Cambridge A Level Mathematics 9709 — 2013 Oct/Nov Paper 4 · Variant 2

6 A B 1.2 m Particles A of mass 0.4 kg and B of mass 1.6 kg are attached to the ends of a light inextensible string which passes over a fixed smooth pulley. A is held at rest and B hangs freely, with both straight parts of the string vertical and both particles at a height of 1.2 m above the floor (see diagram). A is released and both particles start to move. (i) Find the work done on B by the tension in the string, as B moves to the floor. [5] When particle B reaches the floor it remains at rest. Particle A continues to move upwards. (ii) Find the greatest height above the floor reached by particle A. [4] [Question 7 is printed on the next page.]

Mark scheme: 6 (ii) [1.6 × 10 × 1.2 = ½ 1.6 v2 + 7.68] M1 For using PE loss = KE gain + WD by T to find v2 v2 = 14.4 A1 14.4 = 2 × 10 × h For using PCE for A’s motion after h = 0.72 B reaches the ground or H = 2 × 1.2 + h 0 = u2 – 2gh M1 and H = 2 × 1.2 + h Greatest height is 3.12 m A1 4 First Alternative Marking Scheme for 6 (ii) [v2 = 2 × 6 × 1.2] M1 For using v2 = 2as to find v2 v2 = 14.4 A1 14.4 = 2 × 10 × h For using PCE for A’s motion after h = 0.72 B reaches the ground or H = 2 × 1.2 + h 0 = u2 – 2gh M1 and H = 2 × 1.2 + h Greatest height is 3.12 m A1 4 Second Alternative Marking Scheme for 6 (ii) WD by T = Increase in PE For applying WD by T to particle 7.68 = 0.4 × g × s M1 A’s complete motion s = 1.92 A1 H = 1.2 + s M1 For adding 1.2 to s H = 1.2 +1.92 = 3.12 Height = 3.12 m A1 4 GCE AS/A LEVEL – October/November 2013 9709 42

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7 v (m s –1) 0.4 t (s) O 5 24 28 An elevator is pulled vertically upwards by a cable. The velocity-time graph for the motion is shown above. Find (i) the distance travelled by the elevator, [2] (ii) the acceleration during the first stage and the deceleration during the third stage. [2] The mass of the elevator is 800 kg and there is a box of mass 100 kg on the floor of the elevator. (iii) Find the tension in the cable in each of the three stages of the motion. [3] (iv) Find the greatest and least values of the magnitude of the force exerted on the box by the floor of the elevator. [3]

Mark scheme: 7 (i) [s = ½ 5 × 0.4 + 19 × 0.4 + ½ 4 × 0.4] For using the area property for M1 distance Distance = 9.4 A1 2 (ii) Acceleration is 0.08 ms–2 B1 Deceleration is 0.1ms–2 B1 2 (iii) [T – (800 + 100) g = (800 +100)a] For applying Newton’s 2nd law to M1 the elevator and box T – 900g = 900a A1 T = 9072 N in 1st stage T = 9000 N in 2nd stage T = 8910 N in 3rd stage A1 3 (iv) [R – 100g = 100a] For applying Newton’s 2nd law to M1 the box For obtaining the greatest value of R = 1008 N A1 the force on the box For obtaining the least value of the R = 990 N A1 3 force on the box

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