Cambridge A Level Mathematics 9709 — 2020 Oct/Nov Paper 4 · Variant 2

1 Two particles P and Q, of masses 0.2 kg and 0.5 kg respectively, are at rest on a smooth horizontal plane. P is projected towards Q with speed 2 m s−1. (a) Write down the momentum of P. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) After the collision P continues to move in the same direction with speed 0.3 m s−1. Find the speed of Q after the collision. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 1(a) B1 1 1(b) 0.4 = 0.2 × 0.3 + 0.5v M1 Apply conservation of momentum, 3 terms v = 0.68 ms–1 A1 FT FT on answer in 1(a) 2

More questions on Momentum

2 A car of mass 1800 kg is travelling along a straight horizontal road. The power of the car’s engine is constant. There is a constant resistance to motion of 650 N. (a) Find the power of the car’s engine, given that the car’s acceleration is 0.5 m s−2 when its speed is 20 m s−1. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the steady speed which the car can maintain with the engine working at this power. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(a) DF – 650 = 1800 × 0.5 [DF = 1550] M1 Apply Newton’s second law, 3 terms 650 1800 0.5 20 P − = × B1 [Power P = 1550 × 20 =] 31 000 W or 31 kW A1 3 2(b) 31000 650 0 v − = M1 Use P = Fv with F = 650 v = 47.7 ms–1 A1 FT FT on their P ≠ 13 000 Allow 620 13 2

More questions on Kinematics of motion in a straight line

3 45Å 60Å T N 20 N m kg A block of mass m kg is held in equilibrium below a horizontal ceiling by two strings, as shown in the diagram. One of the strings is inclined at 45Å to the horizontal and the tension in this string is T N. The other string is inclined at 60Å to the horizontal and the tension in this string is 20 N. Find T and m. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 20 cos 60 = T cos 45 M1 Resolve forces horizontally, 2 terms T = 10√2 or T = 14.1 A1 20 sin 60 + T sin 45 = mg or W M1 Resolve forces vertically, 3 terms 20 sin 60 + T sin 45 = mg A1 m = 2.73 [= √3 + 1] A1 Alternative method for question 3 or 20 sin150 sin75 sin135 T mg W   = =     M1 Attempt at one pair of terms using Lami’s Method 20 sin150 sin75 sin135 = = T mg A1 All terms correct in Lami’s Method Attempt to solve for either T or m or W M1 T = 10√2 or T = 14.1 A1 m = 2.73 [= √3 + 1] A1 5 Question Answer Mark Guidance 3 Alternative method for question 3 or 20 sin30 sin105 sin 45 T mg W   = =     M1 Attempt the triangle of forces method and state one equation which involves any two of the forces T, m and 20. 20 sin30 sin105 sin45 = = T mg A1 All correct Attempt to solve for either T or m or W M1 T = 10√2 or T = 14.1 A1 m = 2.73 [= √3 + 1] A1 5

More questions on Forces and equilibrium

4 v (m s−1) 20 V 00 T t (s) The diagram shows a velocity-time graph which models the motion of a car. The graph consists of four straight line segments. The car accelerates at a constant rate of 2 m s−2 from rest to a speed of 20 m s−1 over a period of T s. It then decelerates at a constant rate for 5 seconds before travelling at a constant speed of V m s−1 for 27.5 s. The car then decelerates to rest at a constant rate over a period of 5 s. (a) Find T. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Given that the distance travelled up to the point at which the car begins to move with constant speed is one third of the total distance travelled, find V. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) 20 2 T   =     → T = 10 B1 1 4(b) Distance travelled before constant speed = ½ × 10 × 20 + ½ × (20 + V) × 5 ½ × 10 × 20 + ½ × (20 – V) × 5 + 5V [= 150 + 2.5V] B1 FT May be implied if seen within total distance FT on T value from 4(a) Distance travelled after constant speed = 27.5V + ½ × 5V [= 30V] B1 May be implied if seen within total distance ½ × 10 × 20 + ½ × (20 + V) × 5 = ⅓ [½ × 10 × 20 + ½ × (20 + V) × 5 + 27.5V + ½ × 5V] M1 For attempting to use 1 2 or 1 3 correctly and for obtaining an equation for V which includes all parts of the journey. or ½ × 10 × 20 + ½ × (20 + V) × 5 = ½ [27.5V + ½ × 5V] V = 12 A1 4

More questions on Kinematics of motion in a straight line

5 A particle is projected vertically upwards with speed 40 m s−1 alongside a building of height h m. (a) Given that the particle is above the level of the top of the building for 4 s, find h. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) One second after the first particle is projected, a second particle is projected vertically upwards from the top of the building with speed 20 m s−1. Denoting the time after projection of the first particle by t s, find the value of t for which the two particles are at the same height above the ground. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) 40 – gt = 0 [t = 4] M1 Using v = u + at with u = 40, v = 0 and a = –g to find the time taken to reach the highest point. Time to top of building = 4 – ½(4) = 2 A1 May see t = 4 + 2 = 6 for A1 h = 40 × 2 – ½ × 10 × 22 h = 40 × 6 – ½ × 10 × 62 M1 Using s = ut + ½ at2 with u = 40, a = -g and t = 2 or t = 6 to set up an equation which enables the value of h, the height of the building, to be found. h = 60 A1 Alternative method for question 5(a) 0 = 402 + 2 × (–10) × H M1 For using v2 = u2 + 2as with u = 40, v = 0 and a = –g in order to find H, the greatest height achieved H = 80 A1 s = ½ × 10 × 22 M1 Use either s = vt – ½ at2 with v = 0, a = -g, t = 2 or use s = ut + ½ at2 with u = 0, a = g, t = 2 to find the distance travelled either in the final 2 seconds going up or the first 2 seconds going down s = 20 and so h = 80 – 20 = 60 A1 4 Question Answer Mark Guidance 5(b) Height of first particle above ground = 40t – ½ × 10t2 B1 Height of second particle above top of building = 20(t – 1) – ½ × 10 × (t – 1)2 B1 60 + 20(t – 1) – ½ × 10 × (t – 1)2 = 40t – ½ × 10t2 M1 Set up an equation involving expressions for displacement to enable the time at which the particles reach the same height to be found. t = 3.5 seconds A1 Alternative method for question 5(b) h1 = 40 × 1 – 5 × 12 [= 35] and v1 = 40 – 10 × 1 [= 30] B1 Distance travelled and speed of first particle after 1 second H1 = 30T – 5 × T2, H2 = 20T – 5 × T2 B1 Distance travelled by both particles, T seconds after the second particle is projected. 30T – 5 × T2 = 20T – 5 × T2 + (60 – 35) M1 Set up an equation in T involving expressions for displacement to enable the time at which the particles are at the same height to be found. T = 2.5 and so time to meet = 2.5 + 1 = 3.5 seconds A1 4

More questions on Kinematics of motion in a straight line

6 A block of mass 5 kg is placed on a plane inclined at 30Å to the horizontal. The coefficient of friction between the block and the plane is -. (a) 40 N 5 kg 30Å Fig. 6.1 When a force of magnitude 40 N is applied to the block, acting up the plane parallel to a line of greatest slope, the block begins to slide up the plane (see Fig. 6.1). Show that - < 1 3. [4] 5 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) 5 kg 40 N 30Å Fig. 6.2 When a force of magnitude 40 N is applied horizontally, in a vertical plane containing a line of greatest slope, the block does not move (see Fig. 6.2). Show that, correct to 3 decimal places, the least possible value of - is 0.152. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) R = 5g cos 30 [= 25√3] B1 40 – 5g sin 30 – F > 0 M1 State that the net force up the plane is positive, 3 terms F = µ × 5g cos 30 M1 For using F = µR with R as a component of 5g to obtain an equality/inequality in µ only with 3 terms µ < 1 3 5 A1 AG Alternative scheme for question 6(a) R = 5g cos 30 [= 25√3] B1 40 – 5g sin 30 – F = 5a M1 Acceleration a > 0 F = µ × 5g cos 30 [40 – 5g sin 30 – µ × 5g cos 30 = 5a] M1 For using F = µR with R as a component of 5g to obtain an equality in µ and a µ < 1 3 5 A1 AG. From µ = 1 3 cos30 5 a g = with a > 0 4 Question Answer Mark Guidance 6(b) Attempt to resolve forces parallel to or perpendicular to the inclined plane, 3 relevant terms in either direction M1 R = 5g cos 30 + 40 sin 30 [= 20 + 25√3 = 63.3] A1 F = 40 cos 30 – 5g sin 30 [= 20√3 – 25 = 9.64] A1 µ ⩾ 0.152 B1 AG. Using F ⩽ µR Alternative method for question 6(b) Attempt to resolve forces horizontally or vertically with 3 relevant terms in either direction M1 40 = R sin 30 + F cos 30 [40 = ½R + √3/2F] A1 5g = R cos 30 – F sin 30 [5g = √3/2R – ½F] A1 µ ⩾ 0.152 B1 AG. Solve for R and F and use F ⩽ µR

More questions on Forces and equilibrium

7 A particle P moves in a straight line, starting from a point O with velocity 1.72 m s−1. The acceleration 3 a m s−2 of the particle, t s after leaving O, is given by a = 0.1t 2. (a) Find the value of t when the velocity of P is 3 m s−1. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the displacement of P from O when t = 2, giving your answer correct to 2 decimal places. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) 3/2 0.1  t dt 5/2 0.04 = v t +1.72 A1 5/2 0.04 1.72 3 + = t DM1 For attempting to solve the equation v = 3, to obtain t t = 4 A1 4 Question Answer Mark Guidance 7(b) ( ) 5/2 0.04 1.72  + t dt [ ( ) 7/2 2 1.72 ' 175 = + + s t t C ] *M1 For integrating v which itself has come from integration For using correct limits correctly DM1 Displacement when t = 2 is 3.57 m A1 3

More questions on Kinematics of motion in a straight line

8 A 0.3 kg B 0.5 kg 3.5 N 30Å Two particles A and B, of masses 0.3 kg and 0.5 kg respectively, are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley which is attached to a horizontal plane and to the top of an inclined plane. The particles are initially at rest with A on the horizontal plane and B on the inclined plane, which makes an angle of 30Å with the horizontal. The string is taut and B can move on a line of greatest slope of the inclined plane. A force of magnitude 3.5 N is applied to B acting down the plane (see diagram). (a) Given that both planes are smooth, find the tension in the string and the acceleration of B. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) It is given instead that the two planes are rough. When each particle has moved a distance of 0.6 m from rest, the total amount of work done against friction is 1.1 J. Use an energy method to find the speed of B when it has moved this distance down the plane. [You should assume that the string is sufficiently long so that A does not hit the pulley when it moves 0.6 m.] [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 8(a) For A: T = 0.3a For B: 3.5 + 0.5g sin 30 – T = 0.5a System: 3.5 + 0.5g sin 30 = (0.3 + 0.5)a or to the system. Correct number of terms. A1 Two correct equations For solving either for T or for a M1 a = 7.5 ms–2 A1 T = 2.25 N A1 5 8(b) 0.5g sin 30 × 0.6 [= 1.5] B1 PE loss by B Apply the work-energy equation to the system M1 5 relevant terms, their PE for 0.5 kg, WD by 3.5 N, WD against friction and two relevant KE terms. 0.5g sin 30 × 0.6 + 3.5 × 0.6 = ½ × 0.8 × v2 + 1.1 A1 v = 2.5 ms–1 A1 4

More questions on Newton’s laws of motion