Mathematics 9709/42 October/November 2021: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, October/November 2021. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
v (m s−1 )
20
6
0 t (s)
0 5 25 30 T 50 60
The diagram shows a velocity-time graph which models the motion of a car. The graph consists of six straight line segments. The car accelerates from rest to a speed of 20 m s−1 over a period of 5 s, and then travels at this speed for a further 20 s. The car then decelerates to a speed of 6 m s−1 over a period of 5 s. This speed is maintained for a further T − 30 s. The car then accelerates again to a speed of 20 m s−1 over a period of 50 − T s, before decelerating to rest over a period of 10 s.
(a) Given that during the two stages of the motion when the car is accelerating, the accelerations are equal, find the value of T . [2]
(b) Find the total distance travelled by the car during the motion. [2]
Mark scheme
1(a) 20 - 6 20 20 M1 Equate the accelerations and set up an equation in T.
= or 20 = 6 +
50 - T 5 5(50 - T ) 20
Allow correct use of their incorrect .
5
T = 46.5 A1
2
1(b) Distance = 12 × 5 × 20 + 20 × 20 + 12 × 5 × (20 + 6) + M1 Attempt to find the total distance travelled using areas.
Allow with T not yet substituted.
+ 6 × (T – 30) + 12 × (50 – T) × (20 + 6) + 12 × 10 × 20
Allow one error in use of area formulae or omission of only
[= 50 + 400 + 65 + 99 + 45.5 + 100] one of the areas: 0–5, 5–25, 25–30, 30–T, T–50, 50–-60.
OR
Distance = 12 × 20 × (60 + 45) – 12 × 14 × (25 + T – 30)
[= 1050 – 290.5]
Total distance travelled = 759.5 m A1 FT FT their T value:
Provided 30 < T < 50 and distance = 1085 − 7T
2
A van of mass 3600 kg is towing a trailer of mass 1200 kg along a straight horizontal road using a light horizontal rope. There are resistance forces of 700 N on the van and 300 N on the trailer.
(a) The driving force exerted by the van is 2500 N.
Find the tension in the rope. [4]
The driving force is now removed and the van driver applies a braking force which acts only on the van. The resistance forces remain unchanged.
(b) Find the least possible value of the braking force which will cause the rope to become slack. [2]
Mark scheme
2(a) For van: 2500 – 700 – T = 3600a M1 Apply Newton’s 2nd law to the van or to the trailer or to the
For trailer: T – 300 = 1200a system of van and trailer. Correct number of terms.
For system: 2500 – 700 – 300 = (3600 + 1200)a
A1 For any two correct.
Obtain an equation in T only M1
5
a = 16 = 0.3125
Tension in the rope = T = 675 N A1
4
2(b) For van: –F – 700 = 3600a M1 Apply Newton’s 2nd law to any two of the van, the trailer and
For trailer: – 300 = 1200a the system with braking force F and with T = 0.
System: − F − 700 − 300 = ( 3600 + 1200 ) a
Least possible value of braking force = F = 200 N A1 Allow F = –200
2
A C
1.8 m
B
The diagram shows a semi-circular track ABC of radius 1.8 m which is fixed in a vertical plane. The points A and C are at the same horizontal level and the point B is at the bottom of the track. The section AB is smooth and the section BC is rough. A small block is released from rest at A.
(a) Show that the speed of the block at B is 6 m s−1 . [2]
The block comes to instantaneous rest for the first time at a height of 1.2 m above the level of B. The work done against the resistance force during the motion of the block from B to this point is 4.5 J.
(b) Find the mass of the block. [3]
Mark scheme
3(a) mg × 1.8 = 12 mv2 M1 Use of conservation of energy, 2 terms.
Must NOT use constant acceleration equations. Use of
equations such as v 2 = u 2 + 2as scores M0 A0.
Speed of block at B = v = 6 ms–1 A1 AG
2
3(b) Attempt the work-energy equation M1 In the form: ± KE lost = ± PE gain ± WD against Resistance
1
2
× m × 62 = 4.5 + mg × 1.2 If using motion from A to final point
A1 mg × 1.8 = mg × 1.2 + 4.5
Mass of the block = m = 0.75 kg A1
3
A cyclist starts from rest at a point A and travels along a straight road AB, coming to rest at B. The displacement of the cyclist from A at time t s after the start is s m, where
s = 0.004 75t2 − t3 .
(a) Show that the distance AB is 250 m. [4]
(b) Find the maximum velocity of the cyclist. [3]
Mark scheme
4(a) For differentiation of s *M1
v = 0.004(150t – 3t2) [ = 0.6t – 0.012t2] A1
v = 0 when t = 50. At t = 50, DM1 Solve v = 0 for t and substitute this value into s.
s = 0.004(75 × 502 – 503) = 0.3 × 502 – 0.004 ×503
Distance AB = 250 m A1 AG
4
4(b) Attempt to determine stationary points for v by differentiation or by use *M1 If symmetry used then an attempt to find the required mid-
of symmetry [a = 0.004(150 – 6t) = 0.6 – 0.024t] point must be seen.
or using symmetry attempt to find the mid-point between t = 0 and their
t value at v = 0
Maximum v when a = 0 so t = 25 DM1 Attempt to solve a = 0 or use symmetry to find the relevant t
Or finding the mid-point if symmetry is used value.
e.g. v = 0.004(150 × 25 – 3 × 252) = 0.6 × 25 – 0.012 × 252 [= 7.5 ms–1]
Maximum velocity = 7.5 ms–1 A1
Alternative method for question 4(b)
Attempt to velocity as v = −0.012 ( t − 25) − 252
2 M1* Attempt to complete the square for their velocity as far as
k ( t − a ) − a 2
2
v = −0.012 ( t − 25 ) + 0.012 × 252
2 DM1 Or select the 0.012 × 252 term as the maximum velocity.
and select t = 25 as the maximum point.
Maximum = [ 0.012 × 625 = ] 7.5 ms–1 A1
3
A railway engine of mass 75 000 kg is moving up a straight hill inclined at an angle ! to the horizontal, where sin ! = 0.01. The engine is travelling at a constant speed of 30 m s−1 . The engine is working at 960 kW. There is a constant force resisting the motion of the engine.
(a) Find the resistance force. [3]
The engine comes to a section of track which is horizontal. At the start of the section the engine is travelling at 30 m s−1 and the power of the engine is now reduced to 900 kW. The resistance to motion is no longer constant, but in the next 60 s the work done against the resistance force is 46 500 kJ.
(b) Find the speed of the engine at the end of the 60 s. [4]
Mark scheme
5(a) 960 000 B1 Allow for 960 000 = DF × 30
Driving force = DF =
30
DF − 75000 g × sin α − R = 0 M1 Resolve forces along the slope.
Must use a value for either sin α or α .
Resistance force = R = 24 500 N A1 Allow correct work with 24500 to 3 sf.
3
5(b) WD by engine in 60 s = 900 000 × 60 [= 54000000] B1
1 1 B1 For either correct expression for KE.
KEinit = × 75000 × 302 KE final = × 75000 × v 2
2 2
1 1 M1 For use of the work-energy equation with 4 terms, correct
900 000 × 60 + × 75000 × 302 = 46 500 000 + × 75000 × v 2 dimensions.
2 2
Speed of engine after 60 s = v = 33.2 ms–1 A1 Allow v = 1100 = 10 11
4
60Å
XN
5 kg
60Å
A block of mass 5 kg is held in equilibrium near a vertical wall by two light strings and a horizontal force of magnitude X N, as shown in the diagram. The two strings are both inclined at 60Å to the vertical.
(a) Given that X = 100, find the tension in the lower string. [4]
(b) Find the least value of X for which the block remains in equilibrium in the position shown. [4]
Mark scheme
6(a) Horizontal: 100 – TU sin 60 – TL sin 60 = 0 M1 Resolve horizontally or vertically or perpendicular to the
Vertical: TU cos 60 – TL cos 60 – 5g = 0 upper string to reach an equation. Correct number of terms,
Perp to TU TL cos 30 + 5g cos 30 = 100 cos 60 Allow X for 100 in horizontal equation.
A1 Either horizontal and vertical equations correct or
perpendicular correct.
Must see X = 100 used for A1.
Solve for either TL or TU using equation(s) with no missing term. M1 May see TU = 107.74
TL = 7.74 N A1 Allow 7.73
4
6(b) Horizontal: X − Tup sin 60 = 0 M1 Resolve either horizontally or vertically or perpendicular to
Vertical: Tup cos 60 − 5 g = 0 the upper string. Must be using the tension Tlow = 0 .
Equivalent to Lami as:
Perp to Tup 5 g cos30 = X cos 60
5g X Tup
= =
sin150 sin120 sin 90
A1 Either horizontal and vertical equations correct or
perpendicular correct.
Eliminate Tup and/or solve for X M1 Tup = 100
Least value of X = 86.6 A1 Allow X = 50 3
4
P m kg
6.4 m
Q
2m kg
!
Particles P and Q have masses m kg and 2m kg respectively. The particles are initially held at rest
6.4 m apart on the same line of greatest slope of a rough plane inclined at an angle ! to the horizontal, where sin ! = 0.8 (see diagram). Particle P is released from rest and slides down the line of greatest slope. Simultaneously, particle Q is projected up the same line of greatest slope at a speed of 10 m s−1 .
The coefficient of friction between each particle and the plane is 0.6.
(a) Show that the acceleration of Q up the plane is −11.6 m s−2 . [4]
(b) Find the time for which the particles are in motion before they collide. [5]
(c) The particles coalesce on impact.
Find the speed of the combined particle immediately after the impact. [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
7(a) For Q: Apply Newton’s 2nd law along or perpendicular to the plane
–2mg sin α – F = 2ma [–16m – 7.2m = 2ma] to particle Q.
R = 2mg cos α [= 12m] M1 Must use values for α or sin α or cos α .
A1 Both correct.
F = 0.6 × 2mg cos α = 0.6 × 0.6 × 20m [= 7.2m] M1 Using F = 0.6 R where R is a component of 2mg only
[2(m)a = –2(m)g (0.8) – 0.6 × 2(m)g (0.6)]
Acceleration of Q up the plane while moving up the plane is A1 AG
a = –11.6 ms–2
4
7(b) For P: mg sin α – 0.6R = ma, leading to 8m – 3.6m = ma M1 Apply Newton’s 2nd law to attempt to find the acceleration of
[ R = mg cos α = 6m, a = 4.4 ms −2 ] particle P. Must use values for α or sin α .
25 M1 For using constant acceleration equations to attempt to
Q comes to rest when 10 − 11.6T1 = 0, T1 = 29 = 0.862 determine when vQ = 0 .
1 M1 Use constant acceleration equations to attempt to find either
For P sP( down ) = × 4.4 × T12 [ = 1.635]
2 sP ( down ) or sQ( up ) at time T1 .
1
For Q sQ( up ) = 10T1 + × ( −11.6 ) × T12 [ = 4.31]
2
d = 6.4 − s P ( down ) − sQ ( up ) [ = 0.455] M1 For attempting to find the extra distance d [ = 0.455] needed
and to find T2 [ = 0.12] by using d = sP 2 − sQ 2 = ( 4.4T1 ) × T2 to reach 6.4 m and using uP = 4.4T1 at T1 to find T2 as
1 1
[ sP 2 and sQ 2 are distances travelled by P and Q in time T2 ] d = ( 4.4T1 ) T2 + × 4.4T22 − × 4.4T22 .
2 2
Time before collision = [t = T1 + T2 = 0.862 + 0.12 =] 0.982 A1 t = 0.98194357 …
7(b) Alternative method for Question 7(b)
For P: mg sin α – 0.6R = ma, leading to 8m – 3.6m = ma M1 Apply Newton’s 2nd law to attempt to find the acceleration of
[ R = mg cos α = 6m, a = 4.4 ms −2 ] particle P. Must use values for α or sin α
25 M1 For using constant acceleration equations to attempt to
Q comes to rest when 10 − 11.6T1 = 0, T1 = 29 = 0.862 determine when vQ = 0
1 M1 Use constant acceleration equations to attempt to find either
For P sP( down ) = × 4.4 × t 2 sP ( down ) or sQ( up ) at time t where t is the total time before
2
1 1 collision.
sQ( up ) = 10T1 + × ( −11.6 ) T12 − × 4.4 ( t − T1 )
2
For Q
2 2
1 1 1 M1 For using sP ( down ) + sQ ( up ) = 6.4 and solving for t
× 4.4t 2 + 10T1 + × ( −11.6 ) T12 − × 4.4 ( t − T1 ) = 6.4
2
2 2 2
Time before collision is t = 0.982 s A1 t = 0.98194357 …
5
Special case for those who do not take into account the fact that Q comes to rest and then changes its direction
For P: mg sin α – 0.6R = ma, leading to 8m – 3.6m = ma M1 Apply Newton’s 2nd law to attempt to find the acceleration of
[ R = mg cos α = 6m, a = 4.4 ms −2 ] particle P. Must use values for α or sin α .
For P sp(down) = (±) 12 × 4.4t2 M1 For using constant acceleration equations to attempt to find
either sp(down) or sq(up).
For Q sq(up) = (±) 10t + 12 × (–11.6)t2
sp + sq = 6.4 leading to 1
2
× 4.4t2 + 10t + 12 × (–11.6)t2 = 6.4 M1 For applying (±) sp + (±) sq = 6.4 using their expressions for sp
and sq to set up and solve a 3-term quadratic equation in t to
obtain at least 1 solution.
7(b) Time that particles are in motion before collision = t = 1 s A1 Must reject t = 16/9
Maximum mark 4 out of 5
4
7(c) up(down) = 0 + 4.4 × 0.982 [= 4.3208] B1 FT Allow ±4.4. FT on their 4.4 and their 0.982
uq(down) = 4.4 × 0.12 [= 0.528] B1 FT Allow ±4.4. FT on their 4.4 and their 0.12
±m × 4.3208 ± 2m × 0.528 = ± (m + 2m)v M1 Apply conservation of momentum, 4 terms, using their up and
[Correct equation is m × 4.3208 + 2m × 0.528 = ± (m + 2m)v] uq values with m and 2m respectively. Velocity of P and Q
after impact must be equal.
Speed of combined particle immediately after impact = v = 1.79 ms–1 A1 Must be positive
Special case for those who do not take into account the fact that Q comes to rest and then changes its direction
up(down) = 0 + 4.4 × 1 [= 4.4] B1 FT Allow ±4.4, FT on their 1 and their 4.4
uq(up) = 10 – 11.6 × 1 [= –1.6] so uq(down) = 1.6 B1 FT Allow ± (10 – 11.6 × 1), FT on their 1
± m × 4.4 ± 2m × 1.6 = ± (m + 2m)v M1 Apply conservation of momentum, 4 terms, using their up and
uq values with m and 2m respectively. Velocity of P and Q
after impact must be equal.
Speed of combined particle immediately after impact = v = 2.53 ms–1 38
A1 Allow v = 15 . Must be positive.
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 2021 mark scheme?+
On this page, free: the mark scheme for AS & A Level Mathematics 9709/42, October/November 2021 (9709_w21_ms_42), and the question paper (9709_w21_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 2021 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/21 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.