Mathematics 9709/42 October/November 2020: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, October/November 2020. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
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) Momentum = 0.2 × 2 = 0.4 kg ms–1 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
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
P B1
− 650 = 1800 × 0.5
20
[Power P = 1550 × 20 =] 31 000 W or 31 kW A1
3
2(b) 31000 M1 Use P = Fv with F = 650
− 650 = 0
v
v = 47.7 ms–1 A1 FT 620
FT on their P ≠ 13 000 Allow
13
2
45Å 60Å
TN 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
T mg or W 20 M1 Attempt at one pair of terms using Lami’s Method
sin150 = sin 75 = sin135
T mg 20 A1 All terms correct in Lami’s Method
= =
sin150 sin 75 sin135
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
3 Alternative method for question 3
T mg or W 20 M1 Attempt the triangle of forces method and state one equation which
sin 30 = sin105 = sin 45 involves any two of the forces T, m and 20.
T mg 20 A1 All correct
= =
sin 30 sin105 sin 45
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
v (m s−1 )
20
V
0 t (s)
0 T
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 B1
2 = T → T = 10
1
4(b) Distance travelled before constant speed = B1 FT May be implied if seen within total distance
½ × 10 × 20 + ½ × (20 + V) × 5 FT on T value from 4(a)
½ × 10 × 20 + ½ × (20 – V) × 5 + 5V
[= 150 + 2.5V]
Distance travelled after constant speed B1 May be implied if seen within total distance
= 27.5V + ½ × 5V [= 30V]
½ × 10 × 20 + ½ × (20 + V) × 5 M1 1 1
For attempting to use or correctly and for obtaining an
= ⅓ [½ × 10 × 20 + ½ × (20 + V) × 5 + 27.5V + ½ × 5V] 2 3
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
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 M1 Using s = ut + ½ at2 with u = 40, a = -g and t = 2 or t = 6 to set up
h = 40 × 6 – ½ × 10 × 62 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
5(b) Height of first particle above ground = 40t – ½ × 10t2 B1
Height of second particle above top of building B1
= 20(t – 1) – ½ × 10 × (t – 1)2
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
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 - < 15 3.
[4]
(b)
40 N 5 kg
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
1 A1 AG
µ< 3
5
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 M1 For using F = µR with R as a component of 5g to obtain an equality
[40 – 5g sin 30 – µ × 5g cos 30 = 5a] in µ and a
1 A1 1 a
µ< 3 AG. From µ = 3 = cos30 with a > 0
5 5 g
4
6(b) Attempt to resolve forces parallel to or perpendicular to the inclined M1
plane, 3 relevant terms in either direction
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 M1
terms in either direction
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
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) 0.1t 3/2 dt *M1 For integrating a
v = 0.04t 5/ 2 +1.72 A1
0.04t 5/2 + 1.72 = 3 DM1 For attempting to solve the equation v = 3, to obtain t
t=4 A1
4
7(b) ( )
0.04t 5/2 + 1.72 dt *M1 For integrating v which itself has come from integration
2 7/2
[s = t + 1.72t ( +C ') ]
175
For using correct limits correctly DM1
Displacement when t = 2 is 3.57 m A1
3
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]
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
8(a) For A: T = 0.3a M1 For applying Newton’s 2nd law for either particle A or to particle B
For B: 3.5 + 0.5g sin 30 – T = 0.5a or to the system. Correct number of terms.
System: 3.5 + 0.5g sin 30 = (0.3 + 0.5)a
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
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 9709/42 October/November 2020 mark scheme?+
On this page, free: the mark scheme for AS & A Level Mathematics 9709/42, October/November 2020 (9709_w20_ms_42), and the question paper (9709_w20_qp_42), to read or download, with each question's marking points written out underneath.
How many marks is 9709/42 out of?+
Mathematics 9709/42 for October/November 2020 is out of 50 marks, with 1 hour 15 minutes to answer it.
Can I get my 9709/42 answers marked?+
Yes. Upload your answers to 9709/42/O/N/20 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.