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Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, May/June 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_s20_qp_419709_s20_ms_419709/41/M/J/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
50 N A ! 100 N ! 50 N Three coplanar forces of magnitudes 100 N, 50 N and 50 N act at a point A, as shown in the diagram. The value of cos ! is 54 . Find the magnitude of the resultant of the three forces and state its direction. [3]
Mark scheme
1 Resultant = 100 – 2 × 50cos α M1 20 N A1 Direction is to the left (or equivalent) B1 3
Question
A car of mass 1800 kg is towing a trailer of mass 400 kg along a straight horizontal road. The car and trailer are connected by a light rigid tow-bar. The car is accelerating at 1.5 m s−2 . There are constant resistance forces of 250 N on the car and 100 N on the trailer. (a) Find the tension in the tow-bar. [2] (b) Find the power of the engine of the car at the instant when the speed is 20 m s−1 . [3]
Mark scheme
2(a) [T – 100 = 400 × 1.5] M1 T = 700 N A1 2 2(b) F – 250 – 100 = 2200 × 1.5 (F = 3650 N) M1 (M1 for using Newton’s second law for the system or for the car using the result from 2(a)) For use of power =Fv M1 73 000 W or 73 kW A1 3
Question
A particle P is projected vertically upwards with speed 5 m s−1 from a point A which is 2.8 m above horizontal ground. (a) Find the greatest height above the ground reached by P. [3] (b) Find the length of time for which P is at a height of more than 3.6 m above the ground. [4]
Mark scheme
3(a) 0 = 52 – 2gs M1 s = 1.25 A1 [Height above ground =] 4.05 m A1 3 3(b) Use of s = ut + ½ at2 M1 0.8 = 5t – 5t2 A1 t = 0.2 or 0.8 M1 Length of time = 0.6 s A1 4
Question
30Å TN The diagram shows a ring of mass 0.1 kg threaded on a fixed horizontal rod. The rod is rough and the coefficient of friction between the ring and the rod is 0.8. A force of magnitude T N acts on the ring in a direction at 30Å to the rod, downwards in the vertical plane containing the rod. Initially the ring is at rest. (a) Find the greatest value of T for which the ring remains at rest. [4] (b) Find the acceleration of the ring when T = 3. [3]
Mark scheme
4(a) Resolving forces in either direction M1 R = T sin30 + 0.1g, F = T cos30 A1 T cos30 = 0.8 (T sin30 + 0.1g) M1 T = 1.72 (1.7166...) A1 4 4(b) R = 3sin30 + 0.1g B1 3 cos30 – 0.8(3sin30 + 0.1g) = 0.1a M1 a = 5.98 ms–2 (5.9807...) A1 3
Question
45Å 4m P A child of mass 35 kg is swinging on a rope. The child is modelled as a particle P and the rope is modelled as a light inextensible string of length 4 m. Initially P is held at an angle of 45Å to the vertical (see diagram). (a) Given that there is no resistance force, find the speed of P when it has travelled half way along the circular arc from its initial position to its lowest point. [4] (b) It is given instead that there is a resistance force. The work done against the resistance force as P travels from its initial position to its lowest point is X J. The speed of P at its lowest point is 4 m s−1 . Find X . [3]
Mark scheme
5(a) Attempt at finding PE lost M1 PE lost = 35g (4cos22.5 – 4cos45) A1 1 M1 × 35v2 = 35g (4cos22.5 – 4cos45) 2 Speed = 4.16 ms–1 (4.1643...) A1 4 5(b) Use of the work-energy equation in the form: PE lost = KE gain + WD against resistance M1 1 A1 × 35 × 42 = 35g (4 – 4cos45) – X 2 X = 130 (130.05...) A1 3
Question
A particle moves in a straight line AB. The velocity v m s−1 of the particle t s after leaving A is given by v = k t2 − 10t + 21, where k is a constant. The displacement of the particle from A, in the direction towards B, is 2.85 m when t = 3 and is 2.4 m when t = 6. (a) Find the value of k. Hence find an expression, in terms of t, for the displacement of the particle from A. [7] (b) Find the displacement of the particle from A when its velocity is a minimum. [4]
Mark scheme
6(a) k ( t − 10t + 21) dt M1 2 1 A1 s = k t 3 + 5t 2 + 21t + C 3 1 1 M1 2.85 = k × 33 − 5 × 32 + 21 × 3 + C or 2.4 = k × 63 − 5 × 6 2 + 21 × 6 + C 3 3 2.85 = 27k + C, 2.4 = 18k + C A1 (A1 for both) Solving for k M1 k = 0.05 A1 1 A1 s = 0.05 t 3 − 5t 2 + 21t + 1.5 3 7 6(b) Differentiating v or completing the square for v M1 a = 0.05(2t –10) A1 Min value of v is at t = 5. M1 Displacement at t = 5 is 2.58 m (2.5833...) A1 4
Question
0.3 kg P 2 .5 m Q 0.2 kg 30Å 1.5 m A particle P of mass 0.3 kg, lying on a smooth plane inclined at 30Å to the horizontal, is released from rest. P slides down the plane for a distance of 2.5 m and then reaches a horizontal plane. There is no change in speed when P reaches the horizontal plane. A particle Q of mass 0.2 kg lies at rest on the horizontal plane 1.5 m from the end of the inclined plane (see diagram). P collides directly with Q. (a) It is given that the horizontal plane is smooth and that, after the collision, P continues moving in the same direction, with speed 2 m s−1 . Find the speed of Q after the collision. [5] (b) It is given instead that the horizontal plane is rough and that when P and Q collide, they coalesce and move with speed 1.2 m s−1 . Find the coefficient of friction between P and the horizontal plane. [5] 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) 0.3gsin 30 = 0.3a (a = 5) M1 (M1 for applying Newton’s second law parallel to the plane) v2 = 0 + 2 × 2.5 × a M1 v=5 A1 0.3 × 5 + 0 = 0.3 × 2 + 0.2 w M1 Velocity of Q = 4.5 ms–1 A1 5 7(b) 0.3 × z + 0 = 0.5 × 1.2 M1 Velocity of P before collision z = 2 A1 Friction force on P after reaches horizontal plane F = μ × 0.3 g B1 1 1 M1 μ × 0.3g × 1.5 = × 0.3 × 52 – × 0.3 × 22 2 2 Coefficient μ = 0.7 A1 Alternative method for question 7(b) 0.3 × z + 0 = 0.5 × 1.2 M1 Velocity of P before collision z = 2 A1 Friction force on P after reaches horizontal plane F = μ × 0.3 g B1 a = (52 – 22) / (2 × 1.5) = 7, F = 0.3 × 7 M1 Coefficient μ = 0.7 A1 5
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