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Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, October/November 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_w20_qp_439709_w20_ms_439709/43/O/N/20
Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.
Question
A particle P is projected vertically upwards with speed v m s−1 from a point on the ground. P reaches its greatest height after 3 s. (a) Find v. [1] (b) Find the greatest height of P above the ground. [2]
Mark scheme
1(a) v = 30 B1 Use v = u + at (or equivalent suvat) with v = 0, a = –g and t = 3 1 1(b) [0 = 302 + 2(–10)s] M1 Using v2 = u2 + 2as with a = –g, v = 0 and u = value from 1(a), or equivalent suvat method Greatest height is 45 m A1 2
Question
A box of mass 5 kg is pulled at a constant speed a distance of 15 m up a rough plane inclined at an angle of 20Å to the horizontal. The box moves along a line of greatest slope against a frictional force of 40 N. The force pulling the box is parallel to the line of greatest slope. (a) Find the work done against friction. [1] (b) Find the change in gravitational potential energy of the box. [2] (c) Find the work done by the pulling force. [1]
Mark scheme
2(a) WD = 40 × 158 = 600 J B1 1 2(b) [PE = 5 × 10 × 15 sin 20] M1 Attempt PE gain 257 J (256.5151... J) A1 2 2(c) WD = 40 × 15 + 5 × 10 × 15 sin 20 = 857 J B1 FT FT 600 + ‘PE’(> 0) from 2(b) 1
Question
A string is attached to a block of mass 4 kg which rests in limiting equilibrium on a rough horizontal table. The string makes an angle of 24Å above the horizontal and the tension in the string is 30 N. (a) Draw a diagram showing all the forces acting on the block. [1] (b) Find the coefficient of friction between the block and the table. [5]
Mark scheme
3(a) B1 4 forces, labelled 1 3(b) For resolving horizontally or vertically M1 30 cos 24 = F (F = 27.406…) A1 R + 30 cos 24 = 40 (R = 27.797…) A1 30cos 24 M1 Using µ = F/R μ= 40 − 30sin 24 μ = 0.986 (0.9859…) A1 5
Question
Two small smooth spheres A and B, of equal radii and of masses 4 kg and m kg respectively, lie on a smooth horizontal plane. Initially, sphere B is at rest and A is moving towards B with speed 6 m s−1 . After the collision A moves with speed 1.5 m s−1 and B moves with speed 3 m s−1 . Find the two possible values of the loss of kinetic energy due to the collision. [6]
Mark scheme
4 For using conservation of momentum (either case) M1 6 × 4 = 3m + 4 × 1.5 or A1 6 × 4 = 3m – 4 × 1.5 m = 6 and m = 10 A1 KEA initial = ½ × 4 × 62 (72 J) B1 FT KE = ½ × m × v2 FT 4.5m for KEB or KEA after = ½ × 4 × 1.52 (4.5 J) or KEB after = ½ × 6 × 32 (27 J) or KEB after = ½ × 10 × 32 (45 J) KE loss = [½ × 4 × 62 – ½ × 4 × 1.52 – ½ × 6 × 32] M1 Uses KE loss = KE before – KE after or [½ × 4 × 62 – ½ × 4 × 1.52 – ½ × 10 × 32] Loss of KE = 40.5 J or 22.5 J A1 6
Question
A particle P moves in a straight line. It starts at a point O on the line and at time t s after leaving O it has velocity v m s−1 , where v = 4t2 − 20t + 21. (a) Find the values of t for which P is at instantaneous rest. [2] (b) Find the initial acceleration of P. [2] (c) Find the minimum velocity of P. [2] (d) Find the distance travelled by P during the time when its velocity is negative. [4]
Mark scheme
5(a) 4t2 – 20t + 21 = (2t – 3)(2t – 7) = 0 → t = … M1 For setting v = 0 and attempting to solve v = 0 t = 1.5 and t = 3.5 A1 2 5(b) a = 8t – 20, a(0) = … M1 For using a = dv/dt and evaluating for t = 0 a = –20 A1 2 5(c) 8t – 20 = 0, t = 2.5 → v = … or M1 For setting a = 0, attempting to solve for t and substituting to obtain v, or for attempting to complete the square on the v = (2t – 5)2 – 4, vmin = … expression for v vmin = −4 ms–1 A1 2 5(d) s = ∫(4t2 – 20t + 21) dt M1 For using s = ∫v dt and attempting integration 4 A1 Correct integration s = t 3 − 10t 2 + 21t ( + c ) 3 49 27 M1 Substitute their limits (1.5 and 3.5) into their integral − 6 2 16 A1 Distance = = 5.33 m 3 4
Question
A car of mass 1600 kg is pulling a caravan of mass 800 kg. The car and the caravan are connected by a light rigid tow-bar. The resistances to the motion of the car and caravan are 400 N and 250 N respectively. (a) The car and caravan are travelling along a straight horizontal road. (i) Given that the car and caravan have a constant speed of 25 m s−1 , find the power of the car’s engine. [2] (ii) The engine’s power is now suddenly increased to 39 kW. Find the instantaneous acceleration of the car and caravan and find the tension in the tow-bar. [5] (b) The car and caravan now travel up a straight hill, inclined at an angle of sin−1 0.05 to the horizontal, at a constant speed of v m s−1 . The car’s engine is working at 32.5 kW. Find v. [3]
Mark scheme
6(a)(i) P = 650 × 25 M1 Use P = Fv with F = total resistance P = 16 250 W = 16.25 kW A1 Accept 16 300 W or 16.3 kW (3sf) 2 6(a)(ii) 39000 B1 For using DF = P/v DF = (= 1560) 25 For applying Newton’s 2nd law to the system to form an equation in M1 [1560 – 650 = 2400 × a] a, or to the caravan or the car to form an equation in T and a 1560 – 650 = 2400a A1 Two correct equations T – 250 = 800a 1560 – 400 – T = 1600a (1560 − 650 ) M1 For solving for a or for T a = 2400 a = 0.379 ms–2 (0.37916…) A1 T = 553 N (553.33…) 5 6(b) [DF = 650 + 2400 × 10 × 0.05] M1 Newton’s 2nd law 32 500 = (650 + 24 000 × 0.05)v M1 For using P = Fv v = 17.6 A1 650 Allow v = 37 3
Question
2 kg A B 3 kg P Q 10Å 20Å As shown in the diagram, particles A and B of masses 2 kg and 3 kg respectively are attached to the ends of a light inextensible string. The string passes over a small fixed smooth pulley which is attached to the top of two inclined planes. Particle A is on plane P, which is inclined at an angle of 10Å to the horizontal. Particle B is on plane Q, which is inclined at an angle of 20Å to the horizontal. The string is taut, and the two parts of the string are parallel to lines of greatest slope of their respective planes. (a) It is given that plane P is smooth, plane Q is rough, and the particles are in limiting equilibrium. Find the coefficient of friction between particle B and plane Q. [5] (b) It is given instead that both planes are smooth and that the particles are released from rest at the same horizontal level. Find the time taken until the difference in the vertical height of the particles is 1 m. [You should assume that this occurs before A reaches the pulley or B reaches the bottom of plane Q.] [6] Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
7(a) [T = 2g sin 10] or [3g sin 20 = F + T] M1 Resolve forces parallel to plane P for particle A or parallel to plane Q for Particle B T = 2g sin 10 and 3g sin 20 = F + T A1 R = 30 cos 20 (= 28.19...) B1 Resolving forces perpendicular to plane Q for particle B 3g sin 20 − 2 g sin10 M1 Using µ = F/R μ= 30cos 20 µ = 0.241 (=0.2407…) A1 5 7(b) 3g sin 20 – T = 3a or T – 2g sin 10 = 2a M1 For applying Newton’s second law to either A or to B or to the or System: 3g sin 20 – 2g sin 10 = 5a system a= ( 3g sin 20 − 2 g sin10 ) M1 For applying Newton’s second law to the second particle and/or solving for a 5 a = 1.3575… A1 h1 = x sin 20 B1 Using expressions for height change of each particle after each h2 = x sin 10 moves a distance x along the plane, to obtain equation in x x sin 20 + x sin 10 = 1 1 1 M1 For using s = ut + ½at2 for either particle with s = x, u = 0 and = 0 + × 1.3575 × t 2 using their a (= 1.3575) sin10 + sin 20 2 t = 1.69 A1 6
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Mathematics 9709/43 for October/November 2020 is out of 50 marks, with 1 hour 15 minutes to answer it.
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