Cambridge A Level Mathematics 9709 — 2015 Oct/Nov Paper 4 · Variant 3

7 A straight hill AB has length 400 m with A at the top and B at the bottom and is inclined at an angle of 4Å to the horizontal. A straight horizontal road BC has length 750 m. A car of mass 1250 kg has a speed of 5 m s−1 at A when starting to move down the hill. While moving down the hill the resistance to the motion of the car is 2000 N and the driving force is constant. The speed of the car on reaching B is 8 m s−1. (i) By using work and energy, find the driving force of the car. [5] On reaching B the car moves along the road BC. The driving force is constant and twice that when the car was on the hill. The resistance to the motion of the car continues to be 2000 N. Find (ii) the acceleration of the car while moving from B to C, [3] (iii) the power of the car’s engine as the car reaches C. [3]

Mark scheme: 7 (i) Gain in KE = 1 2 1250(82 – 52) B1 Loss in PE = 1250g × 400sin4o B1 For using WD by DF = Gain in KE – Loss in PE + WD by resistance M1 400(DF) = 1 2 1250 (82 – 52) – 1250g × 400sin4o + 2000 × 400 A1 Driving force is 1189 N or 1190 N A1 5 SR for using Newton’s second law (max 2 / 5) DF + 1250gsin4o – 2000 = 1250a B1 a = (82–52) / 2 × 400 → DF = 1190 N B1 (ii) For using Newton’s second law to find acceleration or for finding vC and using M1 v2 = u2 + 2as to find acceleration 1189 × 2 – 2000 = 1250a or 22.752 = 82 + 2a × 750 A1 DF from part (i) Acceleration is 0.302 ms–2 A1 3 (iii) vc 2 = 64 + 2 × 0.302 × 750 B1 acceleration from part (ii) [P / 22.75 – 2000 = 1250 × 0.302] M1 Power is 54.1 kW or 54100 W A1 3

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