Mathematics 9709/41 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 particle P is projected vertically upwards with speed 25 m s−1 from a point 3 m above horizontal ground.
(i) Find the time taken for P to reach its greatest height. [2]
(ii) Find the length of time for which P is higher than 23 m above the ground. [3]
(iii) P is higher than h m above the ground for 1 second. Find h. [2]
Mark scheme
2(i) [0 = 25 – 10t] M1 Use of v = u + at with u = 25, v = 0 and a = –g
or other complete method for finding t to highest point
t = 2.5 A1
3
2(ii) [20 = 25t – ½ gt2] M1 Applying s = ut + ½at2 with s = 20, u = 25
[t = 1 and t = 4] M1 Solve a 3-term quadratic for t, factorising or formula
Required time = 4 – 1 = 3 seconds A1
Alternative method for question 2(ii)
[v2 = 252 + 2 × (-10) × 20 → v = ± 15] M1 Using v2 = u2 + 2as with u = 25, s = 20 and a = –g
[-15 = 15 – 10T] or equivalent M1 Use v at s = 20 to find the time, T, taken to reach the maximum
height and to return to s = 20
Required time = 1.5 + 1.5 = 3 seconds A1
3
2(iii) Max height reached at 2.5 s, hence reaches h after 2 s M1 Using their t from 2(i) – 0.5 in s = ut + ½at2
h – 3 = 25 × 2 – 5 × 22 Allow finding h without taking note of the additional 3 m
h = 33 m A1
Alternative method for question 2(iii)
Maximum height = ½ × (25 + 0) × 2.5 [= 31.25] o.e. M1 For attempting to find both the maximum height and the distance
In 0.5 s it falls distance ½ × 10 × 0.52 [= 1.25] fallen in 0.5 seconds
h = 31.25 – 1.25 + 3 = 33 m A1
2
A lorry has mass 12 000 kg.
(i) The lorry moves at a constant speed of 5 m s−1 up a hill inclined at an angle of 1 to the horizontal, where sin 1 = 0.08. At this speed, the magnitude of the resistance to motion on the lorry is
1500 N. Show that the power of the lorry’s engine is 55.5 kW. [3]
When the speed of the lorry is v m s−1 the magnitude of the resistance to motion is kv2 N, where k is a constant.
(ii) Show that k = 60. [1]
(iii) The lorry now moves at a constant speed on a straight level road. Given that its engine is still working at 55.5 kW, find the lorry’s speed. [3]
Mark scheme
3(i) DF = 1500 + 12 000 × g × 0.08 [DF = 11100] M1 Using DF = Resistance + weight component (3 terms)
Power = DF × 5 M1 Using P = Fv (their 2 term DF × 5)
Power = 11 100 × 5 = 55.5 kW A1 AG
3
3(ii) k × 52 = 1500, k = 60 B1 AG
1
3(iii) DF = 60v2 B1 Using DF = resistance = 60v2
55500 = DF × v = 60v2 × v = 60v3 M1 P = Fv used and attempt to solve a 2-term cubic equation for v
v = 9.74 ms-1 A1
3
A particle of mass 1.3 kg rests on a rough plane inclined at an angle 1 to the horizontal, where tan 1 = 12
5
. The coefficient of friction between the particle and the plane is -.
(i) A force of magnitude 20 N parallel to a line of greatest slope of the plane is applied to the particle and the particle is on the point of moving up the plane. Show that - = 1.6. [4]
The force of magnitude 20 N is now removed.
(ii) Find the acceleration of the particle. [2]
(iii) Find the work done against friction during the first 2 s of motion. [3]
Mark scheme
4(i) R = 13 cos 67.4 = 13 (5/13) [R = 5] B1 Resolve forces perpendicular to plane. Allow 67.4 used
F + 13 sin 67.4 = F + 13(12/13) = 20 [F = 8] B1 Resolve forces parallel to plane. Allow 67.4 used
M1 Use F = µR
µ = 8/5 = 1.6 A1 AG Must be from exact working here
4
4(ii) 13 sin 67.4 – F = 1.3a M1 For applying Newton’s second law along the plane and also
F = µR = 8 → [4 = 1.3a] using F = µR (3 terms)
a = 3.08 ms-2 A1 Allow a = 40/13
2
4(iii) s = 0 + 0.5 × (40/13) × 22 [= 80/13 = 6.15] Use s = ut + ½at2 with u = 0 and their a ≠ ±g
M1 to find the distance moved in the first 2 seconds
WD = 8 × 6.15 M1 WD = F × d
WD = 49.2 J A1 Allow WD = 640/13 J
Alternative method for question 4(iii)
s = 0 + 0.5 × (40/13) × 22 [= 80/13 = 6.15] M1
[v = (40/13) × 2] M1 Finding v after 2 seconds
and [WD = 1.3g(80/13)(12/13) – ½ × 1.3 × (80/13)2] and using WD = PE loss – KE gain
WD = 49.2 J A1 Allow WD = 640/13 J
3
A particle P moves in a straight line from a fixed point O. The velocity v m s−1 of P at time t s is given by v = t2 − 8t + 12 for 0 ≤ t ≤ 8.
(i) Find the minimum velocity of P. [3]
(ii) Find the total distance travelled by P in the interval 0 ≤ t ≤ 8. [7]
Mark scheme
5(i) a = 2t - 8 M1 Differentiate to find a
a=0→t=4 M1 Set a = 0 and solve for t
Minimum v = –4 ms-1 A1 Full marks available for correct use of a v-t graph
or correct use of “t = -b/2a”
Alternative method for question 5(i)
v = (t – 4)2 – 4 M1 Attempt to complete the square for v
[t = 4] M1 Choose the t value which gives minimum v
Minimum v = –4 ms–1 A1
3
5(ii) v = 0 when (t – 2)(t – 6) = 0 M1 Find values of t when v = 0, factorise or formula
t = 2 or t = 6 A1
[s = ⅓ t3 – 4t2 + 12t (+c)] M1 Integrate v to find s
A1 Correct integration
0≤t≤2 s1 = 8/3 – 16 + 24 = 32/3 M1 Attempt to find s1, s2 and s3
2 ≤ t ≤ 6 s2 = (216/3 – 144 + 72) – (8/3 – 16 + 24)= -32/3 Look for consideration of the need for 3 intervals
Allow use of symmetry when finding s1, and s3
6≤t≤8
s3 = (512/3 – 4 × 82 + 12 × 8) – (216/3 – 144 + 72) = 32/3
A1 2 correct values of displacement
Total distance = 32 m A1 All correct
7
P
A 0.2 kg B
0.4 kg
0.5 m
1Å
Two particles A and B, of masses 0.4 kg and 0.2 kg respectively, are connected by a light inextensible string. Particle A is held on a smooth plane inclined at an angle of 1Å to the horizontal. The string passes over a small smooth pulley P fixed at the top of the plane, and B hangs freely 0.5 m above horizontal ground (see diagram). The particles are released from rest with both sections of the string taut.
(i) Given that the system is in equilibrium, find 1. [3]
(ii) It is given instead that 1 = 20. In the subsequent motion particle A does not reach P and B remains at rest after reaching the ground.
(a) Find the tension in the string and the acceleration of the system. [4]
(b) Find the speed of A at the instant B reaches the ground. [2]
[Question 6 continues on the next page.]
(c) Use an energy method to find the total distance A moves up the plane before coming to instantaneous rest. [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
6(i) Particle A: T = 4 sin θ M1 Resolve forces for A and for B
Particle B: T = 2
M1 Eliminate T and solve for θ
θ = 30 A1
3
6(ii)(a) A: T – 4 sin 20 = 0.4a M1 Apply Newton’s second law to A or to B or to the system
B: 2 – T = 0.2a
System: 2 – 4 sin 20 = (0.4 + 0.2)a
A1 Two correct equations
M1 Solve for a or T
T = 1.79 and a = 1.05 A1 Both correct
4
6(ii)(b) v2 = 2 × 1.053 × 0.5 = 1.053 M1 Attempt to find v using their a ≠ ±g
v = 1.03 ms–1 A1
2
6(ii)(c) Loss in KE = ½ × 0.4 × 1.053 = 0.2106 M1 Attempt KE loss or PE gain for particle A only after particle B
Gain in PE = 0.4 × 10 × d sin 20 hits the ground.
A1ft Both correct, d is distance moved up the plane after B hits
ground
½ × 0.4 × 1.053 = 0.4 × 10 × d sin 20 M1 Apply KE loss = PE gain
A1 FT Correct energy equation
Total dist A moves up plane = 0.5 + d = 0.654 m A1
5
Checked your answers against these? Let AI mark the whole paper, on your own working.
Examiners follow the generic marking principles at the front of the scheme, then each question's points: method (M), accuracy (A) and independent (B) marks, error carried forward where the scheme allows it, and positive marking, so you're credited for what you got right.
Check your answers against it
Done this paper? Upload your script and CIE Buddy's AI marker marks it question by question against this exact scheme, puts the ticks on your own working and says what each lost mark needed.
Where can I find the 9709/41 May/June 2019 mark scheme?+
On this page, free: the mark scheme for AS & A Level Mathematics 9709/41, May/June 2019 (9709_s19_ms_41), and the question paper (9709_s19_qp_41), to read or download, with each question's marking points written out underneath.
How many marks is 9709/41 out of?+
Mathematics 9709/41 for May/June 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
Can I get my 9709/41 answers marked?+
Yes. Upload your answers to 9709/41/M/J/19 as a PDF or photos, and CIE Buddy marks every question against this paper's official mark scheme, using Cambridge's marking rules. You get your score, each lost mark explained on your own script, and the topics to revise. It's a guide, not an official Cambridge grade.