Cambridge A Level Mathematics 9709 — 2018 May/June Paper 5 · Variant 3

9709/53/M/J/18 · 6 questions · 50 marks · ≈56 min

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Mark scheme9 pages

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Questions as text

Q1 · A small ball B is projected from a point O on horizontal ground

1 A small ball B is projected from a point O on horizontal ground. The initial velocity of B has horizontal and vertically upwards components of 18 m s−1 and 25 m s−1 respectively. For the instant 4 s after projection, find the speed and direction of motion of B. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Vertical component of velocity = 25 – 4g M1 Use v = u + at 2 2 2 ( 25 − 4 g ) M1 v = 18 + (25 − 4 g ) or tanθ = 18 v = 23.4 ms − 1 A1 θ = 39.8 ° below the horizontal A1 4

More questions on Kinematics of motion in a straight line

Q2 · 12 N 30Å B 0.5 m 30Å A A non-uniform rod AB of length 0.5 m and weight 8 N is freely…

2 12 N 30Å B 0.5 m 30Å A A non-uniform rod AB of length 0.5 m and weight 8 N is freely hinged to a fixed point at A. The rod makes an angle of 30° with the horizontal with B above the level of A. The rod is held in equilibrium by a force of magnitude 12 N acting in the vertical plane containing the rod at an angle of 30° to AB applied at B (see diagram). Find the distance of the centre of mass of the rod from A. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 M1 Attempt to take moments about A 8xcos30 = 0.5 × 12sin30 A1 Correct equation x = 0.433 m A1 3

More questions on Forces and equilibrium

Q3 · A particle P of mass 0.4 kg is projected horizontally along a smooth horizontal plane…

3 A particle P of mass 0.4 kg is projected horizontally along a smooth horizontal plane from a point O. At time t s after projection the velocity of P is v m s−1. A force of magnitude 0.8t N directed away from O acts on P and a force of magnitude 2e−t N opposes the motion of P. dv (i) Show that = 2t −5e−t. 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(ii) Given that v = 8 when t = 1, express v in terms of t. 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(iii) Find the speed of projection of P. 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Mark scheme: 3(i) dv t M1 Use Newton’s Second Law 0.4 = 0.8t – 2 e− horizontally dt dv t A1 AG = 2t – 5e− dt 2 3(ii) ∫ dv = ∫ (2t − 5e− t )dt M1 Attempt to integrate the 2 t equation from part (i) v = t + 5e− ( + c) t = 1 and v = 8 so c = 5.16 M1 Attempt to find the constant of integration, c v = t 2 + 5e − t + 5.16 or v = t 2 + 5e − t + 7 − 5e − 1 A1 3 3(iii) Evaluates v for t = 0 M1 V = 10.2 ms − 1 A1 2

More questions on Newton’s laws of motion

Q5 · A particle P of mass 0.7 kg is attached by a light elastic string to a fixed point O on a…

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. 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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

More questions on Energy, work and power

Q6 · H P 0.4 m Q A A particle P of mass 0.2 kg is attached to one end of a light inextensible…

6 H P 0.4 m Q A A particle P of mass 0.2 kg is attached to one end of a light inextensible string of length 0.6 m. The other end of the string is attached to a particle Q of mass 0.3 kg. The string passes through a small hole H in a smooth horizontal surface. A light elastic string of natural length 0.3 m and modulus of elasticity 15 N joins Q to a fixed point A which is 0.4 m vertically below H. The particle P moves on the surface in a horizontal circle with centre H (see diagram). (i) Calculate the greatest possible speed of P for which the elastic string is not extended. 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(ii) Find the distance HP given that the angular speed of P is 8 rad s−1. 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Mark scheme: 6(i) r [= 0.6 – (0.4-0.3)]= 0.5 B1 T = 0.3g B1 Resolve vertically for Q 0.2v 2 / 0.5 = 0.3 g M1 Use Newton’s Second Law horizontally for P v = 2.74 ms− 1 A1 4 6(ii) r = 0.5 + e B1 e = extension of the string 15e B1 Use T = λx/l T = = 50e 0.3 0.2 × 8 2 (5 + e ) = 50e + 0.3g M1 Use Newton’s Second Law horizontally with a = rω2 ( 6.4 − 3 ) A1 e = ( = 0.0914) ( 50 − 12.8 ) HP = 0.591 m A1 5

More questions on Newton’s laws of motion

Q7 · 1.2 m 0.2 m 0.5 m A uniform solid cone has height 1.2 m and base radius 0.5 m

7 1.2 m 0.2 m 0.5 m A uniform solid cone has height 1.2 m and base radius 0.5 m. A uniform object is made by drilling a cylindrical hole of radius 0.2 m through the cone along the axis of symmetry (see diagram). (i) Show that the height of the object is 0.72 m and that the volume of the cone removed by the drilling is 0.03520 m3. [4] [The volume of a cone is 30r2h.]1 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the distance of the centre of mass of the object from its base. 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Mark scheme: 7(i) 0.2 M1 Use ratio of corresponding Height of conical tip = 1.2 × = 0.48 sides, similar figures 0. 5 Cylindrical height = 1.2 – 0.48 = 0.72 A1 AG 2 0.48 2 M1 Volume removed = π 0.2 × + π0.2 × 0.72 3 ( = 0.0064 π + 0.0288 π) Volume removed = 0.0352π A1 AG 4 7(ii) Moment of cone removed about the base B1 0.48 = 0.0064π(0.72 + ) = 0.0064 π× 0.84 4 Moment of cylinder removed about the base B1 0.72 = 0.0288 π× = 0.0288 π× 0.36 2 Moment of the original cone about the base B1 2 1.2 = π 0.5 × × 0.3 = 0.1π× 0.3 3 M1 Attempt to take moments about the base 0.1π× 0.3 = 0.0064 π×0.84 + 0.0288π× 0.36 + A1 Note 0.0648π is the volume 0.0648πx of the object x = 0.22 m A1 6

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Cambridge’s own grade thresholds for 2018 May/June, Paper 5 · Variant 3. A higher threshold means an easier paper — the bar moves with how the cohort did.

A33/50
B25/50
C21/50
D15/50
E10/50