Mathematics 9709/42 February/March 2022: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, February/March 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A crane is used to raise a block of mass 600 kg vertically upwards at a constant speed through a height of 15 m. There is a resistance to the motion of the block, which the crane does 10 000 J of work to overcome.
(a) Find the total work done by the crane. [2]
(b) Given that the average power exerted by the crane is 12.5 kW, find the total time for which the block is in motion. [2]
Mark scheme
1(a) 600g × 15 [= 90 000] M1 Attempt potential energy.
Total work done by crane = [90 000 + 10 000 =] 100 000 J A1
2
1(b) 100 000 = 12 500 × t M1 Use of work done = power × time to set up an
equation from which t can be found.
Time = 8 s A1 FT FT on their work done = 100 000
Alternative scheme for question 1(b)
Total WD M1 A complete method, using P = Fv , for setting up
Average force F = an equation from which t can be found.
15
s 15
Average velocity v = =
t t
Total WD 15
P = Fv → 12500 = ×
15 t
Time = 8 s A1 FT
2
A particle P is projected vertically upwards from horizontal ground with speed u m s−1 . P reaches a maximum height of 20 m above the ground.
(a) Find the value of u. [2]
(b) Find the total time for which P is at least 15 m above the ground. [3]
Mark scheme
2(a) 0 = u2 – 2 × 10 × 20 M1 Complete method to set up an equation in u only.
OR Use of v2 = u2 + 2as
u +0 or finding time to reach maximum height (t = 2)
0 = u – 10t and 20 = vt + 12 ´10´t 2 or 20 = ut - 12 ´10´t 2 or 20 = ´t and using this value to set up another equation in u
2
only.
u = 20 A1
2
2(b) 1 M1 1
15 = 20t − × 10 × t 2 Use of s = ut + at 2 and attempt to set up an
2 2
equation from which a relevant t value can be
found.
Must be using their u and a = −10
t = 1 or t = 3 A1
Total time = 2 s A1 CWO
Alternative method for question 2(b)
1 M1 1
5 = × 10 × t 2 Use of s = ut + at 2 and attempt to set up an
2 2
equation from which a relevant t value can be
found.
Must be using u = 0 and a = 10
t=1 A1
Total time = 2 s A1 CWO
3
A car of mass m kg is towing a trailer of mass 300 kg down a straight hill inclined at 3Å to the horizontal at a constant speed. There are resistance forces on the car and on the trailer, and the total work done against the resistance forces in a distance of 50 m is 40 000 J. The engine of the car is doing no work and the tow-bar is light and rigid.
(a) Find the value of m. [3]
The resistance force on the trailer is 200 N.
(b) Find the tension in the tow-bar between the car and the trailer. [2]
Mark scheme
3(a) PE lost in 50 m = (m + 300) g × 50 sin 3 B1
(m + 300) g × 50 sin 3 – 40 000 = 0 M1 Use of the work-energy equation.
m = 1230 to 3 sf A1 m = 1228.6
Alternative method for question 3(a)
40000 B1
Resistance force R = [= 800 N]
50
( m + 300) g sin 3 − R = 0 M1 Apply Newton’s second law to the system, 3 terms.
m = 1230 to 3 sf A1 m = 1228.6
3
3(b) T + 300 g sin 3 – 200 = 0 (Trailer) M1 Apply Newton’s 2nd law either to the trailer or to
or the car using a = 0, three terms in either case.
mg sin 3 = T + 600 (Car)
T = 43[.0] N to 3 sf A1
2
The total mass of a cyclist and her bicycle is 70 kg. The cyclist is riding with constant power of 180 W up a straight hill inclined at an angle ! to the horizontal, where sin ! = 0.05. At an instant when the cyclist’s speed is 6 m s−1 , her acceleration is −0.2 m s−2 . There is a constant resistance to motion of magnitude F N.
(a) Find the value of F. [4]
(b) Find the steady speed that the cyclist could maintain up the hill when working at this power. [2]
Mark scheme
4(a) 180 B1
Forward force exerted by cyclist driving force = [= 30 N]
6
DF – F – 70g sin α = 70 × ‒ 0.2 M1 Attempt Newton’s second law, 4 terms required.
A value must be used for sin α.
30 – F – 70g × 0.05 = 70 × ‒ 0.2 A1 Correct equation
F=9 A1 From exact working only
4
4(b) 180 M1 Apply Newton’s second law up the hill with a = 0.
– F – 70g × sin α = 0 Must have 3 relevant terms using their F from 4(a).
v
A value for sin α must be used.
v = 4.09 m s−1 A1 45
Allow
11
2
Four coplanar forces act at a point. The magnitudes of the forces are 10 N, F N, G N and 2F N. The directions of the forces are as shown in the diagram.
(a) Given that the forces are in equilibrium, find the values of F and G. [5]
(b) Given instead that F = 3, find the value of G for which the resultant of the forces is perpendicular to the 10 N force. [2]
Mark scheme
5(a) Attempt to resolve vertically or horizontally M1 Correct number of terms.
G sin 60° + 2F sin 40° − 10 = 0 A1 Correct resolution vertically.
F + G cos 60° − 2F cos 40° = 0 A1 Correct resolution horizontally.
Attempt to solve simultaneously for F or G M1 From equations with 3 relevant terms in each
F = 4.53, G = 4.82 A1 For both correct.
5
5(b) G sin 60° + 2 × 3 sin 40° − 10 = 0 M1 Resolve forces parallel to the 10 N force and
equate this expression to zero, 3 terms.
G = 7.09 to 3 sf A1
2
A cyclist starts from rest at a fixed point O and moves in a straight line, before coming to rest k seconds
−1 1 later. The acceleration of the cyclist at time t s after leaving O is a m s−2 , where a = 2t 2 − 35 t 2 for
0 < t ≤ k.
(a) Find the value of k. [4]
(b) Find the maximum speed of the cyclist. [3]
(c) Find an expression for the displacement from O in terms of t. Hence find the total distance travelled by the cyclist from the time at which she reaches her maximum speed until she comes to rest. [4]
7
Mark scheme
6(a) For an attempt at integration *M1 Power of at least one term increased by 1 and the
coefficient changed.
1
2
3 A1 Correct v
v = 4t 2 − t 2 [ +C ]
5
1
2
3 DM1 Equating their 2-term v to zero and attempt to
4t 2 − t 2 = 0 solve for t or k.
5
k = 10 A1 Final answer t = 10 is A0
4
6(b) − 3
1 1 M1 Attempt to solve a = 0 and find a value of t.
Max speed when 2t 2 − t 2 = 0
5
10 A1
t=
3
Maximum speed = 4.87 ms−1 to 3 sf B1 8 30
Allow maximum speed as
9
3
6(c) For an attempt at integration of their v *M1 Power of at least one term increased by 1 and the
coefficient changed.
8
3
4
5 A1 Correct s
s = t 2 − t 2 [ +C ]
3 25
10 DM1 10
Substitute their t = and t = 10 Use their t = 10 and their t = ( ≠ 0 ) correctly.
3 3
Distance = 20.7 m A1 Distance = 20.7479...
4
A bead, A, of mass 0.1 kg is threaded on a long straight rigid wire which is inclined at sin−1 25 to the horizontal. A is released from rest and moves down the wire. The coefficient of friction between
A and the wire is -. When A has travelled 0.45 m down the wire, its speed is 0.6 m s−1 .
(a) Show that - = 0.25. [6]
Another bead, B, of mass 0.5 kg is also threaded on the wire. At the point where A has travelled 0.45 m down the wire, it hits B which is instantaneously at rest on the wire. A is brought to instantaneous rest in the collision. The coefficient of friction between B and the wire is 0.275.
(b) Find the time from when the collision occurs until A collides with B again. [6]
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) 0.62 = 0 + 2a × 0.45 M1 Use of constant acceleration equations to find a.
a = 0.4 A1
24 é 24 ù B1 Must use a value for cos α.
R = 0.1g × cos α = 0.1g × = 0.1g × cos 16.3° ê R = = 0.96ú
25 ê
ë 25 úû
7 M1 Newton’s second law, 3 terms.
0.1g × – F = 0.1 × 0.4 [0.28 – F = 0.04 → F = 0.24]
25
24 é 24 μ ù M1 Use of F = µR, where R is a component of 0.1g
F = µ × 0.1g × êF = = 0.96μ ú
25 êë 25 úû
µ = 0.25 A1 AG Must be from exact working µ = 0.25 only
7(a) Alternative scheme for question 7(a)
Attempt PE loss or KE gain M1 Use of either PE = mgh or KE = ½ mv2
7 63 A1 Both correct.
PE loss = 0.1× g × 0.45sin16.3 = 0.1 × g × 0.45 × = 500 = 0.126
25
1 9
KE gain = × 0.1 × 0.62 = = 0.018
2 500
24 é 24 ù B1 Must use a value for cosα.
R = 0.1g × cosα = 0.1g × = 0.1g × cos 16.3° ê R = = 0.96ú
25 ê
ë 25 úû
7 1 M1 Use of work-energy equation as
0.1 × g × 0.45 × = × 0.1 × 0.62 + F × 0.45 PE loss = KE gain + WD against friction
25 2
63 9 54
500 = 500 + μ × 125 or [ 0.126 = 0.018 + μ × 0.432]
24 é 24 μ ù M1 Use of F = µR, where R is a component of 0.1g
F = µ × 0.1g × êF = = 0.96μ ú
25 êë 25 úû
µ = 0.25 A1 AG Must be from exact working µ = 0.25 only
6
7(b) 0.1 × 0.6 = 0.5v M1 Use of conservation of momentum, 2 terms.
v = 0.12 A1
7 24 B1 Apply Newton’s second law for particle B, 3 terms.
For B 0.5g × – 0.275 × 0.5g × = 0.5a [leading to a = 0.16] Allow correct unsimplified expression in a only.
25 25
1 1 *M1 Attempt an expression for either sA or sB.
sA = 0 + × 0.4t 2 sB = 0.12t + × 0.16t 2 Must see uA = 0 and uB ≠ 0 but uB must have been
2 2
found from a momentum equation.
For both sA and sB and attempt to solve sA = sB to find t DM1 Must be from 3 terms leading to a 2-term
quadratic.
If energy used in 7(a) then must find a = 0.4 for A.
Their working must be leading to a positive t value.
Required time is t = 1 s A1
6
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 February/March 2022 mark scheme?+
On this page, free: the mark scheme for AS & A Level Mathematics 9709/42, February/March 2022 (9709_m22_ms_42), and the question paper (9709_m22_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 February/March 2022 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/F/M/22 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.