conservation of energy 1.75λ = 10.5 A1 ft For any correct equation in λ , ft only if initial and final EPE are used λ = 6 A1 8 GCE A/AS LEVEL – November 2005 9709, 8719 5

This question in 9709/51 Oct/Nov 2005

5 One end of a light elastic string, of natural length 0.5 m and modulus of elasticity 140 N, is attached to a fixed point O. A particle of mass 0.8 kg is attached to the other end of the string. The particle is released from rest at O. By considering the energy of the system, find (i) the speed of the particle when the extension of the string is 0.1 m, [4] (ii) the extension of the string when the particle is at its lowest point. [4]

8 marks

Mark scheme: 5 (i) Gain in Elastic PE = 140(0.1)²/(2 x 0.5) B1 Loss in GPE = 0.8 g(0.5 + 0.1) B1 ½0.8v² + 140 x 0.1² = M1 For using Gain in KE + Gain in 0.8 g (0.5+0.1) EPE = Loss in GPE Speed is 2.92 ms −1 A1 4 (ii) M1 For using Gain in EPE = Loss in GPE 140x² = 0.8 g (0.5 + x) A1 (5x – 1)(28x + 4) = 0 M1 For solving the resulting 3 term quadratic equation Extension is 0.2 m A1 4 8

This question in 9709/51 May/June 2007

6 A and B are fixed points on a smooth horizontal table. The distance AB is 2.5 m. An elastic string of natural length 0.6 m and modulus of elasticity 24 N has one end attached to the table at A, and the other end attached to a particle P of mass 0.95 kg. Another elastic string of natural length 0.9 m and modulus of elasticity 18 N has one end attached to the table at B, and the other end attached to P. The particle P is held at rest at the mid-point of AB (see diagram). (i) Find the tensions in the strings. [3] The particle is released from rest. (ii) Find the acceleration of P immediately after its release. [2] (iii) P reaches its maximum speed at the point C. Find the distance AC. [4]

9 marks

Mark scheme: 6 (i) 24 x 0.65/0.6 or 18 x 0.35/0.9 M1 For using T = λ x/L Tension in AP is 26N A1 Tension in BP is 7N A1 3 (ii) 26 – 7 = 0.95a M1 For using Newton’s second law (3 terms) Acceleration is 20 ms −2 A1 2 ft T AP − T BP = 0.95a (iii) M1 For using T AP = T BP 24x/0.6 = 18(1 – x)/0.9 A1 x = 1/3 DM1 For attempting to solve for x Distance is 0.933 m A1 4 9

This question in 9709/51 May/June 2007

5 Each of two light elastic strings, S1 and S2, has modulus of elasticity 16 N. The string S1 has natural length 0.4 m and the string S2 has natural length 0.5 m. One end of S1 is attached to a fixed point A of a smooth horizontal table and the other end is attached to a particle P of mass 0.5 kg. One end of S2 is attached to a fixed point B of the table and the other end is attached to P. The distance AB is 1.5 m. The particle P is held at A and then released from rest. (i) Find the speed of P at the instant that S2 becomes slack. [4] (ii) Find the greatest distance of P from A in the subsequent motion. [3]

7 marks

Mark scheme: 5 (i) M1 For using EE = λ x²/2L M1 For using EE S 2 (initial) = ½mv² + EE S 1 (S2 just slack) ½(16x1²/0.5) = ½0.5v²+½(16x0.6²/0.4) A1 Speed is 5.93ms −1 A1 4 (ii) M1 For using EE S 2 (initial) = EE S 1 ½(16x1²/0.5) = ½(16x²/0.4) A1 (x = 0.894) Distance is 1.29m A1 3 7

This question in 9709/51 Oct/Nov 2007

6 One end of a light elastic string of natural length 1.25 m and modulus of elasticity 20 N is attached to a fixed point O. A particle P of mass 0.5 kg is attached to the other end of the string. P is held at rest at O and then released. When the extension of the string is x m the speed of P is v m s−1. (i) Show that v2 = −32x2 + 20x + 25. [4] (ii) Find the maximum speed of P. [3] (iii) Find the acceleration of P when it is at its lowest point. [4]

11 marks

Mark scheme: 6 (i) EE gain = 20x2/(2x1.25) B1 PE loss = 0.5g(1.25 + x) B1 [ 1 0.5v2 = 6.25 + 5x – 8x2] M1 For using KE gain = PE loss – EE gain 2 v2 = -32x2 + 20x + 25 A1 4 AG ALTERNATIVE For using Newton’s second law with [5.0 v ( dv / dx ) = − 20 x / .1 25 + 5.0 g ] M1 A=v(dv/dx) and T = λx / L 1 For integrating and using [ 0.5v2 = -10x2/1.25 + 5x + c] M1 1 2 0.5v(0)2 = 0.5gx1.25 2 c = 6.25 A1 v2 = -32x2 + 20x + 25 A1 (4) AG 2 For obtaining v2 in the form (ii) [v2=-32(x-5/16) + 28.125] M1 a(x – b)2 + c [ vmax = 28.125 ] M1 For substituting v max = c Maximum speed is 5.30ms-1 A1 3 ALTERNATIVE 1 [-64x + 20 = 0 ⇒ x = 5/16] M1 For solving d(v2)/dx for x [vmax 2= -32(5/16)2 + 20(5/16) + 25] M1 For substituting x found into v2 (x) Maximum speed is 5.30ms-1 A1 (3) ALTERNATIVE 2 [T = mg = 5, T = λ x/L = 20x/1.25 ⇒ M1 Using a = 0 at maximum speed x = 5/16 ] M1 For substituting x = 5/16 in v2 Maximum speed is 5.30ms-1 A1 (3) (iii) [-32x2 + 20x + 25 = 0 ⇒ x = 1.25] M1 For attempting to solve v = 0 [a = v(dv)/dx = 1 d(v2)/dx = -32x + 10] 2 M1 For using a = 12 d(v2)/dx a = -32x1.25 + 10 A1ft Acceleration is 30ms-2 (upwards) A1 4 ALTERNATIVE 1 [-32x2 + 20x + 25 = 0 ⇒ x = 1.25] M1 For attempting to solve v = 0 [0.5g – 20x/1.25 = 0.5a] M1 For using Newton’s second law a = g – 2(20x1.25/1.25) A1ft Acceleration is 30ms-2 (upwards) A1 (4) ALTERNATIVE 2 [-32x2 + 20x + 25 = 0 ⇒ x = 1.25] M1 For attempting to solve v = 0 M1 Using Newton’s 2nd Law 20 – 5 = 0.5a or 5 – 20 = 0.5a A1ft If a = -30 then the direction should be Acceleration is 30ms-2 A1 (4) 11 explained GCE A/AS LEVEL – May/June 2008 9709 05 For using Newton’s second law and

This question in 9709/51 May/June 2008

6 A light elastic string has natural length 4 m and modulus of elasticity 2 N. One end of the string is attached to a fixed point O of a smooth plane which is inclined at 30◦to the horizontal. The other end of the string is attached to a particle P of mass 0.1 kg. P is held at rest at O and then released. The speed of P is v m s−1 when the extension of the string is x m. (i) Show that v2 = 45 −5(x −1)2. [5] Hence find (ii) the distance of P from O when P is at its lowest point, [2] (iii) the maximum speed of P. [2]

9 marks

Mark scheme: 6 (i) EE gain = 2x 2 /(2x4) B1 PE loss = 0.1g(4 + x)sin30 o B1 [ 12 0.1v 2 = 2 + 0.5x – 0.25x 2 ] M1 For using KE gain=PE loss-EE gain v 2 = 40 + 10x – 5x 2 = 45 – (5x 2 – 10 x + 5) M1 For attempting to express v 2 in the required form v 2 = 45 – 5(x – 1) 2 A1 5 AG (ii) 5(x – 1) 2 = 45 M1 For substituting v = 0 Distance is 8 m A1 2 (iii) M1 For using x = 1 for maximum v Maximum speed is 6.71 ms −1 A1 2 [9] GCE A/AS LEVEL – October/November 2008 9709 05

This question in 9709/51 Oct/Nov 2008

6 P A M B 2 m A particle P of mass 1.6 kg is attached to one end of each of two light elastic strings. The other ends of the strings are attached to fixed points A and B which are 2 m apart on a smooth horizontal table. The string attached to A has natural length 0.25 m and modulus of elasticity 4 N, and the string attached to B has natural length 0.25 m and modulus of elasticity 8 N. The particle is held at the mid-point M of AB (see diagram). (i) Find the tensions in the strings. [2] (ii) Show that the total elastic potential energy in the two strings is 13.5 J. [2] P is released from rest and in the subsequent motion both strings remain taut. The displacement of P from M is denoted by x m. Find (iii) the initial acceleration of P, [2] (iv) the non-zero value of x at which the speed of P is zero. [4]

10 marks

Mark scheme: 6 (i) [TA = 4x0.75/0.25 and TB = 8x0.75/0.25] M1 For using T = λ x/L Tensions are 12 N and 24 N A1 2 (ii) [Total EE = 4x0.752/(2x0.25) + 8x0.752/(2x0.25)] M1 For using T = λ x2/2L Total EE = 13.5J A1 2 AG (iii) [TB – TA = ma] M1 For using Newton’s second law Acceleration is 7.5 ms–2 A1√ 2 Ft 0.625(TB – TA) (iv) M1 For attempting to set up an equation using EE 4(0.75 + x) 2/(2x0.25) + 8(0.75 – x) 2/(2x0.25) = 13.5 A1 [–12x(1–2x) = 0 ⇒ x = 0, ½ ] M1 For attempting to solve the correct quadratic equation Value of x is 0.5 A1 4 [10]

This question in 9709/51 May/June 2009

2 A particle of mass 0.2 kg is attached to one end of a light elastic string of natural length 0.6 m and modulus of elasticity 4 N. The other end of the string is attached to a fixed point O. The particle is held at a point which is (0.6 + x) m vertically below O. The particle is released from rest. In the subsequent motion the speed of the particle is 3 m s−1 when the string becomes slack. By considering energy, find the value of x. [5]

5 marks

Mark scheme: 2 Gain in KE = ½ 0.2 × 32 (= 0.9 J) B1 Gain in GPE = 0.2gx (= 2x J) B1 Loss in EPE = ½ 4x2/0.6 (= 10x2/3 J) B1 [10x2/3 = 9/10 + 2x] M1 For using gain in KE + Gain in GPE = Loss in EPE x = 0.9 A1 5 [5]

This question in 9709/51 Oct/Nov 2009

2 20 cm 46 cm 2 cm 32 cm A bucket that consists of three parts stands on horizontal ground. The base is in the form of a uniform circular disc of diameter 32 cm and thickness 2 cm. The body is in the form of a uniform hollow cylinder of outer diameter 32 cm and height 46 cm. The handle is in a vertical plane, attached at opposite ends of an outer diameter at the top of the cylinder. The handle is in the form of a uniform circular arc of radius 20 cm. The diagram shows the cross-section of the bucket in the plane of the handle. (i) Show that the centre of mass of the handle is 53.25 cm above the ground, correct to 4 significant figures. [3] The weights of the base, body and handle are 50 N, 100 N and 25 N respectively. (ii) Find the height of the centre of mass of the bucket above the ground. [2]

5 marks

Mark scheme: 2 (i) [ y handle is 20 × 0.8/0.927.. from centre M1 For using y = rsin α ÷ α (17.25)] Centre of arc is 12 cm below top of B1 cylinder Height = 2 + 46 – 12 + 17.25 = 53.25 cm A1 3 AG (ii) [(50 + 100 + 25) y = M1 For taking moments about the base 50 × 1 + 100 × 25 + 25 × 53.25] Height is 22.2 cm A1 2 5

This question in 9709/52 Oct/Nov 2009

4 One end of a light elastic string of natural length 3 m and modulus of elasticity 15m N is attached to a fixed point O. A particle P of mass m kg is attached to the other end of the string. P is released from rest at O and moves vertically downwards. When the extension of the string is x m the velocity of P is v m s−1. (i) Show that v2 = 5(12 + 4x −x2). [4] (ii) Find the magnitude of the acceleration of P when it is at its lowest point, and state the direction of this acceleration. [3]

7 marks

Mark scheme: 4 (i) EE = 1 (15 m)x2/3 B1 2 M1 For using Loss of PE = Gain in KE + EE 1 mv2 + 1 (15 m)x2/3 = mg(3 + x2) A1ft Ft error in EE 2 2 v2 = 5(12 + 4x – x) A1 4 AG (ii) [a = –2.5(2x – 4) or a = g – 15x/3] M1 For using a = v(dv/dx) or a = (mg – λx/L)/m [v = 0 → x = 6 → a = –20] M1 For finding x at the lowest point and substituting Magnitude is 20 ms–2; direction is A1 upwards 3 7

This question in 9709/52 Oct/Nov 2009

6 4.8 m M A B 0.7 m P A particle P of mass 0.35 kg is attached to the mid-point of a light elastic string of natural length 4 m. The ends of the string are attached to fixed points A and B which are 4.8 m apart at the same horizontal level. P hangs in equilibrium at a point 0.7 m vertically below the mid-point M of AB (see diagram). (i) Find the tension in the string and hence show that the modulus of elasticity of the string is 25 N. [4] P is now held at rest at a point 1.8 m vertically below M, and is then released. (ii) Find the speed with which P passes through M. [6]

10 marks

Mark scheme: 6 (i) [0.35g = 2T{0.7/ (2.42 + 0.72)1/2}] M1 For resolving forces on P vertically Tension is 6.25N A1 [6.25 = λ × ¼] M1 For using T = λx/L Modulus is 25N A1 AG [4] (ii) M1 For using EE = λx2/2L EE on release = 25×22/(2×4) A1 EE when P is at M = 25×0.82/(2×4) A1 M1 For using EE on release = mgh + EE when P is at M + 12 mv2 25×22/(2×4) = 0.35g×1.8+25×0.82/(2×4) + 1 2 0.35v2 A1 Speed is 4.90ms–1 A1 [6]

This question in 9709/51 May/June 2010

6 4.8 m M A B 0.7 m P A particle P of mass 0.35 kg is attached to the mid-point of a light elastic string of natural length 4 m. The ends of the string are attached to fixed points A and B which are 4.8 m apart at the same horizontal level. P hangs in equilibrium at a point 0.7 m vertically below the mid-point M of AB (see diagram). (i) Find the tension in the string and hence show that the modulus of elasticity of the string is 25 N. [4] P is now held at rest at a point 1.8 m vertically below M, and is then released. (ii) Find the speed with which P passes through M. [6]

10 marks

Mark scheme: 6 (i) [0.35g = 2T{0.7/ (2.42 + 0.72)1/2}] M1 For resolving forces on P vertically Tension is 6.25N A1 [6.25 = λ × ¼] M1 For using T = λx/L Modulus is 25N A1 AG [4] (ii) M1 For using EE = λx2/2L EE on release = 25×22/(2×4) A1 EE when P is at M = 25×0.82/(2×4) A1 M1 For using EE on release = mgh + EE when P is at M + 12 mv2 25×22/(2×4) = 0.35g×1.8+25×0.82/(2×4) + 1 2 0.35v2 A1 Speed is 4.90ms–1 A1 [6]

This question in 9709/52 May/June 2010

7 One end of a light elastic string of natural length 3 m and modulus of elasticity 24 N is attached to a fixed point O. A particle P of mass 0.4 kg is attached to the other end of the string. P is projected vertically downwards from O with initial speed 2 m s−1. When the extension of the string is x m the speed of P is v m s−1. (i) Show that v2 = 64 + 20x −20x2. [4] (ii) Find the greatest speed of the particle. [3] (iii) Calculate the greatest tension in the string. [4]

11 marks

Mark scheme: 7 (i) 0.4v2/2 + 24x2/(2×3) M1 PE, EE, KE terms 0.4g(3 + x) + 0.4×22/2 A2 –1 each error to zero v2 = 64 + 20x – 20x2 AG A1 [4] (ii) 2vdv/dx = 20 – 40x = 0 M1 0.4g = 24x/3 x = 0.5 A1ft v = 8.31 A1 [3] (iii) 20x2 – 20x – 64 = 0 M1 And attempts to solve x = 2.357 A1 T = 24×2.357/3 M1 T = 18.9 A1 [4]

This question in 9709/53 May/June 2010

2 A 0.8 m O 23 p rad B A bow consists of a uniform curved portion AB of mass 1.4 kg, and a uniform taut string of mass m kg which joins A and B. The curved portion AB is an arc of a circle centre O and radius 0.8 m. Angle AOB is 23π radians (see diagram). The centre of mass of the bow (including the string) is 0.65 m from O. Calculate m. [6]

6 marks

Mark scheme: 2 OG = 0.8sin(π /3)/(π /3) B1 0.66159 OM = 0.8cos(π /3) B1 0.4 M1 For taking moments about O 0.65(m + 1.4) = 0.4m + 0.66159x1.4 A1 0.25m = 0.01159 x 1.4 M1 For collecting like terms m = 0.0649 A1 OR OG = 0.8sin(π /3)/(π /3) B1 0.66159 OM = 0.8cos(π /3) B1 0.4 M1 Taking moments about M (1.4 + m) × 0.25 = 1.4 × 0.26159 A1 0.25m = 1.4x0.01159 M1 For collecting like terms m = 0.0649 A1 [6] 2

This question in 9709/51 Oct/Nov 2010

5 A particle P of mass 0.28 kg is attached to the mid-point of a light elastic string of natural length 4 m. The ends of the string are attached to fixed points A and B which are at the same horizontal level and 4.8 m apart. P is released from rest at the mid-point of AB. In the subsequent motion, the acceleration of P is zero when P is at a distance 0.7 m below AB. (i) Show that the modulus of elasticity of the string is 20 N. [4] (ii) Calculate the maximum speed of P. [3]

7 marks

Mark scheme: 5 (i) 2Tcosθ = 0.28g M1 Tension component = weight 2T x 0.7/2.5 = 2.8, T = 5 A1 5 = λ x 0.5/2 M1 Hookes Law λ = 20 N A1 [4] (ii) 0.28v2/2 + 2x20x0.52 /(2x2) = M1 PE/EE/KE conservation with 4 terms 0.28gx0.7 +2x20x0.42/(2x2) A1 v = 2.75 ms–1 A1 [3] GCE A LEVEL – October/November 2010 9709 51

This question in 9709/51 Oct/Nov 2010

2 A 0.8 m O 23 p rad B A bow consists of a uniform curved portion AB of mass 1.4 kg, and a uniform taut string of mass m kg which joins A and B. The curved portion AB is an arc of a circle centre O and radius 0.8 m. Angle AOB is 23π radians (see diagram). The centre of mass of the bow (including the string) is 0.65 m from O. Calculate m. [6]

6 marks

Mark scheme: 2 OG = 0.8sin(π /3)/(π /3) B1 0.66159 OM = 0.8cos(π /3) B1 0.4 M1 For taking moments about O 0.65(m + 1.4) = 0.4m + 0.66159x1.4 A1 0.25m = 0.01159 x 1.4 M1 For collecting like terms m = 0.0649 A1 OR OG = 0.8sin(π /3)/(π /3) B1 0.66159 OM = 0.8cos(π /3) B1 0.4 M1 Taking moments about M (1.4 + m) × 0.25 = 1.4 × 0.26159 A1 0.25m = 1.4x0.01159 M1 For collecting like terms m = 0.0649 A1 [6] 2

This question in 9709/52 Oct/Nov 2010

5 A particle P of mass 0.28 kg is attached to the mid-point of a light elastic string of natural length 4 m. The ends of the string are attached to fixed points A and B which are at the same horizontal level and 4.8 m apart. P is released from rest at the mid-point of AB. In the subsequent motion, the acceleration of P is zero when P is at a distance 0.7 m below AB. (i) Show that the modulus of elasticity of the string is 20 N. [4] (ii) Calculate the maximum speed of P. [3]

7 marks

Mark scheme: 5 (i) 2Tcosθ = 0.28g M1 Tension component = weight 2T x 0.7/2.5 = 2.8, T = 5 A1 5 = λ x 0.5/2 M1 Hookes Law λ = 20 N A1 [4] (ii) 0.28v2/2 + 2x20x0.52 /(2x2) = M1 PE/EE/KE conservation with 4 terms 0.28gx0.7 +2x20x0.42/(2x2) A1 v = 2.75 ms–1 A1 [3] GCE A LEVEL – October/November 2010 9709 52

This question in 9709/52 Oct/Nov 2010

1 0.9 m D C 0.9 m A B 1.8 m ABCD is a uniform lamina with AB = 1.8 m, AD = DC = 0.9 m, and AD perpendicular to AB and DC (see diagram). (i) Find the distance of the centre of mass of the lamina from AB and the distance from AD. [4] The lamina is freely suspended at A and hangs in equilibrium. (ii) Calculate the angle between AB and the vertical. [2]

6 marks

Mark scheme: 1 (i) 2mx0.45 + mx0.3 = 3mv M1 Table of values idea v = 0.4m (from AB) A1 2mx0.45 + mx(0.9+0.3) = 3mh M1 Table of values idea h = 0.7m (from AD) A1 [4] (ii) tanα = 0.4/0.7 M1 α = 29.7° A1ft Accept 0.519 radians [2]

This question in 9709/53 Oct/Nov 2010

5 2.4 m A B 0.5 m P A light elastic string has natural length 2 m and modulus of elasticity λ N. The ends of the string are attached to fixed points A and B which are at the same horizontal level and 2.4 m apart. A particle P of mass 0.6 kg is attached to the mid-point of the string and hangs in equilibrium at a point 0.5 m below AB (see diagram). (i) Show that λ = 26. [4] P is projected vertically downwards from the equilibrium position, and comes to instantaneous rest at a point 0.9 m below AB. (ii) Calculate the speed of projection of P. [5] [Question 6 is printed on the next page.]

9 marks

Mark scheme: 5 (i) T = λ ( 2.1 2 + 5.0 2 – 1)/1 B1 T = 0.3λ or T = 0.3x26 2xTx0.5/1.3 = 6 B1 T = 0.3λ = 7.8 M1 λ = 26 AG A1 [4] (ii) EE1 = 2x26x0.32/2x1 M1 (= 2.34) Use of EPE formula, either EE2 = 2x26( 2.1 2 + 9.0 2 – 1) 2/2x1 A1 (= 6.5) Both expressions correct M1 Conservation of energy (including KE/GPE/EPE) 0.6v2/2 + 0.6x10x(0.9 – 0.5) = 6.5 – 2.34 A1 V = 2.42ms–1 A1 [5] GCE A LEVEL – October/November 2010 9709 53

This question in 9709/53 Oct/Nov 2010

3 0.5 m s–1 0.6 m 0.6 m A P B A light elastic string of natural length 1.2 m and modulus of elasticity 24 N is attached to fixed points A and B on a smooth horizontal surface, where AB = 1.2 m. A particle P is attached to the mid-point of the string. P is projected with speed 0.5 m s−1 along the surface in a direction perpendicular to AB (see diagram). P comes to instantaneous rest at a distance 0.25 m from AB. (i) Show that the mass of P is 0.8 kg. [3] (ii) Calculate the greatest deceleration of P. [3]

6 marks

Mark scheme: 3 (i) EE gain B1 EE gain = 0.1 = 2 × 24[√ (0.62 + 0.252) – 0.6]2/(2 × 0.6) m × 0.52/2 = 0.1 M1 KE loss = EE gain m = 0.8 (kg) AG A1 [3] (ii) T = 24 × (0.65 – 0.6)/0.6 ( = 2) B1 2 × 2 × 0.25/0.65 = 0.8a M1 Newton’s Second Law with attempt to resolve 2T a = 1.92 A1 [3]

This question in 9709/51 May/June 2011

5 B P 0.61 m 0.22 m C 0.61 m A ABC is a uniform triangular lamina of weight 19 N, with AB = 0.22 m and AC = BC = 0.61 m. The plane of the lamina is vertical. A rests on a rough horizontal surface, and AB is vertical. The equilibrium of the lamina is maintained by a light elastic string of natural length 0.7 m which passes over a small smooth peg P and is attached to B and C. The portion of the string attached to B is horizontal, and the portion of the string attached to C is vertical (see diagram). (i) Show that the tension in the string is 10 N. [3] (ii) Calculate the modulus of elasticity of the string. [2] (iii) Find the magnitude and direction of the force exerted by the surface on the lamina at A. [3]

8 marks

Mark scheme: 5 (i) M1 Moments about A, 3 terms 19 × 0.6/3 + T × 0.22 = T × 0.6 A1 T = 10 AG A1 [3] (ii) 10 = λ (0.11 + 0.6 – 0.7)/0.7 M1 λ = 700 A1 [2] (iii) F 2 = 10 2 + (19 – 10) 2 M1 F = 13.5 A1 α = tan −(9/10)1 = 42.(0) o (with horizontal) B1 Or for a = tan −(10/9)1 = 48 o (with vertical) [3] 2

This question in 9709/51 May/June 2011

4 The ends of a light elastic string of natural length 0.8 m and modulus of elasticity λ N are attached to fixed points A and B which are 1.2 m apart at the same horizontal level. A particle of mass 0.3 kg is attached to the centre of the string, and released from rest at the mid-point of AB. The particle descends 0.32 m vertically before coming to instantaneous rest. (i) Calculate λ. [4] (ii) Calculate the speed of the particle when it is 0.25 m below AB. [4]

8 marks

Mark scheme: 4 (i) e = √(0.62 + 0.322) – 0.4 ( = 0.28) B1 Extension of half string = 0.28 m 0.3g × 0.32 = 2[λ (0.28 2 – 0.2 2 )] / (2 × 0.4) M1, A1 PE loss = EE gain λ = 10 A1 [4] (ii) e = √(0.62 + 0.252) – 0.4 B1 Extension of half string = 0.25 m 0.3g × 0.25 = 0.3v 2 / 2 + M1 PE loss = KE gain + EE gain 2[10(0.25 2 – 0.2 2 ) / (2 × 0.4)] A1ft N.B. 0.25 is extension of half string v = 1.12 A1 ft on candidates λ only [4]

This question in 9709/53 May/June 2011

3 One end of a light elastic string of natural length 0.4 m and modulus of elasticity 20 N is attached to a fixed point O. The other end of the string is attached to a particle P of mass 0.25 kg. P hangs in equilibrium below O. (i) Calculate the distance OP. [2] The particle P is raised, and is released from rest at O. (ii) Calculate the speed of P when it passes through the equilibrium position. [3] (iii) Calculate the greatest value of the distance OP in the subsequent motion. [3]

8 marks

Mark scheme: 3 (i) 0.25g = 20e/0.4 M1 Uses T = λx/L OP ( = 0.05 + 0.4) = 0.45 m A1 [2] (ii) 20 × 0.052/(2 × 0.4) + 0.25v2/2 M1 = 0.25g × 0.45 A1 v = 2.92 ms–1 A1 [3] (iii) 20(d – 0.4)2 /(2 × 0.4) = 0.25gd M1 Hence d2 – (0.8 + 0.1)d + 0.16 = 0 d = [0.9 ± √(0.92 – 4 × 0.16)]/2 M1 Solves a 3 term quadratic equation d = 0.656 A1 [3] Ignore d = 0.244 if seen

This question in 9709/51 Oct/Nov 2011

3 One end of a light elastic string of natural length 0.4 m and modulus of elasticity 20 N is attached to a fixed point O. The other end of the string is attached to a particle P of mass 0.25 kg. P hangs in equilibrium below O. (i) Calculate the distance OP. [2] The particle P is raised, and is released from rest at O. (ii) Calculate the speed of P when it passes through the equilibrium position. [3] (iii) Calculate the greatest value of the distance OP in the subsequent motion. [3]

8 marks

Mark scheme: 3 (i) 0.25g = 20e/0.4 M1 Uses T = λx/L OP ( = 0.05 + 0.4) = 0.45 m A1 [2] (ii) 20 × 0.052/(2 × 0.4) + 0.25v2/2 M1 = 0.25g × 0.45 A1 v = 2.92 ms–1 A1 [3] (iii) 20(d – 0.4)2 /(2 × 0.4) = 0.25gd M1 Hence d2 – (0.8 + 0.1)d + 0.16 = 0 d = [0.9 ± √(0.92 – 4 × 0.16)]/2 M1 Solves a 3 term quadratic equation d = 0.656 A1 [3] Ignore d = 0.244 if seen

This question in 9709/52 Oct/Nov 2011

4 A light elastic string has natural length 2.4 m and modulus of elasticity 21 N. A particle P of mass m kg is attached to the mid-point of the string. The ends of the string are attached to fixed points A and B which are 2.4 m apart at the same horizontal level. P is projected vertically upwards with velocity 12 m s−1 from the mid-point of AB. In the subsequent motion P is at instantaneous rest at a point 1.6 m above AB. (i) Find m. [4] (ii) Calculate the acceleration of P when it first passes through a point 0.5 m below AB. [4]

8 marks

Mark scheme: 4 (i) 2 2 B1 Use of EE formula (= 5.6 J) EE = 21( 1.2 + 1.6 – 1.2)2/(2 × 1.2) m122/2 = mg × 1.6 + M1 KE/EE/PE conservation 2 2 A1 2 × 21( 1.2 + 1.6 – 1.2)2/(2 × 1.2) m = 0.2 A1 [4] (ii) T = 21( 1.2 2 + 0.5 2 – 1.2)/1.2 B1 2 2 5.0 M1 Newton’s Second Law with ma = 2 × 21( 1.2 + 0.5 – 1.2)/1.2 × 3.1 A1 component of T or reversed signs – mg a = (–)3.27 ms–2 A1 [4] [8] GCE AS/A LEVEL – May/June 2012 9709 51

This question in 9709/51 May/June 2012

3 A light elastic string has natural length 2.2 m and modulus of elasticity 14.3 N. A particle P of mass m kg is attached to the mid-point of the string. The ends of the string are attached to fixed points A and B which are 2.4 m apart at the same horizontal level. P is released from rest at the mid-point of AB. In the subsequent motion P has its greatest speed at a point 0.5 m below AB. (i) Find m. [4] (ii) Calculate the greatest speed of P. [3]

7 marks

Mark scheme: 3 (i) Length = 1.2 2 + 0.5 2 = 1.3 B1 Pythagoras on 12 string 2 × [14.3 × (1.3 – 1.1)/1.1] × [0.5/1.3] M1* Uses T = λ x/L = mg D* M1 Component(s) T equated to weight m = 0.2 A1 [4] (ii) M1 KE/EE/PE balance (4 terms) 0.2v2/2 = 0.2g × 0.5 – A1 candidate’s value of m from (i) [14.3 × 0.22/(2 × 1.1) – 14.3 × 0.12/(2 × 1.1)] × 2 v = 2.47 ms–1 A1 [3] [7] GCE AS/A LEVEL – May/June 2012 9709 52

This question in 9709/52 May/June 2012

3 A particle P of mass 0.2 kg is projected horizontally from a fixed point O, and moves in a straight line on a smooth horizontal surface. A force of magnitude 0.4x N acts on P in the direction PO, where x m is the displacement of P from O. (i) Given that P comes to instantaneous rest when x = 2.5, find the initial kinetic energy of P. [4] (ii) Find the value of x on the first occasion when the speed of P is 2 m s−1. [2]

6 marks

Mark scheme: 3 (i) 0.2vdv/dx = –0.4x M1 Newton’s Second Law, – sign essential v2/2 = –2x2/2 (+ c) A1 Accept uncancelled form 0 = –2 × 2.52/2 + c →c = 6.25 M1 KE = 0.2 × 6.25 = 1.25 J A1 [4] v = 3.54 ms–1 (ii) 22/2 = –2x2/2 + 6.25 M1 v = 2 in accurate integral attempt at limits or finding arbitrary constant e.g. in (i) x = 2.06 A1 [2] [6]

This question in 9709/53 May/June 2012

6 B A 0.6 m O C D A uniform lamina OABCD consists of a semicircle BCD with centre O and radius 0.6 m and an isosceles triangle OAB, joined along OB (see diagram). The triangle has area 0.36 m2 and AB = AO. (i) Show that the centre of mass of the lamina lies on OB. [4] (ii) Calculate the distance of the centre of mass of the lamina from O. [4]

8 marks

Mark scheme: 6 (i) Height of triangle = 0.36 / 0.3(= 1.2 m) B1 Semi-circle C of M = 2 × 0.6 / (3π / 2) B1 Centre of mass lamina from BOD 0.36 × (1.2 / 3) = π × 0.62 / 2 × 2 × 0.6 / (3π / 2) M1 Equating moments idea 0.144 = 0.144 A1 [4] Evidence of checking equality OR 0.36 × (1.2 / 3) – π × 0.62 / 2 ×2 ×0.6 /(3π/2) = distance × total area M1 Table of moments idea Distance = 0 A1 (ii) 0.36 × 0.3 A1 Correct sum of parts = (0.36 + π 0.62 / 2) × OG A1 Correct moment of whole OG = 0.117 m A1 [4] GCE A LEVEL – October/November 2012 9709 52

This question in 9709/52 Oct/Nov 2012

7 A light elastic string has natural length 3 m and modulus of elasticity 45 N. A particle P of weight 6 N is attached to the mid-point of the string. The ends of the string are attached to fixed points A and B which lie in the same vertical line with A above B and AB = 4 m. The particle P is released from rest at the point 1.5 m vertically below A. (i) Calculate the distance P moves after its release before first coming to instantaneous rest at a point vertically above B. (You may assume that at this point the part of the string joining P to B is slack.) [4] (ii) Show that the greatest speed of P occurs when it is 2.1 m below A, and calculate this greatest speed. [5] (iii) Calculate the greatest magnitude of the acceleration of P. [3]

12 marks

Mark scheme: 7 (i) M1 Energy conservation, no KE, 2 EE terms 45 × 12 / (2 × 1.5) + 0.6 gh = 45 h2 / (2 × 1.5) A1 5h2 – 2h – 5 = 0 M1 Simplifies, tries to solve a 3 term quadratic equation h = 1.22 m A1 [4] (ii) 45e / 1.5 = 45(1 – e) / 1.5 + 6 M1 Finds equilibrium position (e = 0.6) AP = (1.5 + 0.6) = 2.1 AG A1 0.6 v2 / 2 = 0.6 g × 0.6 + 45 (1)2 / (2 × 1.5) M1 Energy conservation with KE/PE/EE – 4.5(0.6)2 / (2 × 1.5) – 45(0.4)2 / (2 × 1.5) A1 terms v = 6 ms–1 A1 [5] (iii) 0.6 a = ± (0.6g + 45 × 1 / 1.5) M1* Top a = ± 60 ms–2 0.6 a = ± (0.6g – 45 × 1.22 / 1.5) M1* Bottom a = ± 51 ms–2 | a | = 60 ms–2 A**1 [3] Needs acceleration at both extreme positions considered.

This question in 9709/52 Oct/Nov 2012

2 A particle P of mass 0.4 kg is attached to one end of a light elastic string of natural length 1.2 m and modulus of elasticity 19.2 N. The other end of the string is attached to a fixed point A. The particle P is released from rest at the point 2.7 m vertically above A. Calculate (i) the initial acceleration of P, [3] (ii) the speed of P when it reaches A. [4]

7 marks

Mark scheme: 2 (i) T = 19.2 ×(2.7 – 1.2)/1.2 B1 T = 24 N 0.4a = 0.4g + T M1 Newton’s Second Law with 3 terms a = 70 ms − 2 A1 [3] (ii) 19.2(2.7 – 1.2) 2 /(2 × 1.2) B1 Initial EE = 18 M1 For a 3 term energy equation 0.4v 2 /2 = 0.4g × 2.7 A1 + 19.2 × (2.7 – 1.2) 2 /(2 × 1.2 ) v = 12 ms − 1 A1 [4] [7]

This question in 9709/51 May/June 2013

3 A particle P of mass 0.2 kg is attached to one end of a light elastic string of natural length 1.6 m and modulus of elasticity 18 N. The other end of the string is attached to a fixed point O which is 1.6 m above a smooth horizontal surface. P is placed on the surface vertically below O and then projected horizontally. P moves with initial speed 1.5 m s−1 in a straight line on the surface. Show that, when OP = 1.8 m, (i) P is at instantaneous rest, [3] (ii) P is on the point of losing contact with the surface. [4]

7 marks

Mark scheme: 3 (i) EE = 18 × (1.8 – 1.6) 2 /(2 × 1.6) B1 0.2 × 1.5 2 /2 = M1 Energy equation, 3 terms 18(1.8 – 1.6) 2 /(2 × 1.6) + KE B KE B = 0 leads to v B = 0 A1 [3] (ii) T = 18 × (1.8 – 1.6)/1.6 B1 T = 2.25 Tcosθ + R = 0.2g M1 2.25 × 1.6/1.8 + R = 0.2g A1 R = 0 A1 [4] Needs g = 10 [7]

This question in 9709/52 May/June 2013

5 B 0.9 m 0.8 m P A block B of mass 3 kg is attached to one end of a light elastic string of modulus of elasticity 70 N and natural length 1.4 m. The other end of the string is attached to a particle P of mass 0.3 kg. B is at rest 0.9 m from the edge of a horizontal table and the string passes over a small smooth pulley at the edge of the table. P is released from rest at a point next to the pulley and falls vertically. At the first instant when P is 0.8 m below the pulley and descending, B is in limiting equilibrium with the part of the string attached to B horizontal (see diagram). (i) Calculate the speed of P when B is first in limiting equilibrium. [5] (ii) Find the coefficient of friction between B and the table. [3]

8 marks

Mark scheme: 5 (i) Ext = 0.8 + 0.9 – 1.4 (= 0.3 m) B1 Ext when in limiting equilibrium EE = 70 × 0.30 2 /(2 × 1.4) (= 2.25 J) B1 EE in limiting equilibrium M1 EE/PE/KE balance GCE AS/A LEVEL – May/June 2013 9709 52 0.3v 2 /2 = 0.3gx0.8 – 2.25 A1 v = 1 ms −1 A1 [5] (ii) T = 70 × 0.3/1.4 (= 15N) B1 Uses ext from part (i) 15 = µ (3g) M1 F = µ R, using mass of B µ = 0.5 A1 [3] [8]

This question in 9709/52 May/June 2013

5 One end of a light elastic string S1 of modulus of elasticity 20 N and natural length 0.5 m is attached to a fixed point O. The other end of S1 is attached to a particle P of mass 0.4 kg. P hangs in equilibrium vertically below O. (i) Find the distance OP. [2] The opposite ends of a light inextensible string S2 of length l m are now attached to O and P respectively. The elastic string S1 remains attached to O and P. The particle P hangs in equilibrium vertically below O. (ii) Find the tension in the inextensible string S2 for each of the following cases: (a) l < 0.5; (b) l > 0.6; (c) l = 0.54. [4] In the case l = 0.54, the inextensible string S2 suddenly breaks and P begins to descend vertically. (iii) Calculate the greatest speed of P in the subsequent motion. [3]

9 marks

Mark scheme: 5 (i) 0.4g = 20e/0.5 M1 Weight = λ ext/L (e = 0.1) OP = 0.6 m A1ft [2] 0.5 + cv(e) (iia) 4 N B1 (iib) 0 N B1 (iic) T = 0.4g – 20 × 0.04/0.5 M1 Weight(P) – λ ext/L T = 2.4 N A1 [4] (iii) M1 PE/KE/EE energy conservation 0.4v2/2 = 0.4g(0.6-0.54) A1 EE change (0.168 J) –[20(0.1)2/(2 × 0.5) – 20(0.04)2 /(2 × 0.5)] v = 0.6 ms −1 A1 [3] 9 GCE AS/A LEVEL – May/June 2013 9709 53

This question in 9709/53 May/June 2013

7 A small ball B of mass 0.2 kg moves in a narrow fixed smooth cylindrical tube OA of length 1 m, closed at the end A. When the ball has displacement x m from O, it has velocity v m s−1 in the k direction OA and experiences a resisting force of magnitude N. 1 −x (i) O A B 1.2 m s–1 1 m The tube is fixed in a horizontal position and B is projected from O towards A with velocity 1.2 m s−1 (see diagram). Given that B comes to instantaneous rest after travelling 0.55 m, show that k = 0.1803, correct to 4 significant figures. [6] (ii) The tube is now fixed in a vertical position with O above A. The ball B is released from rest at O. Calculate the speed of B after it has descended 0.1 m. [4]

10 marks

Mark scheme: 7 (i) 0.2a = – k/(1 – x) M1 N2L, single force a = – 5k/(1 – x) M1 Attempts ∫, accept use of dv/dt ∫vdv = – 5k∫1/(1 – x)dx v 2 /2 = 5kln(1 – x) (+c) A1 x = 0, v = 1.2, hence c = 0.72 M1 2 0 0 .55 [v /2] 1. 2 = [5kln(1 – x)] 0 5kln(1 – 0.55) + 0.72 = 0 DM1 k = 0.1803 AG A1 6 (ii) 0.2vdv/dx = 0.2 g – 0.1803/(1 – x) M1 N2L, difference of 2 forces 0.2v 2 /2 = 0.2 gx + 0. (1 – x) (+c) A1 Accept omission of c 0.2v 2 /2 = 0.2 gx 0.1 + 0.1803 ln(1 – 0.1) M1 nb c = 0, so can be omitted/lost v = 1.35 ms −1 A1 4 1.345 10

This question in 9709/53 May/June 2013

7 A particle P of mass 0.4 kg is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 32 N. The other end of the string is attached to a fixed point O. The particle is released from rest at O. (i) Calculate the distance OP at the instant when P first comes to instantaneous rest. [4] A horizontal plane is fixed at a distance 1 m below O. The particle P is again released from rest at O. (ii) Calculate the speed of P immediately before it collides with the plane. [3] (iii) In the collision with the plane, P loses 96% of its kinetic energy. Calculate the distance OP at the instant when P first comes to instantaneous rest above the plane, given that this occurs when the string is slack. [3]

10 marks

Mark scheme: 7 (i) M1 PE/EE balance 0.4gd = 32(d–0.8) 2 /(2 × 0.8) A1 20d 2 – 36d + 12.8 = 0 M1 Solves 3 term quadratic d = 1.31 m only A1 [4] Other value 0.4876.. OR 0.4g(0.8 + e) = 32e 2 /(2 × 0.8) M1 PE/EE balance 20e 2 –4e + 3.2 = 0 A1 e = 0.5(1) (also –3.12) M1 Solves 3 term quadratic d = 1.31 m only A1 M1 EE/KE/PE balance (ii) 0.4v 2 /2 A1 = 0.4g × 1 – 32(1–0.8) 2 /(2 × 0.8) v = 4 ms − 1 A1 [3] B1ft ftcv(v(ii) × (1–0.96) = 0.2v(ii) (iii) Rebound v = 0.8 0 = 0.4 × 0.8 2 / 2 + 32 × 0.2 2 / 1.6–0.4gh M1 EE/PE/KE balance, h = 0.232 OP (=1–h) = 0.768 m A1 [3] [10]

This question in 9709/51 Oct/Nov 2013

7 A particle P of mass 0.4 kg is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 32 N. The other end of the string is attached to a fixed point O. The particle is released from rest at O. (i) Calculate the distance OP at the instant when P first comes to instantaneous rest. [4] A horizontal plane is fixed at a distance 1 m below O. The particle P is again released from rest at O. (ii) Calculate the speed of P immediately before it collides with the plane. [3] (iii) In the collision with the plane, P loses 96% of its kinetic energy. Calculate the distance OP at the instant when P first comes to instantaneous rest above the plane, given that this occurs when the string is slack. [3]

10 marks

Mark scheme: 7 (i) M1 PE/EE balance 0.4gd = 32(d–0.8) 2 /(2 × 0.8) A1 20d 2 – 36d + 12.8 = 0 M1 Solves 3 term quadratic d = 1.31 m only A1 [4] Other value 0.4876.. OR 0.4g(0.8 + e) = 32e 2 /(2 × 0.8) M1 PE/EE balance 20e 2 –4e + 3.2 = 0 A1 e = 0.5(1) (also –3.12) M1 Solves 3 term quadratic d = 1.31 m only A1 M1 EE/KE/PE balance (ii) 0.4v 2 /2 A1 = 0.4g × 1 – 32(1–0.8) 2 /(2 × 0.8) v = 4 ms − 1 A1 [3] B1ft ftcv(v(ii) × (1–0.96) = 0.2v(ii) (iii) Rebound v = 0.8 0 = 0.4 × 0.8 2 / 2 + 32 × 0.2 2 / 1.6– M1 EE/PE/KE balance, h = 0.232 0.4gh OP (=1–h) = 0.768 m A1 [3] [10]

This question in 9709/52 Oct/Nov 2013

6 One end of a light elastic string of natural length 0.8 m and modulus of elasticity 50 N is attached to a fixed point O. A particle P of mass 0.4 kg is attached to the other end of the string. P is projected downwards with speed 1.5 m s−1 from a point 0.82 m vertically below O. (i) Find the greatest speed of P. [5] (ii) Show that P cannot reach O. [3]

8 marks

Mark scheme: 6 (i) 0.4 g = 50e/0.8 M1 Uses T = λ × /L (e = 0.064) Moves down = 0.044 m A1 (0.8 + 0.064 – 0.82) 0.4 × 1.5 2 /2 + 0.4 g × 0.044 + M1 Sets up 2EE/2KE/PE equation 50(0.82–0.8) 2 /(2 × 0.8) 2 2 A1 =0.4v /2 + 50 × 0.064 /(2 × 0.8) v = 1.6(0) ms − 1 A1 [5] (ii) PE gain to reach O = 0.4 g × 0.82 B1 From initial position, (3.28J) KE + EE = 0.4 × 1.5 2 /2 M1 At initial position, (0.4625J) 2 A1 +50(0.82 – 0.8) /(2 × 0.8) Shows by evaluation that insufficient [3] 8 energy GCE A LEVEL – October/November 2013 9709 53

This question in 9709/53 Oct/Nov 2013

3 A light elastic string has natural length 0.8 m and modulus of elasticity 16 N. One end of the string is attached to a fixed point O, and a particle P of mass 0.4 kg is attached to the other end of the string. The particle P hangs in equilibrium vertically below O. (i) Show that the extension of the string is 0.2 m. [2] P is projected vertically downwards from the equilibrium position. P first comes to instantaneous rest at the point where OP = 1.4 m. (ii) Calculate the speed at which P is projected. [3] (iii) Find the speed of P at the first instant when the string subsequently becomes slack. [2]

7 marks

Mark scheme: 3 (i) 0 .4 g = 16 e/ 0 .8 M1 Uses mg = 16 ext / 8.0 e = 0.2 AG A1 2 (ii) EE at C = 16 × 6.0 2 / (2 × 8.0 ) B1 4.0u 2 / 2 + 16 × 2.0 2 / (2 × 8.0 ) M1 KE/EE/PE balance attempted with 4 2 terms. + 4.0 g (4.1 − 0.1 ) = 16 × 6.0 / ( 2 × 8.0 ) u = 2.83 ms–1 A1 3 8 not allowed (iii) 16 × 6.0 2 / (2 × 8.0 ) M1 KE/EE/PE balance attempted with 3 2 terms. = 4.0 v / 2 + 4.0 g (4.1 − 8.0 ) v =2.45 ms–1 A1 2

This question in 9709/51 May/June 2014

3 A light elastic string has natural length 0.8 m and modulus of elasticity 16 N. One end of the string is attached to a fixed point O, and a particle P of mass 0.4 kg is attached to the other end of the string. The particle P hangs in equilibrium vertically below O. (i) Show that the extension of the string is 0.2 m. [2] P is projected vertically downwards from the equilibrium position. P first comes to instantaneous rest at the point where OP = 1.4 m. (ii) Calculate the speed at which P is projected. [3] (iii) Find the speed of P at the first instant when the string subsequently becomes slack. [2]

7 marks

Mark scheme: 3 (i) 0 .4 g = 16 e/ 0 .8 M1 Uses mg = 16 ext / 8.0 e = 0.2 AG A1 2 (ii) EE at C = 16 × 6.0 2 / (2 × 8.0 ) B1 4.0u 2 / 2 + 16 × 2.0 2 / (2 × 8.0 ) M1 KE/EE/PE balance attempted with 4 2 terms. + 4.0 g (4.1 − 0.1 ) = 16 × 6.0 / ( 2 × 8.0 ) u = 2.83 ms–1 A1 3 8 not allowed (iii) 16 × 6.0 2 / (2 × 8.0 ) M1 KE/EE/PE balance attempted with 3 2 terms. = 4.0v / 2 + 4.0 g (4.1 − 8.0 ) v =2.45 ms–1 A1 2

This question in 9709/52 May/June 2014

3 One end of a light elastic string of natural length 1.6 m and modulus of elasticity 28 N is attached to a fixed point O. The other end of the string is attached to a particle P of mass 0.35 kg which hangs in equilibrium vertically below O. The particle P is projected vertically upwards from the equilibrium position with speed 1.8 m s−1. Calculate the speed of P at the instant the string first becomes slack. [5]

5 marks

Mark scheme: 3 28 e = 0.35g M1 Equates λext/l and weight 1.6 e = 0.2 A1 OP = 1.8 m 0.35 v 2 0.2 2 1.8 2 = 28 × × 1.6 + 0.35 × − 0.35 g × 0.2 M1 EE/KE/PE balance 2 2 2 A1 All correct terms with candidate’s value of e v = 1.11 m s–1 A1 [5]

This question in 9709/51 Oct/Nov 2014

7 A 2 m R 0.4 m P rad s−1 One end of a light elastic string with modulus of elasticity 15 N is attached to a fixed point A which is 2 m vertically above a fixed small smooth ring R. The string has natural length 2 m and it passes through R. The other end of the string is attached to a particle P of mass m kg which moves with constant angular speed rad s−1 in a horizontal circle which has its centre 0.4 m vertically below the ring. PR makes an acute angle with the vertical (see diagram). 3 (i) Show that the tension in the string is N and hence find the value of m. [4] cos (ii) Show that the value of does not depend on . [4] It is given that for one value of the elastic potential energy stored in the string is twice the kinetic energy of P. (iii) Find this value of . [4]

12 marks

Mark scheme:  cos θ  λ ext 7 (i) T = M1 Uses T = 2 2 3 T = AG A1 cos θ Tcosθ = mg M1 Resolves vertically for P m = 0.3 A1 [4] (ii) r = 0.4tanθ B1 0.3v 2 = T sin θ OR 0.3ω2r = Tsinθ M1 Newton’s 2nd law with correct r expression for radial accn, ft cv(m(i)) 3 0.3ω2(0.4tanθ) = × sinθ A1 cosθ ω = 5 A1 SC [4] Candidates who choose at least two specific values of θ: Calculation of r twice B1 Both calculations give ω = 5 B1  0.4  2 15    cos θ  (iii) EPE = B1 2 × 2 0.3 ( 5 × 0.4 tan θ ) 2 KE = B1 ft candidate’s value of ω 2 Award if × 2 is with wrong term  0.4  2 15   2  cos θ   0.3(2 tan θ )  =   × 2 M1 2 × 2  2  cos2θ tan2θ = 0.5 OR sin2θ = 0.5 θ = 45 A1 www [4]

This question in 9709/51 Oct/Nov 2014

4 B C 1.8 m D T N 0.4 m E 1.6 m A F 0.4 m ABCDEF is the cross-section through the centre of mass of a uniform solid prism. ABCF is a rectangle in which AB = CF = 1.6 m, and BC = AF = 0.4 m. CDE is a triangle in which CD = 1.8 m, CE = 0.4 m, and angle DCE = 90Å. The prism stands on a rough horizontal surface. A horizontal force of magnitude T N acts at B in the direction CB (see diagram). The prism is in equilibrium. (i) Show that the distance of the centre of mass of the prism from AB is 0.488 m. [4] (ii) Given that the weight of the prism is 100 N, find the greatest and least possible values of T. [3]

7 marks

Mark scheme: 4 (i) ABCF area = 0.64 and CDE = 0.36 B1 Both areas correct 0.4 1.8 (0.64 + 0.36)d = 0.64× + 0.36×(0.4 + ) M1 Table of moments idea 2 3 A1 All terms correct d = 0.488 m AG A1 [4] (ii) 0.488 × 100 = 1.6T M1 Either limiting case T = 30.5 N A1 (no turning about A) (0.488 – 0.4) × 100 = 1.6T T = 5.5 A1 [3] (no turning about F)

This question in 9709/52 Oct/Nov 2014

5 A particle P of mass 0.3 kg is attached to one end of a light elastic string of natural length 0.9 m and modulus of elasticity 18 N. The other end of the string is attached to a fixed point O which is 3 m above the ground. (i) Find the extension of the string when P is in the equilibrium position. [2] P is projected vertically downwards from the equilibrium position with initial speed 6 m s−1. At the instant when the tension in the string is 12 N the string breaks. P continues to descend vertically. (ii) (a) Calculate the height of P above the ground at the instant when the string breaks. [2] (b) Find the speed of P immediately before it strikes the ground. [4]

8 marks

Mark scheme: 5 (i) 18e λ x 0.3g = M1 Uses T = 0.9 l e = 0.15 m A1 [2] (ii) (a) 18ext 12 = and ht = 3 – 0.9 – ext M1 Both ideas needed, ext = 0.6 0.9 ht = 1.5 m A1 [2] (ii) (b) 0.3 × 6 2 0.3u 2 – + 0.3g(0.6 – 0.15) M1 KE/PE/EE balance up to string 2 2 A1 breaking 18 × 0.6 2 18 × 0.15 2 = – 2 × 0.9 2 × 0.9  3.0u 2   = .3 375  u2 = 22.5  2  0.3v2 = 0.3u2 + 0.3g(3 – 0.6 – 0.9) M1 KE/PE balance after string breaks or OR v2 = u2 + 2g(3 – 0.6 – 0.9) v2 = u2 + 2g(ht) using ht from (ii)(a) v = 7.25 ms–1 A1 4 7.2456

This question in 9709/51 May/June 2015

2 One end of a light elastic string of natural length 0.5 m and modulus of elasticity 30 N is attached to a fixed point O. The other end of the string is attached to a particle P which hangs in equilibrium vertically below O, with OP = 0.8 m. (i) Show that the mass of P is 1.8 kg. [2] The particle is pulled vertically downwards and released from rest from the point where OP = 1.2 m. (ii) Find the speed of P at the instant when the string first becomes slack. [3]

5 marks

Mark scheme: 2 (i) mg = 30(0.8 – 0.5)/0.5 M1 m = 1.8 kg AG A1 2 (ii) EE = 30 2.1( − 5.02) /(2 × 0.5) B1 1.8v2/2 = 30(1.2–0.5)2/(2 × 0.5) M1 KE/EE/PE equation, 3 terms – 1.8 × (1.2 – 0.5)g RHS = 2.1 v = 1.53 ms–1 A1 3

This question in 9709/52 May/June 2015

6 y 1.2 m x A O B A uniform circular disc has centre O and radius 1.2 m. The centre of the disc is at the origin of x- and y-axes. Two circular holes with centres at A and B are made in the disc (see diagram). The point A is on the negative x-axis with OA = 0.5 m. The point B is on the negative y-axis with OB = 0.7 m. The hole with centre A has radius 0.3 m and the hole with centre B has radius 0.4 m. Find the distance of the centre of mass of the object from (i) the x-axis, [4] (ii) the y-axis. [3] The object can rotate freely in a vertical plane about a horizontal axis through O. (iii) Calculate the angle which OA makes with the vertical when the object rests in equilibrium. [2]

9 marks

Mark scheme: 6 (i) Mass of disc = π (1.22 – 0.42 – 0.32) B1 1.19π (or in (ii)) 0 = π (1.22 – 0.42 – 0.32)y – M1 LHS = π (1.22 – 0.32)×0 (0.42) × 0.7 A1 y = 0.0941 m A1 4 (ii) 0 = π (1.22 – 0.42 – 0.32)x–π(0.32).5 M1 LHS = π (1.22 – 0.42)×0 A1 x = 0.0378 m A1 3 0.0941176 (iii) tanθ = M1 0.0378151 θ = 68.1° A1 2

This question in 9709/51 Oct/Nov 2015

6 y 1.2 m x A O B A uniform circular disc has centre O and radius 1.2 m. The centre of the disc is at the origin of x- and y-axes. Two circular holes with centres at A and B are made in the disc (see diagram). The point A is on the negative x-axis with OA = 0.5 m. The point B is on the negative y-axis with OB = 0.7 m. The hole with centre A has radius 0.3 m and the hole with centre B has radius 0.4 m. Find the distance of the centre of mass of the object from (i) the x-axis, [4] (ii) the y-axis. [3] The object can rotate freely in a vertical plane about a horizontal axis through O. (iii) Calculate the angle which OA makes with the vertical when the object rests in equilibrium. [2]

9 marks

Mark scheme: 6 (i) Mass of disc = π (1.22 – 0.42 – 0.32) B1 1.19π (or in (ii)) 0 = π (1.22 – 0.42 – 0.32)y – M1 LHS = π (1.22 – 0.32)×0 (0.42) × 0.7 A1 y = 0.0941 m A1 4 (ii) 0 = π (1.22 – 0.42 – 0.32)x–π(0.32).5 M1 LHS = π (1.22 – 0.42)×0 A1 x = 0.0378 m A1 3 0.0941176 (iii) tanθ = M1 0.0378151 θ = 68.1° A1 2

This question in 9709/52 Oct/Nov 2015

7 A particle P of mass M kg is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 12.5 N. The other end of the string is attached to a fixed point A. The particle is released from rest at A and falls vertically until it comes to instantaneous rest at the point B. The greatest speed of P during its descent is 4.4 m s−1 when the extension of the string is e m. (i) Show that e = 0.64M. [2] (ii) Find a second equation in e and M, and hence find M. [6] (iii) Calculate the distance AB. [3]

11 marks

Mark scheme: 125.e 7 (i) Mg = M1 Uses T = λe/l 8.0 e = 0.64M AG A1 2 (ii) Mg(0.8 + e) = M1 PE/KE/EE conservation M × 44 2 12 5.e 2 + A1 2 ( 2 × 8.0) 10M(0.8 + 0.64M) = M1 12 5.( .064 M ) 2 2 9.68M + A1 8M+6.4M 2= 9.68M +3.2M 6.1 8+6.4M = 9.68 + 3.2M M1 Attempt to solve equation in M M = 0.525 A1 6 12 5.( d − 8.0) 2 (iii) 0.525gd = M1 PE/EE balance ( 2 × 8.0) 0.672d = d 2 – 1.6d + 0.64 M1 d = 1.94 A1 3

This question in 9709/53 Oct/Nov 2015

5 A particle P of mass 0.6 kg is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 24 N. The other end of the string is attached to a fixed point A, and P hangs in equilibrium. (i) Calculate the extension of the string. [2] P is projected vertically downwards from the equilibrium position with speed 4.5 m s−1. (ii) Find the distance AP when the speed of P is 3.5 m s−1 and P is below the equilibrium position. [4] (iii) Calculate the speed of P when it is 0.5 m above the equilibrium position. [3]

9 marks

Mark scheme: 5 (i) 24e/0.8 = 0.2g M1 e = 0.2 A1 2 (ii) 24 × 0.2 2 / (2 × 0.8) (= 0.6) B1 ft(cv0.2) Initial EE 0.6 × 4.5 2 / 2 + 0.6gd + 24 × 0.2 2 / (2 × 0.8) M1 PE/EE/KE balance attempt = 0.6 × 3.5 2 / 2 + 24 × (0.2 + d 2) / (2 × 0.8) A1 d = distance particle falls d = 0.4 so AP ( = 0.8 + 0.2 + 0.4) = 1.4m A1 4 (iii) 24 × 0.2 2 / (2 × 0.8) + 0.6 × 4.5 2 / 2 = M1 PE/EE/KE balance, 4 terms. Award A1 B1ft for initial KE if not already 0.6 v 2 /2 + 0.6g × 0.5 seen in part ii v = 3.5 m −s1 A1 3

This question in 9709/52 Feb/March 2016

7 A particle P is attached to one end of a light elastic string of natural length 1.2 m and modulus of elasticity 12 N. The other end of the string is attached to a fixed point O on a smooth plane inclined at an angle of 30Å to the horizontal. P rests in equilibrium on the plane, 1.6 m from O. (i) Calculate the mass of P. [2] A particle Q, with mass equal to the mass of P, is projected up the plane along a line of greatest slope. When Q strikes P the two particles coalesce. The combined particle remains attached to the string and moves up the plane, coming to instantaneous rest after moving 0.2 m. (ii) Show that the initial kinetic energy of the combined particle is 1 J. [4] The combined particle subsequently moves down the plane. (iii) Calculate the greatest speed of the combined particle in the subsequent motion. [5]

11 marks

Mark scheme: 7 (i) 12(1.6–1.2)/1.2 = mgsin30 M1 Uses T = λext/l m = 0.8 kg A1 2 (ii) PE change = 1.6 B1 2 × ans(i) B1 Both EE terms correct IKE + 12 × 0.4 2 /2.4 = M1 KE/PE/EE balance 2 Both EE terms correct 1.6 × 0.2gsin30 + 12 × 0.2 /2.4 IKE = 1 J AG A1 4 (iii) 12e/1.2 = 1.6 g sin30 M1 λe × t/l = new weight component e = 0.8 A1 May be stated without explanation 1.6 v 2 /2 + 12 × 0.8 2 /2.4 = M1 Must use new equilibrium position 2 A1 1.6g × 0.6sin30 + 12 × 0.2 /2.4 v = 1.5 m s−1 A1 5

This question in 9709/51 May/June 2016

7 A particle P is attached to one end of a light elastic string of natural length 1.2 m and modulus of elasticity 12 N. The other end of the string is attached to a fixed point O on a smooth plane inclined at an angle of 30Å to the horizontal. P rests in equilibrium on the plane, 1.6 m from O. (i) Calculate the mass of P. [2] A particle Q, with mass equal to the mass of P, is projected up the plane along a line of greatest slope. When Q strikes P the two particles coalesce. The combined particle remains attached to the string and moves up the plane, coming to instantaneous rest after moving 0.2 m. (ii) Show that the initial kinetic energy of the combined particle is 1 J. [4] The combined particle subsequently moves down the plane. (iii) Calculate the greatest speed of the combined particle in the subsequent motion. [5]

11 marks

Mark scheme: 7 (i) 12(1.6–1.2)/1.2 = mgsin30 M1 Uses T = λext/l m = 0.8 kg A1 2 (ii) PE change = 1.6 B1 2 × ans(i) B1 Both EE terms correct IKE + 12 × 0.4 2 /2.4 = M1 KE/PE/EE balance 2 Both EE terms correct 1.6 × 0.2gsin30 + 12 × 0.2 /2.4 IKE = 1 J AG A1 4 (iii) 12e/1.2 = 1.6 g sin30 M1 λe × t/l = new weight component e = 0.8 A1 May be stated without explanation 1.6 v 2 /2 + 12 × 0.8 2 /2.4 = M1 Must use new equilibrium position 2 A1 1.6g × 0.6sin30 + 12 × 0.2 /2.4 v = 1.5 m s−1 A1 5

This question in 9709/53 May/June 2016

6 A O 0.9 m P The diagram shows a smooth narrow tube formed into a fixed vertical circle with centre O and radius 0.9 m. A light elastic string with modulus of elasticity 8 N and natural length 1.2 m has one end attached to the highest point A on the inside of the tube. The other end of the string is attached to a particle P of mass 0.2 kg. The particle is released from rest at the lowest point on the inside of the tube. By considering energy, calculate (i) the speed of P when it is at the same horizontal level as O, [4] (ii) the speed of P at the instant when the string becomes slack. [3]

7 marks

Mark scheme: 6 (i) EE = 8(0.9π – 1.2)2/(2 x 1.2) B1 Initial EE = 8.83 J 8.83 = 0.2g x 0.9 + 0.2v2/2 + M1 8(0.9π/2 – 1.2)2/(2 x 1.2) A1 v = 8.29 m s–1 A1 4 (ii) θ = 1.2/0.9 = 4/3 rad (=76.4°) B1 8.83 = 0.2g x 0.9 + 0.2g x 0.9cosθ M1 0.2 x 8.292/2 = 0.2g x 0.9cosθ + 0.2v2/2 + 0.2v2/2 v = 8.13 m s–1 A1 3 2

This question in 9709/51 Oct/Nov 2016

2 A particle P of mass 0.5 kg is attached to one end of a light elastic string with modulus of elasticity 24 N and natural length 0.6 m. The other end of the string is attached to a fixed point A. The particle P hangs in equilibrium vertically below A. (i) Find the distance AP. [2] The particle P is raised to A and released from rest. (ii) Calculate the greatest speed of P in the subsequent motion. [3]

5 marks

Mark scheme: 2 (i) 5 = 24e /0.6 M1 Hence e = 0.125 AP = 0.725 m A1 [2] (ii) 24 x 0.1252/2 x 0.6 B1 EE at eqm (= 0.3125) 0.5g x 0.725 = M1 KE/EE/PE conservation 24 x 0.1252/ 2 x 0.6 + 0.5v2 /2 v = 3.64 m s–1 A1 [3]

This question in 9709/52 Oct/Nov 2016

2 0.6 m 0.6 m A B C A uniform wire is bent to form an object which has a semicircular arc with diameter AB of length 1.2 m, with a smaller semicircular arc with diameter BC of length 0.6 m. The end C of the smaller arc is at the centre of the larger arc (see diagram). The two semicircular arcs of the wire are in the same plane. (i) Show that the distance of the centre of mass of the object from the line ACB is 0.191 m, correct to 3 significant figures. [3] The object is freely suspended at A and hangs in equilibrium. (ii) Find the angle between ACB and the vertical. [4]

7 marks

Mark scheme: 2 (i) CoM(large) = 0.6/(π/2) or B1 CoM(small) = 0.3/(π/2) (π x 0.6 + π x 0.3)D = M1 OR (2+1)D = 2(1.2/π) – 1(0.6/π) π x 0.6(1.2/π) – π x 0.3(0.6/π) Moments about ACB D = 0.191 m AG A1 3 (ii) (π x 0.6 + π x 0.3)H = M1 OR 3H = 2 x 0.6 + 1 x 0.9 π x 0.6 x 0.6 + π x 0.3 x 0.9 Moments about A H = 0.7 A1 tanθ = 0.191/0.7 M1 θ = 15.3° A1 4 2

This question in 9709/53 Oct/Nov 2016

6 A O 0.9 m P The diagram shows a smooth narrow tube formed into a fixed vertical circle with centre O and radius 0.9 m. A light elastic string with modulus of elasticity 8 N and natural length 1.2 m has one end attached to the highest point A on the inside of the tube. The other end of the string is attached to a particle P of mass 0.2 kg. The particle is released from rest at the lowest point on the inside of the tube. By considering energy, calculate (i) the speed of P when it is at the same horizontal level as O, [4] (ii) the speed of P at the instant when the string becomes slack. [3]

7 marks

Mark scheme: 6 (i) EE = 8(0.9π – 1.2)2/(2 x 1.2) B1 Initial EE = 8.83 J 8.83 = 0.2g x 0.9 + 0.2v2/2 + M1 8(0.9π/2 – 1.2)2/(2 x 1.2) A1 v = 8.29 m s–1 A1 4 (ii) θ = 1.2/0.9 = 4/3 rad (=76.4°) B1 8.83 = 0.2g x 0.9 + 0.2g x 0.9cosθ M1 0.2 x 8.292/2 = 0.2g x 0.9cosθ + 0.2v2/2 + 0.2v2/2 v = 8.13 m s–1 A1 3 2

This question in 9709/53 Oct/Nov 2016

7 One end of a light elastic string of natural length 0.6 m and modulus of elasticity 24 N is attached to a fixed point O. The other end of the string is attached to a particle P of mass 0.4 kg which hangs in equilibrium vertically below O. (i) Calculate the extension of the string. [2] … … … … … P is projected vertically downwards from the equilibrium position with speed 5 m s−1. (ii) Calculate the distance P travels before it is first at instantaneous rest. [4] … … … … … … … … … … … … … … … … When P is first at instantaneous rest a stationary particle of mass 0.4 kg becomes attached to P. (iii) Find the greatest speed of the combined particle in the subsequent motion. [4] … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 7(i) 0.4g = 24e/0.6 M1 Uses T = λx/L e = 0.1 m A1 Total: 2 7(ii) Initial EE = 24 x 0.12/(2 x 0.6) (= 0.2 J) B1 Uses EE = λx2/2L 0.4 x 52/2 + 0.4gd=24(0.1 + d)2/(2 x 0.6) –24 M1 A1 Set up a 4 term energy equation involving x 0.12/(2 x 0.6) EE, PE and KE d = 0.5 m A1 Total: 4 7(iii) e = 0.2 B1 0.8v2/2=24 x 0.62/(2 x 0.6)– 24 x 0.22/(2 x M1 A1 Set up a 4 term energy equation in EE, PE 0.6) – 0.8g x 0.4 and KE v = 2 2 = 2.83 ms–1 A1 Total: 4

This question in 9709/52 Feb/March 2017

6 A particle P of mass 0.15 kg is attached to one end of a light elastic string of natural length 0.4 m and modulus of elasticity 12 N. The other end of the string is attached to a fixed point A. The particle P moves in a horizontal circle which has its centre vertically below A, with the string inclined at 1Å to the vertical and AP = 0.5 m. (i) Find the angular speed of P and the value of 1. [5] … … … … … … … … … … … … … … … … … … … … … … … (ii) Calculate the difference between the elastic potential energy stored in the string and the kinetic energy of P. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(i) B1 Uses T = λx / L 3sinθ = 0.15 2 ω (0.5sinθ) M1 Uses Newton's Second Law horizontally ω = 6.32 rad 1 s− A1 Tcosθ = 0.15g (cosθ = 0.5) M1 Resolves vertically θ = 60 A1 Total: 5 6(ii) v = 6.32 × 0.5sin60 B1 FT Uses v = rω and r = 0.5sin60 KE = 0.15(6.32 × 0.5sin60 2) / 2 (=0.5625J) B1 Difference = 0.5625 – 12 × 0. 21 / (2 × 0.4) M1 Uses EE = λ 2x / (2L) Difference = 0.4125 J A1 Total: 4 B1 Uses F = µ R

This question in 9709/51 May/June 2017

2 A 7 N B P 0.6 m Fig. 1 One end of a light inextensible string is attached to a fixed point A. The other end of the string is attached to a particle P of mass m kg which hangs vertically below A. The particle is also attached to one end of a light elastic string of natural length 0.25 m. The other end of this string is attached to a point B which is 0.6 m from P and on the same horizontal level as P. Equilibrium is maintained by a horizontal force of magnitude 7 N applied to P (see Fig. 1). (i) Calculate the modulus of elasticity of the elastic string. [2] … … … … … … … … … … … … … … … … A P 0.3 m 30Å B Fig. 2 P is released from rest by removing the 7 N force. In its subsequent motion P first comes to instantaneous rest at a point where BP = 0.3 m and the elastic string makes an angle of 30Å with the horizontal (see Fig. 2). (ii) Find the value of m. [4] … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 2(i) λ = 5 A1 Total: 2 2(ii) EE = 2 0.35 × 5 / (2 × 0.25) or 2 0.05 × 5 / (2 × 0.05) B1 Uses EE = λ 2x / 2L PE = mg × 0.3sin30 B1 mg × 0.3sin30 = 2 0.35 × 5 / (2 × 0.25) − 2 0.05 × 5 / (2 × 0.25) M1 Sets up a 3 term energy equation involving EE, KE and PE m = 0.8 A1 Total: 4

This question in 9709/53 May/June 2017

6 A particle P of mass 0.15 kg is attached to one end of a light elastic string of natural length 0.4 m and modulus of elasticity 12 N. The other end of the string is attached to a fixed point A. The particle P moves in a horizontal circle which has its centre vertically below A, with the string inclined at 1Å to the vertical and AP = 0.5 m. (i) Find the angular speed of P and the value of 1. [5] … … … … … … … … … … … … … … … … … … … … … … … (ii) Calculate the difference between the elastic potential energy stored in the string and the kinetic energy of P. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(i) B1 Uses T = λx / L 3sinθ = 0.15 2 ω (0.5sinθ) M1 Uses Newton's Second Law horizontally ω = 6.32 rad 1 s− A1 Tcosθ = 0.15g (cosθ = 0.5) M1 Resolves vertically θ = 60 A1 Total: 5 6(ii) v = 6.32 × 0.5sin60 B1 FT Uses v = rω and r = 0.5sin60 KE = 0.15(6.32 × 0.5sin60 2) / 2 (=0.5625J) B1 Difference = 0.5625 – 12 × 0. 21 / (2 × 0.4) M1 Uses EE = λ 2x / (2L) Difference = 0.4125 J A1 Total: 4 B1 Uses F = µ R

This question in 9709/53 May/June 2017

4 2.4 m A B 0.5 m P A light elastic string has natural length 2 m and modulus of elasticity 39 N. The ends of the string are attached to fixed points A and B which are at the same horizontal level and 2.4 m apart. A particle P of mass m kg is attached to the mid-point of the string and hangs in equilibrium at a point 0.5 m below AB (see diagram). (i) Show that m = 0.9. [4] … … … … … … … … … … … … … … … … … … … P is projected vertically downwards from the equilibrium position, and comes to instantaneous rest at a point 1.6 m below AB. (ii) Calculate the speed of projection of P. [5] … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 4(i) 2 2 B1 e = (0.5 + 1.2 ) – 1 = 0.3 T = 39 × 0.3/1 M1 Uses T = λx/L. mg = 2 × (39 × 0.3/1) × 0.5/1.3 M1 Resolves vertically. m = 0.9 A1 AG 4 4(ii) 2 2 B1 E = extension when the particle E = (1.6 + 1.2 ) – 1 = 1 m comes to instantaneous rest. EE = 39 × 21 /(2 × 1) or 39 × 0.32 /(2 × 1) B1 0.9 v 2 /2 + 0.9g(1.6 – 0.5) M1A1 Set up a 4 term energy equation 2 involving EE, KE and PE. = 2[39 × 21 /(2 × 1) – 39 × 0.3 /(2 × 1)] v = 7.54 m −s1 A1 5

This question in 9709/52 Oct/Nov 2017

6 One end of a light elastic string of natural length 0.4 m and modulus of elasticity 8 N is attached to a fixed point O on a smooth horizontal plane. The other end of the string is attached to a particle P of mass 0.2 kg which moves on the plane in a circular path with centre O. The speed of P is v m s−1 and the extension of the string is x m. (i) Given that v = 2.5, find x. [4] … … … … … … … … … … … … … … … … … … … … … … … It is given instead that the kinetic energy of P is twice the elastic potential energy stored in the string. (ii) Form two simultaneous equations and hence find x and v. [5] … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(i) T = 0.2 × 2. 5 2 /(0.4 + e) B1 Uses Newton's Second Law towards the centre of the circle. T = 8e/0.4 B1 Uses T = λx/L. 1.25/(0.4 + e) = 20e→20 2e + 8e – 1.25 = 0 M1 Eliminates T to find e. e = 0.12(0) m A1 4 6(ii) 0.2 v 2 /2 = 2[8 x 2 /(2 × 0.4)] B1 Uses KE = 2EE. 0.2 v 2 /(0.4 + x) = 8x/0.4 B1 Uses T = λx/L and T = m v 2 /r. M1 Attempts to solve the 2 equations to find v or x. x = 0.4 and v = 5.66 or 4 2 A1A1 5

This question in 9709/52 Oct/Nov 2017

6 A solid object consists of a uniform hemisphere of radius 0.4 m attached to a uniform cylinder of radius 0.4 m so that the circumferences of their circular faces coincide. The hemisphere and cylinder each have weight 20 N. The centre of mass of the object lies at the centre O of their common circular face. (i) Show that the height of the cylinder is 0.3 m. [2] … … … … … … … … … … A new object is made by cutting the cylinder in half and removing the half not attached to the hemisphere. The cut is perpendicular to the axis of symmetry, so the new object consists of a hemisphere and a cylinder half the height of the original cylinder. (ii) Find the distance of the centre of mass of the new object from O. [4] … … … … … … … … … … … … … … … … … The new object is placed with its hemispherical part on a rough horizontal surface. The new object is held in equilibrium by a force of magnitude P N acting along its axis of symmetry, which is inclined at 30Å to the horizontal. (iii) Find P. [3] … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(i) 20 × 3 × 0.4/8 = 20 × h/2 M1 Takes moments about the common surface. h = 0.3 m A1 AG 2 6(ii) Cylinder moment = 10 × 0.15/2 B1 20 × 3 × 0.4/8 – 10 × 0.15/2 = 30x M1A1 Takes moments about the base of the cylinder. x = 0.075 m A1 4 6(iii) 30 × 0.075sin60 = P × 0.4sin60 M1A1 Takes moments about point of contact of the cylinder with the surface. P = 5.625 A1 3

This question in 9709/53 Oct/Nov 2017

3 A small ball B is connected to one end of a light elastic string of natural length 0.4 m and modulus of elasticity 12 N. The other end of the string is attached to a fixed point A. The ball is projected with speed 1 m s−1 vertically downwards from a position 0.4 m vertically below A, and reaches its greatest speed at the point 0.7 m below A. (i) Show that the mass of B is 0.9 kg. [2] … … … … … … (ii) Calculate the greatest speed of B. [4] … … … … … … … … … … … … … … …

6 marks

Mark scheme: 3(i) mg = 12(0.7 – 0.4) / 0.4 M1 Use T = λx / L m = 0.9 kg AG A1 2 Question Answer Marks Guidance 3(ii) EPE = 12(0.7 – 0.4)2 / (2 × 0.4) B1 Correct EPE term 0.9v2 / 2 = 0.9g(0.7 – 0.4) + 0.9 × 12 / 2 – 12(0.7 – 0.4)2 / (2 × 0.4) M1 Attempts a 4 term energy equation A1 Correct equation v = 2 m s–1 A1 4

This question in 9709/52 Feb/March 2018

7 A 48Å 0.6 m D r m B 0.25 m 0.3 m 0.3 m C Fig. 1 ABCD is a uniform square lamina with sides of length 0.6 m. A circular hole of radius r m is made in the lamina. The centre of the hole is 0.3 m from AB and 0.25 m from AD. The lamina is freely suspended at A and hangs with the axis of symmetry making an angle of 48° with the horizontal (see Fig. 1). (i) Show that r = 0.214, correct to 3 significant figures. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … 15 N 60Å A D B C Fig. 2 The lamina is held in equilibrium with AD horizontal by a force of magnitude 15 N acting in the plane of the lamina applied at D. The line of action of this force makes an angle of 60° with the vertical (see Fig. 2). (ii) Find the weight of the original square lamina, before the hole was made. [4] … … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(i) tan48 = x /0.3 M1 x is the distance of the centre of mass from AD x = 0.3332 A1 0.62 × 0.3 = πr 2 × 0.25 + (0.62 – πr 2) x OR 0.62 × 0.3 = πr 2 × 0.35 + (0.62 – πr 2) × (0.6 – x ) M1 Take moments about AD Take moments about BC πr 2 × (0.3332 – 0.25) = 0.62 × (0.3332 – 0.3) OR πr 2(0.6 – 0.3332 – 0.35) = 0.62 (0.6 – 0.3332 – 0.3) A1 r = 0.214 AG A1 5 Question Answer Marks Guidance 7(ii) 0.3W = 0.6 × 15cos60 M1 Take moments about C.(W = weight of the lamina) W = 15 A1 Square = 15 × 0.62 / (0.62 – π × 0.2142) M1 Recognise that the ratio of weights = ratio of areas Square = 25(.0) N A1 4

This question in 9709/52 Feb/March 2018

5 A particle P of mass 0.7 kg is attached by a light elastic string to a fixed point O on a smooth plane inclined at an angle of 30° to the horizontal. The natural length of the string is 0.5 m and the modulus of elasticity is 20 N. The particle P is projected up the line of greatest slope through O from a point A below the level of O. The initial kinetic energy of P is 1.8 J and the initial elastic potential energy in the string is also 1.8 J. (i) Find the distance OA. [2] … … … … … … … … … … … (ii) Find the greatest speed of P in the motion. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(i) 20 e 2 M1 λx 1 . 8 = Use T = ( 2 × 0 . 5 ) l e = 0.3, OA = 0.8 A1 2 5(ii) 20 x M1 Use Newton’s Second Law 0.7gsin30 = up the plane 0.5 x = 0.0875 m A1 20 × 0.0875 2 B1 EPE = ( 2 × 0.5 ) 0.7 v 2 20 × 0.0875 2 M1 Attempt to set up a 5 term = 1.8 + 1.8 − 0.7 g ( 0.3 − 0.0875 ) sin30 − energy equation 2 ( 2 × 0.5 ) A1 Correct equation v = 2.78 ms− 1 A1 6

This question in 9709/51 May/June 2018

2 One end of a light elastic string is attached to a fixed point O. The other end of the string is attached to a particle P of mass 0.4 kg. The string has natural length 0.6 m and modulus of elasticity 24 N. The particle is released from rest at O. Find the two possible values of the distance OP for which the particle has speed 1.5 m s−1. [6] … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 2 2 0.6 − x /(2 × 0.6) 0.4 × 2 1.5 /2 = 0.4gx – 24( ) 2 0.6 − x /(2 × 0.6) [20 2 x – 28x + 7.65 = 0 or equivalent] M1 Attempt to find a 3 term energy equation M1 Attempt to solve the 3 term quadratic equation OP = 1.0279 m, 0.372 m (reject) A1 Correct answer chosen 0.4 × 2 1.5 /2 = 0.4gx M1 Note the particle is moving upwards and the string is slack OP = 0.1125 m A1 Total: 6 Question Answer Marks Guidance 2 Alternative method EPE = 24 2 x /(2 × 0.6) B1 x is the extension 0.4 × 2 1.5 /2 = 0.4g(x + 0.6) – 24 2 x /(2 × 0.6) [20 2 x – 4x – 1.95 = 0 or equivalent ] M1 Attempt to find a 3 term energy equation M1 Attempt to solve the 3 term quadratic equation [ x = 0.42787, – 0.22787 .reject] OP = 0.6 + 0.42787 = 1.0279 A1 0.4 × 2 1.5 /2 = 0.4g( x + 0.6) [x = – 0.4875] M1 Note the particle is moving upwards and the string is slack OP = 0.6 – 0.4875 = 0.1125 A1 Total: 6 d = xsinθ/2 – acosθ or equivalent

This question in 9709/52 May/June 2018

5 0.4 m 0.6 m 0.5 m 0.8 m x m A uniform object is made by joining a solid cone of height 0.8 m and base radius 0.6 m and a cylinder. The cylinder has length 0.4 m and radius 0.5 m. The cylinder has a cylindrical hole of length 0.4 m and radius x m drilled through it along the axis of symmetry. A plane face of the cylinder is attached to the base of the cone so that the object has an axis of symmetry perpendicular to its base and passing through the vertex of the cone. The object is placed with points on the base of the cone and the base of the cylinder in contact with a horizontal surface (see diagram). The object is on the point of toppling. (i) Show that the centre of mass of the object is 0.15 m from the base of the cone. [3] … … … … … … … … … … … … … … (ii) Find x. [4] [The volume of a cone is 30r2h.]1 … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(i) tanθ = x /0.6 M1 x = 0.15 m AG A1 Total: 3 5(ii) (π 2 0.6 × 0.8/3) × (0.8/4) – [π( 2 0.5 – 2 x ) × 0.4] × (0.4/2) = [π 2 0.6 × 0.8/3 + π( 2 0.5 – 2 x ) × 0.4] x M1 A1 Attempts to take moments about the base of the cone using their x Note x =0.15 Correct equation for the A mark. M1 Attempts to solve the equation x = 0.464 A1 Note 2 x = 0.216 Total: 4

This question in 9709/52 May/June 2018

5 A particle P of mass 0.7 kg is attached by a light elastic string to a fixed point O on a smooth plane inclined at an angle of 30° to the horizontal. The natural length of the string is 0.5 m and the modulus of elasticity is 20 N. The particle P is projected up the line of greatest slope through O from a point A below the level of O. The initial kinetic energy of P is 1.8 J and the initial elastic potential energy in the string is also 1.8 J. (i) Find the distance OA. [2] … … … … … … … … … … … (ii) Find the greatest speed of P in the motion. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(i) 20 e 2 M1 λx 1 . 8 = Use T = ( 2 × 0 . 5 ) l e = 0.3, OA = 0.8 A1 2 5(ii) 20 x M1 Use Newton’s Second Law 0.7gsin30 = up the plane 0.5 x = 0.0875 m A1 20 × 0.0875 2 B1 EPE = ( 2 × 0.5 ) 0.7 v 2 20 × 0.0875 2 M1 Attempt to set up a 5 term = 1.8 + 1.8 − 0.7 g ( 0.3 − 0.0875 ) sin30 − energy equation 2 ( 2 × 0.5 ) A1 Correct equation v = 2.78 ms− 1 A1 6

This question in 9709/53 May/June 2018

3 A particle P of mass 0.4 kg is attached to a fixed point O by a light elastic string of natural length 0.5 m and modulus of elasticity 20 N. The particle P is released from rest at O. (i) Find the greatest speed of P in the subsequent motion. [4] … … … … … … … … … … … … … … … … … … … … … … … … (ii) Find the distance below O of the point at which P comes to instantaneous rest. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 3(i) 20e/0.5 = 0.4g M1 Use T = λx/L e = 0.1 A1 0.4v2/2 = 0.4g(0.5 + 0.1) – 20×0.12/(2×0.5) M1 Attempt to set up a 3 term energy equation v = 11 = 3.32 A1 4 3(ii) 0.4g(5 + x) = 20x2/(2×0.5) M1 Attempt to set up a 2 term energy equation [0 = 20x2 – 4x – 2] [ x = 0.432] M1 Attempt to solve a 3 term quadratic equation Distance below O = (0.5 + 0.432) = 0.932 m A1 3

This question in 9709/51 Oct/Nov 2018

2 m 0.3 0.2 m B A A uniform object is made by attaching the base of a solid hemisphere to the base of a solid cone so that the object has an axis of symmetry. The base of the cone has radius 0.3 m, and the hemisphere has radius 0.2 m. The object is placed on a horizontal plane with a point A on the curved surface of the hemisphere and a point B on the circumference of the cone in contact with the plane (see diagram). (i) Given that the object is on the point of toppling about B, find the distance of the centre of mass of the object from the base of the cone. [3] … … … … … … … … … … … … … … … (ii) Given instead that the object is on the point of toppling about A, calculate the height of the cone. [3] [The volume of a cone is 30r2h.1 The volume of a hemisphere is 30r3.]2 … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 2(i) cosθ = 0.2/0.3 B1 Axis makes an angle θ with the horizontal tanθ = x/0.3 M1 x = 0.335(41..) A1 3 2(ii) M1 Attempt to take moments about A (π0.32h/3)×(h/4) = (2π0.23/3)(3×0.2/8) A1 h = 0.231 A1 3

This question in 9709/53 Oct/Nov 2018

3 A particle P of mass 0.4 kg is attached to a fixed point O by a light elastic string of natural length 0.5 m and modulus of elasticity 20 N. The particle P is released from rest at O. (i) Find the greatest speed of P in the subsequent motion. [4] … … … … … … … … … … … … … … … … … … … … … … … … (ii) Find the distance below O of the point at which P comes to instantaneous rest. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 3(i) 20e/0.5 = 0.4g M1 Use T = λx/L e = 0.1 A1 0.4v2/2 = 0.4g(0.5 + 0.1) – 20×0.12/(2×0.5) M1 Attempt to set up a 3 term energy equation v = 11 = 3.32 A1 4 3(ii) 0.4g(5 + x) = 20x2/(2×0.5) M1 Attempt to set up a 2 term energy equation [0 = 20x2 – 4x – 2] [ x = 0.432] M1 Attempt to solve a 3 term quadratic equation Distance below O = (0.5 + 0.432) = 0.932 m A1 3

This question in 9709/53 Oct/Nov 2018

6 E 0.3 m A F 0.3 m 0.6 m D G B C 0.7 m Fig. 1 Fig. 1 shows the cross-section ABCDE through the centre of mass G of a uniform prism. The cross- section consists of a rectangle ABCF from which a triangle DEF has been removed; AB = 0.6 m, BC = 0.7 m and DF = EF = 0.3 m. (i) Show that the distance of G from BC is 0.276 m, and find the distance of G from AB. [5] … … … … … … … … … … … … … … … … … B A G E 2 N C D Fig. 2 The prism is placed with CD on a rough horizontal surface. A force of magnitude 2 N acting in the plane of the cross-section is applied to the prism. The line of action of the force passes through G and is perpendicular to DE (see Fig. 2). The prism is on the point of toppling about the edge through D. (ii) Calculate the weight of the prism. [3] … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(i) 0.375y = 0.42×0.6/2 –0.045(0.6 – 0.3/3) M1 Take moments about BC y = 0.276 m AG A1 0.375x = 0.42×0.7/2 – 0.045(0.7 – 0.3/3) M1 Take moments about AB x = 0.32 m A1 5 6(ii) M1 Attempt to take moments about D 2cos45× (0.7 – 0.32) = 2cos45× (0.3 – 0.276) + W(0.3 – 0.276) A1 W = 21(.0) N A1 3

This question in 9709/53 Oct/Nov 2018

5 A particle P of mass 0.4 kg is attached to one end of a light elastic string of natural length 0.5 m and modulus of elasticity 6 N. The other end of the string is attached to a fixed point O. The particle P is released from rest at the point 0.5 + x m vertically below O. The particle P comes to instantaneous rest at O. (i) Find x. [3] … … … … … … … … … … … … (ii) Find the greatest speed of P. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(i) ( ) ( ) 2 6 0.4 0.5 2 0.5 x g x + = × M1 Set up an energy equation 6x2 – 4x – 2 = 0 or 3x2 – 2x – 1 = 0 M1 Attempt to solve a 3 term quadratic equation x = 1 (ignore 1 3 − if seen) A1 3 5(ii) 6 0.4 0.5 e g = M1 Use x T l λ = to find the extension at the equilibrium position 1 3 e = A1 PE change = 1 0.4 0.5 3 g  +     B1ft Ft for candidate’s e ( ) 2 2 1 6 0.4 1 3 0.4 0.5 2 3 2 0.5 V g         = + −   ×   M1 Set up a three term energy equation V = 3.65 ms–1 A1 5

This question in 9709/51 May/June 2019

4 A particle P of mass 0.5 kg is attached to one end of a light elastic string of natural length 0.8 m and modulus of elasticity 16 N. The other end of the string is attached to a fixed point O. The particle P is released from rest at the point 0.8 m vertically below O. When the extension of the string is x m, the downwards velocity of P is v m s−1 and a force of magnitude 25x2 N opposes the motion of P. (i) Show that, when P is moving downwards, vdv = 10 −40x −50x2. [2] dx … … … … … … … … … … … (ii) For the instant when P has its greatest downwards speed, find the kinetic energy of P and the elastic potential energy stored in the string. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4(i) 2 d 16 0.5 0.5 25 d 0.8 v x v g x x = − − M1 Use Newton’s Second Law vertically 2 d 10 40 50 d v v x x x = − − AG A1 2 Question Answer Marks Guidance 4(ii) ( ) 2 d 10 40 50 d = − − ∫ ∫ v v x x x M1 Attempt to integrate 2 3 2 50 10 20 ( ) 2 3 v x x x c = − − + A1 0 = 10 – 40x – 50x2 M1 Put the acceleration equal to zero x = 0.2 (Ignore x = –1 if seen) A1 2 0.5 8 0.533J 2 15 = = v B1 Use 2 2 = mv KE ( ) 2 0.2 16 0.4J 2 0.8 × = × B1 Use ( ) 2 2 λ = x EE l 6

This question in 9709/52 May/June 2019

5 A particle P of mass 0.4 kg is attached to one end of a light elastic string of natural length 0.5 m and modulus of elasticity 6 N. The other end of the string is attached to a fixed point O. The particle P is released from rest at the point 0.5 + x m vertically below O. The particle P comes to instantaneous rest at O. (i) Find x. [3] … … … … … … … … … … … … (ii) Find the greatest speed of P. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(i) ( ) ( ) 2 6 0.4 0.5 2 0.5 x g x + = × M1 Set up an energy equation 6x2 – 4x – 2 = 0 or 3x2 – 2x – 1 = 0 M1 Attempt to solve a 3 term quadratic equation x = 1 (ignore 1 3 − if seen) A1 3 5(ii) 6 0.4 0.5 e g = M1 Use x T l λ = to find the extension at the equilibrium position 1 3 e = A1 PE change = 1 0.4 0.5 3 g  +     B1ft Ft for candidate’s e ( ) 2 2 1 6 0.4 1 3 0.4 0.5 2 3 2 0.5 V g         = + −   ×   M1 Set up a three term energy equation V = 3.65 ms–1 A1 5

This question in 9709/53 May/June 2019

5 A particle P of mass 0.3 kg is attached to one end of a light elastic string of natural length 0.6 m and modulus of elasticity 9 N. The other end of the string is attached to a fixed point O on a smooth plane inclined at 30Å to the horizontal. OA is a line of greatest slope of the plane with A below the level of O and OA = 0.8 m. The particle P is released from rest at A. (i) Find the initial acceleration of P. [4] … … … … … … … … … … … … … … … … … … … … … … … (ii) Find the greatest speed of P. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 5(i) 9 (0.8 0.6) T 0.6 = M1 Use T = x l . Note 0.4 sin30 = OP T = 3 N A1 0.3a = 3 – 0.3gsin30 M1 Use Newton’s Second Law along the slope a = 5 m 1 −s A1 4 5(ii) 9e 0.3 sin30 0.6 = g M1 Note the maximum speed is at the equilibrium position e = 0.1 A1 2 2 9 (0.8 0.6) 9 EPE or 2 0.6 2 0.6 × − ×0.1 = × × B1 2 2 2 0.3 9 (0.8 0.6) 9 0.1 0.3 0.1sin30 2 2 0.6 2 0.6 × − × = − − × × × v g M1 Set up a 4 term energy equation v = 0.707 m 1 s− A1 5

This question in 9709/51 Oct/Nov 2019

1 A particle of mass 0.3 kg is attached to one end of a light elastic string of natural length 0.6 m and modulus of elasticity 9 N. The other end of the string is attached to a fixed point O on a smooth horizontal surface. The particle is projected horizontally from O with speed 4 m s−1. Find the greatest distance of the particle from O. [3] … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 1 2 2 0.3 9 2 2 0.6 × 4 = × e Note the final velocity is zero. e = 0.566 or 2 2 5 A1 Distance = 1.17 m A1 3

This question in 9709/52 Oct/Nov 2019

5 A particle P of mass 0.3 kg is attached to one end of a light elastic string of natural length 0.6 m and modulus of elasticity 9 N. The other end of the string is attached to a fixed point O on a smooth plane inclined at 30Å to the horizontal. OA is a line of greatest slope of the plane with A below the level of O and OA = 0.8 m. The particle P is released from rest at A. (i) Find the initial acceleration of P. [4] … … … … … … … … … … … … … … … … … … … … … … … (ii) Find the greatest speed of P. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 5(i) 9 (0.8 0.6) T 0.6 = M1 Use T = x l . Note 0.4 sin30 = OP T = 3 N A1 0.3a = 3 – 0.3gsin30 M1 Use Newton’s Second Law along the slope a = 5 m 1 −s A1 4 5(ii) 9e 0.3 sin30 0.6 = g M1 Note the maximum speed is at the equilibrium position e = 0.1 A1 2 2 9 (0.8 0.6) 9 EPE or 2 0.6 2 0.6 × − ×0.1 = × × B1 2 2 2 0.3 9 (0.8 0.6) 9 0.1 0.3 0.1sin30 2 2 0.6 2 0.6 × − × = − − × × × v g M1 Set up a 4 term energy equation v = 0.707 m 1 s− A1 5

This question in 9709/53 Oct/Nov 2019