Cambridge A Level Mathematics 9709 — 2017 Feb/March Paper 4 · Variant 2

1 A particle of mass 0.4 kg is projected with a speed of 12 m s−1 up a line of greatest slope of a smooth plane inclined at 30Å to the horizontal. (i) Find the initial kinetic energy of the particle. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Use an energy method to find the distance the particle moves up the plane before coming to instantaneous rest. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(i) KE = ½ × 0.4 × 122 = 28.8 J B1 Total: 1 1(ii) PE gain = 0.4gh [= 4d sin 30] B1 h = height gained d = distance travelled up the plane 4h = 28.8 M1 Using KE loss = PE gain h = 7.2 h = d sin 30 d = 14.4 m A1 Total: 3

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2 A B 20Å 40Å P A particle P of mass 1.6 kg is suspended in equilibrium by two light inextensible strings attached to points A and B. The strings make angles of 20Å and 40Å respectively with the horizontal (see diagram). Find the tensions in the two strings. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 M1 Resolve forces horizontally and/or vertically TA sin 20 + TB sin 40 = 16 A1 Correct vertical equation TA cos 20 = TB cos 40 A1 Correct horizontal equation M1 Attempt to solve for TA and/or TB TA = 14.2 N A1 TA = 14.1528... TB = 17.4 N A1 TB = 17.3610... Total: 6 Alternative method for Question 2 M1 Attempt to use Lami’s Theorem 16 TA A1 = sin120 sin130 16 TB A1 = sin120 sin110 M1 Attempt to solve for TA and/or TB TA = 14.2 N A1 TB = 17.4 N A1 Total: 6

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3 P N 0.6 kg 21Å A particle of mass 0.6 kg is placed on a rough plane which is inclined at an angle of 21Å to the horizontal. The particle is kept in equilibrium by a force of magnitude P N acting parallel to a line of greatest slope of the plane, as shown in the diagram. The coefficient of friction between the particle and the plane is 0.3. Show that the least possible value of P is 0.470, correct to 3 significant figures, and find the greatest possible value of P. 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Mark scheme: 3 R = 0.6g cos 21 [= 5.60] B1 F = 0.3R = 1.8 cos 21 [= 1.68] M1 Using F = µR P + F = 6 sin 21[ = 2.15] M1 Slipping down P = 2.15 – 1.68 = 0.470 AG A1 Least possible value P – F = 6 sin 21 M1 Slipping up P = 2.15 + 1.68 = 3.83 A1 Greatest possible value Total: 6

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4 A car of mass 900 kg is moving on a straight horizontal road ABCD. There is a constant resistance of magnitude 800 N in the sections AB and BC, and a constant resistance of magnitude R N in the section CD. The power of the car’s engine is a constant 36 kW. (i) The car moves from A to B at a constant speed in 120 s. Find the speed of the car and the distance AB. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The car’s engine is switched offat B. (ii) The distance BC is 450 m. Find the speed of the car at C. 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(iii) The car comes to rest at D. The distance AD is 6637.5 m. Find the deceleration of the car and the value of R. 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Mark scheme: 4(i) 36000 = 800v M1 Using P = Fv v = 45 m s–1 A1 Speed of the car AB = 45 × 120 = 5400 m A1 Total: 3 4(ii) −800 = 900a [a = –8/9] M1 Using Newton’s 2nd law 2 2 16 M1 Using v 2 = u 2 + 2 as v = 45 − × 450 9 v = 35 m s–1 A1 Speed of the car at C Total: 3 Alternative method for Question 4(ii) 0.5 × 900 × (45 – v2) M1 Attempt change in KE 0.5 × 900 × (45 – v2) = 800 × 450 M1 KE loss = WD against Friction v = 35 m s–1 A1 Speed of the car at C Total: 3 4(iii) CD = 6637.5 – 5400 – 450 = 787.5 B1 0 = 352 – 2d × 787.5 M1 Using v 2 = u 2 + 2 as , a = –d d = 7/9 = 0.778 m s–2 A1 d = deceleration P = 900 × (7/9) = 700 A1 Using F = ma Total: 4

More questions on Kinematics of motion in a straight line

5 A particle P moves in a straight line starting from a point O and comes to rest 35 s later. At time t s after leaving O, the velocity v m s−1 of P is given by v = 45t2 0 ≤t ≤5, v = 2t + 10 5 ≤t ≤15, v = a + bt2 15 ≤t ≤35, where a and b are constants such that a > 0 and b < 0. (i) Show that the values of a and b are 49 and −0.04 respectively. 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(ii) Sketch the velocity-time graph. [4] v (m s−1) t (s) 0 5 10 15 20 25 30 35 (iii) Find the total distance travelled by P during the 35 s. 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Mark scheme: 5(i) 0= a + b × 352 M1 For matching velocities at 40 = a + b × 152 t = 15 and using v = 0 at t = 35 [1000b = -40 → b = –0.04] M1 Solve for a and b [a = 0.04 × 352 = 49] a = 49 and b = -0.04 AG A1 Total: 3 5(ii) 0 ⩽ t ⩽ 5 correct B1 Increasing quadratic, from (0,0) to (5,20), concave up 5 ⩽ t ⩽ 15 correct B1 Line from (5,20) to (15,40) 15 ⩽ t ⩽ 35 correct B1 Decreasing quadratic, from (15,40) to (35,0), concave down 20 and 40 seen correct on v-axis B1 Total: 4 5(iii) 5 B1 2 100 0.8t d t = A1 = ∫ 0 3 1 M1 Using trapezium rule or integration for A2 = ( 20 + 40 ) × 10 = 300 t = 5 to t = 15 2 35 M1 Attempt to integrate the quadratic a + bt 2 d t function ) A3 = ∫ ( 15 from t = 15 to t = 35 0.04 3 = 49t − t 3 A3 = 453.3333 = 1360/3 A1 Total Distance = 2360/3 = 787 m A1 Total: 5

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6 1.2 kg 0.8 kg 0.64 m Two particles of masses 1.2 kg and 0.8 kg are connected by a light inextensible string that passes over a fixed smooth pulley. The particles hang vertically. The system is released from rest with both particles 0.64 m above the floor (see diagram). In the subsequent motion the 0.8 kg particle does not reach the pulley. (i) Show that the acceleration of the particles is 2 m s−2 and find the tension in the string. 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(ii) Find the total distance travelled by the 0.8 kg particle during the first second after the particles are released. 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Mark scheme: 6(i) M1 Apply Newton’s law to either of the particles 12 – T = 1.2a and T – 8 = 0.8a A1 Both equations correct M1 Solve for a and T a = 2 m s–2 and T = 9.6 N A1 Total: 4 6(ii) [0.64 = ½ × 2 × t1 2] M1 Attempt to find time t1 taken for 1.2 kg [v = 2t1] particle to reach ground and/or its speed v at the ground t1 = 0.8 A1 v = 2 × 0.8 = 1.6 A1 [0 = 1.6 – 10t2] M1 For attempting to find the time t2 [1.62 = 2 × 10 × s2] and/or distance travelled s2 as 0.8 kg particle comes to rest t2 = 0.16 A1 s2 = 0.128 A1 t3 = 1 – 0.8 – 0.16 = 0.04 B1 Finding the distance s3 travelled s3 = ½ × 10 × 0.042 downwards in t3 seconds Total distance travelled = B1 0.64 + 0.128 + 0.008 = 0.776 m Total: 8

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