Mathematics 9709/43 May/June 2019: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4, May/June 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A bus moves in a straight line between two bus stops. The bus starts from rest and accelerates at
2.1 m s−2 for 5 s. The bus then travels for 24 s at constant speed and finally slows down, with a constant deceleration, stopping in a further 6 s. Sketch a velocity-time graph for the motion and hence find the distance between the two bus stops. [5]
Mark scheme
1 Trapezium B1 Includes (0,0) and (...,0)
(t = 0), t = 5, t = 29, t = 35 B1 Correct trapezium with key time values
vmax = 2.1 × 5 = 10.5 ms–1 B1
[½ × (24 + 35) × 10.5] or [½ × 5 × 10.5 + 24 × 10.5 + ½ × 6 × 10.5] M1 Use of area property to find distance
309.75 m or 310 m A1
5
y
30 N
x
25Å
12 N 24 N
Coplanar forces of magnitudes 12 N, 24 N and 30 N act at a point in the directions shown in the diagram.
(i) Find the components of the resultant of the three forces in the x-direction and in the y-direction.
[4]
Component in x-direction……
Component in y-direction……
(ii) Hence find the direction of the resultant. [2]
Mark scheme
2(i) [24cos25° – 12cos65°] M1 Resolving in x-direction
16.7 N A1 (16.679...)
[30 – 24sin25° – 12sin65°] M1 Resolving in y-direction
8.98 N A1 (8.981...)
4
2(ii) 8.98…. M1 Uses trigonometry to find the angle
[tan–1 ]
16.67 ….
28.3° (anticlockwise) from x-direction A1 (28.300... ) or equivalent
6
A car of mass 1400 kg is travelling up a hill inclined at an angle of 4Å to the horizontal. There is a constant resistance to motion of magnitude 1550 N acting on the car.
(i) Given that the engine of the car is working at 30 kW, find the speed of the car at an instant when its acceleration is 0.4 m s−2 . [4]
(ii) The greatest possible constant speed at which the car can travel up the hill is 40 m s−1 . Find the maximum possible power of the engine. [3]
Mark scheme
3(i) M1 Use of Newton’s Second Law (4 terms)
DF – 1550 – 1400gsin4° = 1400 × 0.4 A1 (DF = 3086.59...)
[30000 = (1400 × 0.4 + 1550 +1400gsin4°)v] M1 Use of P = Fv
v = 9.72 ms–1 A1
4
3(ii) [DF – 1550 – 1400gsin4° = 0] M1 (DF = 2526.59...) Resolving up the hill
[Pmax = (1550 + 1400gsin4°) × 40] M1 Use of P = Fv
P = 101000 W or 101 kW A1 (P = 101063.6...)
3
1.3 kg A
B 0.7 kg
1.75 m
1m
Two particles A and B, of masses 1.3 kg and 0.7 kg respectively, are connected by a light inextensible string which passes over a smooth fixed pulley. Particle A is 1.75 m above the floor and particle B is 1 m above the floor (see diagram). The system is released from rest with the string taut, and the particles move vertically. When the particles are at the same height the string breaks.
(i) Show that, before the string breaks, the magnitude of the acceleration of each particle is 3 m s−2 and find the tension in the string. [4]
(ii) Find the difference in the times that it takes the particles to hit the ground. [6]
Mark scheme
4(i) Particle A: [1.3g – T = 1.3a] or Particle B: [T – 0.7g = 0.7a] M1 Use of Newton’s Second law for A or B or use of a = (mA –
mB)g/(mA + mB)
1.3g – T = 1.3a and T – 0.7g = 0.7a A1 Two correct equations
(1.3 − 0.7 ) g
OR a = and 1.3g – T = 1.3a or T – 0.7g = 0.7a
(1.3 + 0.7 )
1.3 g − T T − 0.7 g M1 Solves for a or for T
[6 = 2a, a=3] or [ = , T = 9.1]
1.3 0.7
a = 3 ms–2 and T = 9.1 N A1 (a = 3)
4
4(ii) Distance while connected = 0.375 m B1
[v 2 = 02 + 2 × 3 × 0.375 → v = ...] M1 Use of suvat to find v at ‘break’ (v 2 = 2as)
v = 1.5 ms–1 A1 Correct value or expression for v
[A: 1.375 = 1.5t + ½ gt 2 → t = 0.395...] M1 Finds one time ‘from break to floor’
[B: 1.375 = –1.5t + ½ gt 2 or –1.375 = 1.5t – ½ gt 2 → t = 0.695...] M1 Finds second time ‘from break to floor’
Difference in times = 0.3 s A1
Alternative Method 1 for 4(ii) (last 3 marks)
[uB = 1.5, vB = 0, a = –g, 0 = 1.5 – gt → t = 0.15] M1 Finds tB from ‘break’ to maximum height
Difference in times = 2 × 0.15 M1
Difference in times = 0.3 s A1
4(ii) Alternative Method 2 for 4(ii) (last 3 marks)
[A: 0.375 = ½ × 3t 2 → t = 0.5 1.375 = 1.5t + ½ gt 2 → t = M1 Use of suvat to find total time for A
0.395... tA total = 0.5 + 0.395... = 0.895... s]
[B: 0.375 = ½ × 3t 2 → t = 0.5; 0 = 1.5 – gt → t = 0.15, M1 Use of suvat to find total time for B
s = 1.5t – 1/2gt2 = 0.1125 1.4875 = ½ × gt 2 → t = 0.545...
tB total = 1.195 s]
Difference in times = 0.3 s A1
6
A particle of mass 18 kg is on a plane inclined at an angle of 30Å to the horizontal. The particle is projected up a line of greatest slope of the plane with a speed of 20 m s−1 .
(i) Given that the plane is smooth, use an energy method to find the distance the particle moves up the plane before coming to instantaneous rest. [4]
(ii) Given instead that the plane is rough and the coefficient of friction between the particle and the plane is 0.25, find the speed of the particle as it returns to its starting point. [8]
Mark scheme
5(i) (PE gain =) 18gdsin30° or (KE loss =) ½ × 18 × 202 B1
(PE gain =) 18gdsin30° and (KE loss =) ½ × 18 × 202 B1
[18gdsin30° = ½ × 18 × 202] or [18gh = ½ × 18 × 202] M1 Energy equation (PE gain = KE loss)
Distance up plane = 40 m A1
4
5(ii) R = 18gcos30° (90√3 or 155.884...) B1
[F = 0.25(18gcos30°)] (45√3/2 or 38.971...) M1 Use of F = µR
[18gsin30° + 0.25(18gcos30°) = –18a → a = ...] (a = –7.165..) M1 Newton’s Second Law (3 term equation)
[02 = 202 + 2 × –7.165.. × s → s = ...] M1 Use of suvat to find s
s = 27.913... A1
5(ii) [18gsin30° – 0.25(18gcos30°) = 18a → a = ....] M1 (a = 2.835..) – Newton’s Second Law (3 term equation)
[v 2 = 02 + 2 × 2.835.. × 27.913.. → v = ....] M1 Use of suvat to find s
v = 12.6 ms–1 A1 (12.580...)
Alternative Method 1 for 5(ii)
R = 18gcos30° (90√3 or 155.884...) B1
[F = 0.25(18gcos30°)] (45√3/2 or 38.971...) M1 Use of F = µR
[KE gain = ½ × 18 × 202 and PE loss = 18gh or 18gs(sin30°)] M1 Use of KE = 1/2 mv 2 and PE = mgh
[½ × 18 × 202 = 18gs(sin30°) + 45cos30° × s ] M1 Work / Energy equation (up plane)
s = 27.913... A1
[WD = 45cos30° × 27.91...] M1 Work done against friction
[½ × 18v 2 = (18gsin30°) × 27.91.. – 45cos30° × 27.91...] M1 Work / Energy equation (down plane)
v = 12.6 ms–1 A1 (12.580...)
Alternative Method 2 for 5(ii) (last 3 marks)
[WD = 2 × 45cos30° × 27.91...] M1 WD against friction (up and down)
[½ × 18 × 202 – ½ × 18v2 = 2 × 45cos30° × 27.91...] M1 Uses KE loss = total WD against friction
v = 12.6 ms–1 A1 (12.580...)
8
A particle P moves in a straight line. The acceleration a m s−2 of P at time t s is given by a = 6t − 12.
The displacement of P from a fixed point O on the line is s m. It is given that s = 5 when t = 1 and s = 1 when t = 3.
(i) Show that s = t3 − 6t2 + pt + q, where p and q are constants to be found. [4]
(ii) Find the values of t when P is at instantaneous rest. [2]
(iii) Find the total distance travelled by P in the interval 0 ≤ t ≤ 4. [4]
Additional Page
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Mark scheme
6(i) [v = 6t2/2 – 12t + C] v = 3t2 – 12t + C *M1 Use of v = ∫ a dt
[s = 3t3/3 – 12t2/2 + Ct + D] s = t3 – 6t2 + Ct + D *M1 Use of s = ∫ v dt
[5 = 1 – 6 + C + D C + D = 10 DM1 Substitutes for s and t and solves equations.
1 = 27 – 54 + 3C + D 3C + D = 28 → C = ....., D = .....] Dependent on both Ms.
s = t 3 – 6t 2 + 9t + 1 or p = 9, q = 1 A1
4
6(ii) [v = 0, 3t 2 – 12t + 9 = 0 (t – 1)(t – 3) = 0 → t = ..... ] M1 Solves v = 0 to find t values
t = 1 or t = 3 A1
2
6(iii) 1 3 4 M1 Attempts to use at least three t intervals
∫ ∫ ∫
[ v dt + v dt + v dt ]
0 1 3
[For 0 ⩽ t ⩽ 1, s = (1 – 6 + 9 + 1) – 1 = 4] M1 Evaluates s for one time interval
[0 ⩽ t ⩽ 1, s = (1 – 6 + 9 + 1) – 1 = 4; 1 ⩽ t ⩽ 3, s = (27 – 54 + 27 + 1) – A1 Correctly finds all at least two distances (ignoring signs)
5 = -4 3 ⩽ t ⩽ 4, s = (64 – 96 + 36 + 1) – 1 = 4 ]
Total distance is 12 m A1
4
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Mathematics 9709/43 for May/June 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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