Cambridge A Level Mathematics 9709 — 2012 May/June Paper 5 · Variant 2

1 A particle P of mass 0.6 kg is projected horizontally with velocity 2 m s−1 from a point O on a smooth horizontal surface. A horizontal force of magnitude 0.3x N acts on P in the direction OP, where x m is the distance of P from O. Calculate the velocity of P when x = 8. [4]

Mark scheme: 1 0.6vdv/dx = 0.3x M1 Newton’s Second Law with a = vdv/dx 0.6v2/2 = 0.3x2/2(+ c) A1 From ∫0.6vdv = ∫0.3xdx [x2/2] 80 = [v2] v2 M1 Uses limits of finds constant v = 6 ms–1 A1 [4] [4]

More questions on Kinematics of motion in a straight line

2 A uniform hemispherical shell of weight 8 N and a uniform solid hemisphere of weight 12 N are joined along their circumferences to form a non-uniform sphere of radius 0.2 m. (i) Show that the distance between the centre of mass of the sphere and the centre of the sphere is 0.005 m. [3] This sphere is placed on a horizontal surface with its axis of symmetry horizontal. The equilibrium of the sphere is maintained by a force of magnitude F N acting parallel to the axis of symmetry applied to the highest point of the sphere. (ii) Calculate F. [3]

Mark scheme: 2 (i) M1 Table of values or moment equation 12 × 3 × 0.2/8 – 8 × 0.2/2 = (8 + 12)d A1 0.9 – 0.8 = 20d d(= 0.1/20) = 0.005 m AG A1 [3] Accept d = –0.005 (ii) M1 Moments about point of contact F × (2 × 0.2) = (12 + 8) × 0.005 A1 F = 0.25 A1 OR M1 Moments about point of contact F × (2 × 0.2) + 8 × 0.1 = 12 × 0.075 A1 F = 0.25 A1 [3] [6] h 1 i

More questions on Forces and equilibrium

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]

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

More questions on Forces and equilibrium

4 A particle P of mass 0.25 kg moves in a straight line on a smooth horizontal surface. At time t s the velocity of P is v m s−1. A variable force of magnitude 3t N opposes the motion of P. (i) Given that P comes to rest when t = 3, find v when t = 0. [4] (ii) Calculate the distance travelled by P in the interval 0 ≤t ≤3. [3]

Mark scheme: 4 (i) 0.25dv/dt = –3t M1 Newton’s Second Law, – sign essential v = –12t2/2 (+ c) A1 Accept uncancelled form 0 = 12 × 32/2 + c M1 Appropriate use of v = 0, t = 3 Initial speed = 54 ms–1 A1 [4] Goes beyond c = 54 (ii) ∫dx = ∫(54 – 6t2)dt M1 Separates variables, integrates v x = [54t – 6t3/3] 30 A1 candidates value [v in (i)] x = 108 m A1 [3] [7] 2 2

More questions on Kinematics of motion in a straight line

5 A ball is projected with velocity 25 m s−1 at an angle of 70◦above the horizontal from a point O on horizontal ground. The ball subsequently bounces once on the ground at a point P before landing at a point Q where it remains at rest. The distance PQ is 17.1 m. (i) Calculate the time taken by the ball to travel from O to P and the distance OP. [3] (ii) Given that the horizontal component of the velocity of the ball does not change at P, calculate the speed of the ball when it leaves P. [4]

Mark scheme: 5 (i) 0 = (25sin70)t – gt2/2 M1 Uses 0 = ut – gt2/2 t = 4.7(0) s A1 OP = (25cos70 × 4.7) = 40.2 m A1 OR OP = 252sin(2 × 70)/g M1 Uses R = v2sin2α/g OP = 40.2 m A1 t[= 40.2/(25cos70)] = 4.7 s A1 OR 0 = 25sin70 – 10t M1 Find time to greatest height and double t = 2.349, 2t = 4.70 A1 it OP = (25cos70 × 4.7) = 40.2 m A1 OR 0 = xtan70 – gx2/(2 × 252cos270) M1 Use trajectory equation x = 40.2 m A1 t = 4.70 A1 [3] (ii) t[=17.1/(25cos70)] = 2 s B1 Finds time of flight –v = v – g × 2 B1ft Finds vertical component of speed V 2 = 102 + (25cos70)2 M1 For squaring components V = 13.2 ms–1 A1 [4] [7] GCE AS/A LEVEL – May/June 2012 9709 52

More questions on Kinematics of motion in a straight line

6 E B F A O C D 1 m r m The diagram shows a uniform lamina ABCDEF, formed from a semicircle with centre O and radius 1 m by removing a semicircular part with centre O and radius r m. (i) Show that the distance in metres of the centre of mass of the lamina from O is 4(1 + r + r2) . [4] 3π(1 + r) The centre of mass of the lamina lies on the arc ABC. (ii) Show that r = 0.494, correct to 3 significant figures. [3] The lamina is freely suspended at F and hangs in equilibrium. (iii) Find the angle between the diameter of the lamina and the vertical. [2]

Mark scheme: 6 (i) 2)1( sin(π / 2) 2 r sin(π / 2) M1 Uses table of values or moment × π (12)/2 = × equation 3π / 2 3π / 2 π (r2)/2 + OG(π /2 – π r2/2) A1 Correct moment equation OG = 4π (1 – r3)/3(1 – r2) A1 OG = 4(1 + r + r2)/3π (1 + r) AG A1 [4] Must use 1 – r as a factor of 1 – r3 and 1 – r2 (ii) r = 4(1 + r + r2)/3π (1 + r) M1* Sets r = answer(i) (3π – 4)r2 + (3π – 4)r – 4 = 0 D* M1 Sets up and starts solving quadratic equation r = 0.494 AG A1 [3] OR OG = 0.494 if G on arc M1* OG = 4(1 + 0.494 + 0.4942)/3π (1 + 0.494) D* M1 Substitutes AG in OG expression = 0.4937 AG A1 Shows value rounding to 0.494 (iii) tan θ = 0.494/1 M1 θ = 26.3° A1 [2] [9]

More questions on Differentiation

7 Particles P and Q, of masses 0.8 kg and 0.5 kg respectively, are attached to the ends of a light inextensible string which passes through a small hole in a smooth horizontal table of negligible thickness. P moves with constant angular speed 6.25 rad s−1 in a circular path on the surface of the table. (i) It is given that Q is stationary and that the part of string attached to Q is vertical. Calculate the radius of the path of P, and find the speed of P. [4] (ii) It is given instead that the part of string attached to Q is inclined at 60◦to the vertical, and that Q moves in a horizontal circular path below the table, also with constant angular speed 6.25 rad s−1. Calculate the total length of the string. [6]

Mark scheme: 7 (i) T = 0.5 g B1 T = 5 T = 0.8 × 6.252 × r M1 r = 0.16 m A1 v = 1 ms–1 B1 [4] (ii) Tcos60 = 0.5 g B1 T = 10 r = 0.32 m B1 (2 × candidate’s value of r) Tsin60 = 0.5 × 6.252 × R M1 Newton’s Second Law with component of T R = 0.443(40) m A1 L = 0.32 + 0.443(4)/sin60 M1 L = 0.832 m A1 [6] [10]

More questions on Newton’s laws of motion