Mathematics 9709/43 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.
A metal post is driven vertically into the ground by dropping a heavy object onto it from above. The mass of the object is 120 kg and the mass of the post is 40 kg (see diagram). The object hits the post with speed 8 m s−1 and remains in contact with it after the impact.
(a) Calculate the speed with which the combined post and object moves immediately after the impact.
[2]
(b) There is a constant force resisting the motion of magnitude 4800 N.
Calculate the distance the post is driven into the ground. [3]
Mark scheme
1(a) 120 × 8 = 120v + 40v M1 Applying conservation of momentum.
v = 6 ms−1 A1
2
1(b) 1600 − 4800 = 160a leading to a = –20 M1 Applying Newton’s 2nd law to the system.
0 = 62 + 2 × (–20) × s M1 Use of constant acceleration equations such as
v2 = u2 + 2as.
Distance travelled by post = 0.9 m A1
Alternative method for question 1(b)
Initial KE = 12 × 160 × 62 M1 Use of KE = 1 mv2 for combined mass.
2
1 2 M1 Forms work/energy equation.
2 × 160 × 6 + 160 × 10 × s = 4800s
Distance travelled by post = 0.9 m A1
3
A particle of mass 8 kg is suspended in equilibrium by two light inextensible strings which make angles of 60Å and 45Å above the horizontal.
(a) Draw a diagram showing the forces acting on the particle. [1]
(b) Find the tensions in the strings. [6]
Mark scheme
2(a) Correct 3 force diagram, including angles shown B1
1
2(b) T1 cos 60 = T2 cos 45 M1 Resolving forces horizontally.
T1 sin 60 + T2 sin 45 = 8 g M1 Resolving forces vertically.
T1 cos 60 = T2 cos 45 and T1 sin 60 + T2 sin 45 = 8 g A1
Attempting to solve for either T1 or T2 M1
T1 = 58.6 N A1
T2 = 41.4 N A1
Alternative method for question 2(b)
T1 T 80 M1 Applies Lami’s Theorem – at least two terms correct.
= 2 =
sin135 sin150 sin 75
A1
80sin135 M1 Solves for T1.
T1 =
sin 75
T1 = 58.6 N A1
80sin150 M1 Solves for T2.
T2 =
sin 75
T2 = 41.4 N A1
6
A ball of mass 1.6 kg is released from rest at a point 5 m above horizontal ground. When the ball hits the ground it instantaneously loses 8 J of kinetic energy and starts to move upwards.
(a) Use an energy method to find the greatest height that the ball reaches after hitting the ground.
[3]
(b) Find the total time taken, from the initial release of the ball until it reaches this greatest height.
[3]
Mark scheme
3(a) PE = 1.6 × 10 × 5 [= 80J] B1 Either finds PE loss
or or
v↓= ( 2 ´10 ´ 5) [= 10] uses v2 = u2 + 2as to find the velocity and hence the
kinetic energy on reaching the ground
KE = 12 × 1.6 × 102 [= 80 J]
1.6 × 10 × 5 = 1.6 × 10 × h + 8 M1 Using Initial PE = Final PE + Loss in KE
or or
1 2
2 × 1.6 × v = 80 – 8, v↑ = 90 , using KE = 12 mv2 to find initial velocity upwards and
0 = 90 + 2× (–10) × h leading to h = … either v2 = u2 + 2as or KE loss = PE gain to form
or equation in h.
1 2
2 × 1.6 × v = 80 – 8, v↑ = 90 ,
1
2 m × 90 = m × 10 × h leading to h = …
h = 4.5 m A1
3
3(b) 5 = 0 + 12 × 10 × t2 leading to t = 1 M1 Use of s = ut + 1 gt2 for downward motion
2
or or
5 = 12 × (0 + 10) × t leading to t = 1 use of s = 12 (u + v)t for downward motion
or or
10 = 10t leading to t = 1 use of v = u + gt for downward motion.
4.5 = 0 – 12 × (–10) × t2 leading to t = 0.9 M1 Use of s = vt – 1 (– g)t2 for upward motion
2
or or
4.5 = 12 × ( 90 + 0) × t leading to t = 0.9 use of s = 12 (u + v)t for upward motion
or or
0 = 90 – 10t leading to t = 0.9 use of v = u – gt for upward motion.
t = 1.95 s A1
3
A car of mass 1400 kg is moving on a straight road against a constant force of 1250 N resisting the motion.
(a) The car moves along a horizontal section of the road at a constant speed of 36 m s−1 .
(i) Calculate the work done against the resisting force during the first 8 seconds. [2]
(ii) Calculate, in kW, the power developed by the engine of the car. [2]
(iii) Given that this power is suddenly increased by 12 kW, find the instantaneous acceleration of the car. [3]
(b) The car now travels at a constant speed of 32 m s−1 up a section of the road inclined at 1Å to the horizontal, with the engine working at 64 kW.
Find the value of 1. [2]
Mark scheme
4(a)(i) [WD = 1250 × 36 × 8] M1 For using Work Done = Force × Distance.
WD = 360000 J A1 or 360 kJ
2
4(a)(ii) 360000 B1 FT a (i )
Power = 1250 × 36 or P = [= 45000 J] FT Work Done from .
8 8
= 45 kW B1
2
4(a)(iii) 57000 M1 Use changed Power in P = DF × v.
DF = [= 1583.3..]
36
57000 M1 For using Newton’s 2nd law applied to the car.
− 1250 = 1400a
36
a = 0.238 ms−2 A1
3
4(b) 64000 M1 For using DF = resistance + component of the weight of
= 1250 + 1400g sin θ the car.
32
θ = 3.1 [3.0708…] A1
2
A particle P moves in a straight line, starting from rest at a point O on the line. At time t s after leaving O the acceleration of P is k 16 − t2 m s−2 , where k is a positive constant, and the displacement from O is s m. The velocity of P is 8 m s−1 when t = 4.
1 t 2 96 − t2 .
(a) Show that s = 64 [5]
(b) Find the speed of P at the instant that it returns to O. [3]
(c) Find the maximum displacement of the particle from O. [3]
Mark scheme
5(a) 1 3 M1 Uses v = ∫a dt.
a = 16k − kt 2 , v = 16kt − kt
3
8 = 16k × 4 – 13 k × 43 leading to k = … M1 Substitutes t = 4, v = 8.
kt 3 3 A1 OE
v = 16kt - and k =
3 16
1 4 24 3 4 M1 Uses s = ∫vdt and attempts to find s in terms of t only.
s = 8kt 2 − kt leading to s = t 2 − t
12 16 192 May be using v = 3t – 161 t 3.
1 2 A1 AG, no errors seen.
s= t (96 − t 2 )
64
5
5(b) s = 0, t2 = 96, t = 4 6 M1 Attempt to find t when s = 0.
3 3 1 M1 Attempt to find v at this t value
v = 16 × × 96 − × × 963
16 16 3
Speed is 29.4 ms-1 A1 Do not condone v = −29.4 .
3
5(c) v = 0, t2 = 48, t = 4 3 M1 Determine the time, t (or t2) at which v = 0
1 M1 Use substitution of the t or t2 value to find s.
s= × 48 × (96 − 48)
64
s = 36 m A1
3
The diagram shows a particle of mass 5 kg on a rough horizontal table, and two light inextensible strings attached to it passing over smooth pulleys fixed at the edges of the table. Particles of masses
4 kg and 6 kg hang freely at the ends of the strings. The particle of mass 6 kg is 0.5 m above the ground. The system is in limiting equilibrium.
(a) Show that the coefficient of friction between the 5 kg particle and the table is 0.4. [2]
The 6 kg particle is now replaced by a particle of mass 8 kg and the system is released from rest.
(b) Find the acceleration of the 4 kg particle and the tensions in the strings. [5]
(c) In the subsequent motion the 8 kg particle hits the ground and does not rebound.
Find the time that elapses after the 8 kg particle hits the ground before the other two particles come to instantaneous rest. (You may assume this occurs before either particle reaches a pulley.)
[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(a) R = 5g, F = 6g – 4g M1 For resolving forces to find F and R.
2g A1 AG
μ= = 0.4
5g
2
6(b) T1 – 4g = 4a or 8g – T2 = 8a M1 For applying Newton’s 2nd law to the 4 kg particle or the
8 kg particle.
T1 – 4g = 4a and 8g – T2 = 8a A1 Both equations correct.
T2 – T1 – F = 5a and F = 0.4 × 5g B1
Adding gives 8g – 4g – 2g = 17a leading to a = …. M1 Attempt to solve for a, T1 or T2.
a = 1.18 ms–2, T1 = 44.7 N, T2 = 70.6 N A1
5
6(c) T – 4g = 4a, -T – F = 5a, F = 2g or –4g – 2g = 9a M1 Applying Newton’s 2nd law to both active particles.
60 A1
a= -
9
20 20 M1 Use of v2 = u2 + 2as or equivalent to find v or v2 when the
v 2 = 2´ ´ 0.5 = leading to v = … [v = 1.0846…] 8 kg particle reaches the ground.
17 17
20 60 M1 Use of v = u + at or equivalent to find t.
0= − t
17 9
t = 0.163 s A1 From t = 0.1626978…
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 9709/43 October/November 2021 mark scheme?+
On this page, free: the mark scheme for AS & A Level Mathematics 9709/43, October/November 2021 (9709_w21_ms_43), and the question paper (9709_w21_qp_43), to read or download, with each question's marking points written out underneath.
How many marks is 9709/43 out of?+
Mathematics 9709/43 for October/November 2021 is out of 50 marks, with 1 hour 15 minutes to answer it.
Can I get my 9709/43 answers marked?+
Yes. Upload your answers to 9709/43/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.