Mathematics 9709/42 May/June 2025: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, May/June 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A crate is being pushed in a straight line along a horizontal surface by a force of magnitude 25 N inclined at 20° above the horizontal. The crate moves a distance of 12 m in 8 seconds with constant speed.
(a) Find the constant speed of the crate. [1]
(b) Find the work done by the 25 N force. [2]
(c) Find the power at which the 25 N force is working. [1]
Mark scheme
1(a) 1.5 m s–1 B1 3 12
OE e.g. , .
2 8
1
1(b) Work done = 25cos 20 ×12 M1 For 25cos 20 ×12 or 25sin 20 ×12 or
25cos 70 ×12 or 25sin 70 ×12 .
282 J A1 281.9077…
2
1(c) 35.2 W B1FT FT their (b) divided by 8
or FT 25cos 20×their(a) .
282
Allow 35.3 (from ).
8
Note: 25 ×1.5 is B0.
Do not accept 35.2 kW.
1
Two particles P and Q, of masses 0.2 kg and 0.1 kg respectively, are free to move in a straight line on a smooth horizontal plane. P is projected towards Q with speed 5 m s -1. At the same instant, Q is projected away from P with speed 2 m s -1. When P collides with Q, the particles coalesce.
Find the kinetic energy lost during the collision. [4]
Mark scheme
2 0.2 × 5 + 0.1× 2 = ( 0.2 + 0.1) v *M1 Attempt at conservation of momentum; correct
number of terms but allow sign errors – use of mg
scores M1A0.
v=4 A1
1 1 1 2 DM1 Allow sign errors only; correct number of terms;
Loss in KE = ± × 0.2 × 52 + × 0.1× 22 − × ( 0.2 + 0.1) × ( their 4 ) dimensionally correct.
2 2 2
= ± ( 2.5 + 0.2 − 2.4 )
[KE lost =] 0.3 J A1 Allow −0.3 .
Use of mg in momentum scores max
M1A0M1A0.
4
2.6 m 35 N
2.4 m
P
A particle P of mass m kg is attached to one end of a light inextensible string of length 2.6 m. The other end of the string is attached to a fixed point on a horizontal ceiling, and the string is taut. The particle
is held in equilibrium by a force of magnitude 35 N, acting in a vertical plane which is perpendicular to the ceiling and contains the string. The force acts in a direction perpendicular to the string (see diagram). The tension in the string is T N and the vertical distance of P from the ceiling is 2.4 m.
Find, in either order, the value of m and the value of T. [4]
Mark scheme
3 12 5
cos θ or sin θ (where θ is the angle
13 13
between the string and vertical).
Candidates may use the complementary angle to
θ .
Resolving in any direction to form an equation M1 Resolving either vertically/horizontally or
parallel/perpendicular to the string. Correct
number of terms, allow sign errors, allow sin/cos
mix; forces that need resolving should be resolved.
M0 for using an angle other than 23 (or better) or
67 (or better).
12 5 A1 Allow T cosθ + 35sin θ = mg with θ = 23 or
T× + 35 × = mg
13 13 better (22.61986495) substituted
OR T = mg cosθ with θ = 23 or better
0.9230T + 35 × 0.3846 = mg (22.61986495) substituted.
0.9230T + 13.461 = mg Allow with their T or m if already found.
12
OR T = mg ×
13
[T = mg × 0.9230]
5 12 A1 Allow T sin θ = 35cos θ with θ = 23 or better
T× = 35 × (22.61986495) substituted
13 13
OR mg sin θ = 35 with θ = 23 or better
T × 0.3846 = 35 × 0.9230 (22.61986495) substituted.
T × 0.3846 = 32.3076
5
OR mg × = 35 [ mg × 0.3846= 35]
13
3 T = 84 and m = 9.1 A1 Allow T = 84.1 and m = 9.11.
Allow AWRT 84.0 for T and 9.10 for m .
Special Case for use of Lami
mg 35 T M1 Attempt at one pair but allow with 2 ‘correct’
= = angles but with the wrong force.
sin 90 sin (180 − 22.6 ) sin ( 90 + 22.6 )
A1 For one correct pair
with θ = 23 or better (22.61986495).
A1 For all three correct
with θ = 23 or better (22.61986495).
T = 84 and m = 9.1 A1 Allow T = 84.1 and m = 9.11.
Allow AWRT 84.0 for T and 9.10 for m .
4
A car is travelling along a straight horizontal road. The car passes through a point A, on the road travelling at a speed of 15 m s -1 , and then accelerates uniformly at 0.4 m s -2 for 30 seconds. The car then moves at constant speed for 3T seconds, where T 1 30 . The car then decelerates uniformly at
0.2 m s -2 and after a further T seconds passes through a point B on the road.
(a) On the given axes, sketch a velocity-time graph for the motion of the car between points A and B. [2]
v (m s–1)
t (s)
O
The distance from A to B is 2750 m.
(b) Find the value of T. [6]
The car continues its journey from B, decelerating uniformly at 0.5 m s -2 until it comes to rest at a point
C on the road.
(c) Find the total distance from A to C. [3]
Mark scheme
4(a) Correct three-line segments for v-t graph B1 First line segment with positive gradient starting
on the positive vertical axis, second line segment
horizontal (parallel to t -axis) and third line
segment with negative gradient, stopping before
the horizontal axis. If the third line continues and
touches the t -axis, there must be some indication
that speed of the car at B is not 0.
Values/expressions labelled on axes correctly B1 15 correct on vertical axis; 30, 30 + 3T and 30 +
4T correctly labelled on horizontal axis.
2
4(b) Speed after 30 seconds = 15 + 0.4 × 30[ = 27] B1
Speed at B is their 27 + ( −0.2) T *M1 Use of v u at using their speed after 30
seconds for u, t = T and a 0.2 .
Attempt at distance from A to B and equate to 2750 *M1 Using the combined area below the three line
segments; using their 27.
1 1 A1 Correct (un-simplified) equation for T.
(15 + 27 ) × 30 + 27 × 3T + ( 27 − 0.2T ) + 27 × T = 2750
2 2
0.1T 2 −108T + 2120 = 0 T = DM1 Re-arranging and attempting to solve their three-
term quadratic equation in T.
If method seen, must be using correct formula OR
if factorising, two terms must be correct for their
three-term quadratic when expanding brackets.
If no method seen, must have at least one correct
value for their three-term quadratic for this mark.
T = 20 only A1 If T = 1060 also stated, then must be rejected.
6
4(c) Speed at B is ( their 27 ) + ( −0.2) × ( their 20 ) [ = 23] *M1 Use of v = u + at using their speed after 30
seconds from (b), their T from (b) and a = ±0.2 .
Distance from B to C is s where 02 = ( their 23) + 2 × ( −0.5) × s and attempt to DM1 Use of v2 = u 2 + 2as with v = 0 and a = −0.5
2
and their speed at B for u (if correct s = 529).
solve for s
Total distance is [529 + 2750 =] 3279 m A1 Condone 3280 m.
3
B
4 kg
5 kg C
A
3 kg
30°
One end of a light inextensible string is attached to a particle A of mass 3 kg. The other end of the string is attached to a particle B of mass 4 kg. Particle A is in contact with a rough plane inclined at 30° to the horizontal, and particle B is in contact with a smooth horizontal plane. A second light inextensible string is attached to B. The other end of this second string is attached to a particle C of mass 5 kg which hangs vertically.
Both strings are taut and pass over small smooth pulleys that are fixed at the ends of the horizontal plane. The part of the string from A to the pulley is parallel to a line of greatest slope of the inclined plane, and A, B and C are in the same vertical plane (see diagram).
The system is released from rest. In the subsequent motion, C moves vertically downwards with acceleration 2 m s -2 , and neither A nor B reach a pulley.
(a) Find the tensions in each of the strings. [3]
(b) Find the coefficient of friction between A and the inclined plane. [4]
When the system has been in motion for 1.5 s, the string attached to A breaks.
(c) Find the total distance that A travels up the plane from the instant that the system is released from rest to the instant that A comes to instantaneous rest. [5]
Mark scheme
5(a) 5 g − TBC = 5 × 2 M1 Attempt at N2L for C – correct number of terms
but allow sign errors (but must be using correct
mass).
TBC − TAB = 4 × 2 M1 Attempt at N2L for B – correct number of terms
but allow sign errors (but must be using correct
mass). Allow with their TBC .
TBC = 40 N and TAB = 32 N A1 Both correct.
3
5(b) TAB − F − 3g sin 30 = 3 × 2 *M1 Attempt at N2L on A – correct number of terms
but allow sign errors; allow sin/cos mix (but must
be using correct mass). For reference: F = 11.
R = 3g cos30 B1 Correct expression for normal contact force at A.
their 32 − μ × 3g cos30 − 3g sin30 = 3× 2 and attempt to solve for μ DM1 Use of F = μ R (where R is a component of
weight) and their TAB to obtain an equation in μ
only and solve for μ .
μ = 0.423 A1 11 3
.
45
4
5(c) 1 B1
Distance travelled by A in first 1.5 seconds is × 2 ×1.52 [ = 2.25]
2
When string breaks A is moving at a speed of 3 (m s–1) B1
26 *M1 Attempt at N2L for A – correct number of terms
− F − 3g sin 30 = 3a a = − but allow sign errors, and cos/sin mix.
3
27 DM1 Attempt at finding the distance travelled by A up
0 = 32 + 2 × ( their a ) × s s = the plane after the string breaks using
52
v2 = u 2 + 2as (or other complete method) with
v = 0,u = 3 and their negative acceleration.
27 A1 36
Total distance travelled by A up the plane is 2.25 + = 2.77 m , 2.769230789.
52 13
5
A particle P moves in a straight line and passes through the point A at time t = 0 . The velocity v m s -1 of P at time t seconds is given by
3 v = (2t + 1) 2 - 2t 2 , where 0 G t G 3.
(a) Find the maximum velocity of P in the interval 0 G t G 3. [5]
It is given that in the interval 0 G t G 3 the velocity of P is always positive.
(b) Find the distance of P from A at the instant when P is moving at this maximum velocity. [4]
Mark scheme
6(a) 3 *M1 Decrease power by 1 and a change in coefficient
Attempt to differentiate ( v = )( 2t + 1) 2 − 2t 2 v
in at least one term; a = is M0.
t
A1 Allow un-simplified.
dv 3 2 −1
3 1
= ( 2t + 1) 2 × 2 − 2 × 2t = 3 ( 2t + 1) 2 − 4t
−1
dt 2
dv 1 DM1 Setting their derivative (of the form
= 0 3 ( 2t + 1) 2 − 4t = 0 9 ( 2t + 1) = 16t 2 1
dt p ( 2t + 1) 2 − qt where p ≠ 0, q ≠ 0 ) equal to
zero and uses a correct method to form an
equation without square roots involved, which
would lead to a quadratic in t .
16t 2 −18t − 9 = 0 t = DM1 Solving correctly their three-term quadratic in t
(must be using correct formula OR if factorising,
two terms must be a correct multiple of their
three-term quadratic when expanding brackets
e.g. 16t 2 −18t − 9 = 0 followed by
3 3
t − t + = 0 is OK for M1), to obtain (at
2 8
least) one positive value of t.
Must get a value of t such that 0 < t < 3 .
3
For reference t = 1.5 t = − .
8
Dependent on both previous M marks.
If no method seen, must have at least one correct
positive value for their three-term quadratic for
this mark.
6(a) Maximum velocity is 3.5 m s–1 ONLY A1 CWO.
3
Using T&I on ( 2t + 1) 2 − 2t 2 to find a maximum
is 0 marks.
Using T&I on 9 ( 2t + 1) = 16t can be awarded
2
full marks for a fully correct solution.
Special case for verification of t = 1.5
3 *M1 Decrease power by 1 and a change in coefficient
Attempt to differentiate ( v = )( 2t + 1) 2 − 2t 2 v
in at least one term; a = is M0.
t
dv 3 3
2 −1
1
A1 Allow un-simplified.
= ( 2t + 1) 2 × 2 − 2 × 2t = 3 ( 2t + 1) 2 − 4t
−1
dt 2
Verification of t = 1.5 DB2 Dependent on previous M1A1.
1 OE.
Must see 3 ( 2 ×1.5 + 1) − 4 ×1.5 = 0 or better
2
Maximum velocity is 3.5 m s–1 ONLY DB1 Dependent on previous M1A1B2.
CWO.
5
6(b) 3 *M1 Increase power by 1 and a change in coefficient in
Attempt to integrate ( v = )( 2t + 1) 2 − 2t 2 at least one term; s = vt is M0.
3
5
A1 Allow un-simplified.
( ) ( )
+1
2t + 1 2 2t 2 +1
2t + 1 2 2t 3
( s =) 3 − ( +c ) = − ( +c )
2 +1 5 3
+ 1 × 2
2
Substitute t = 1.5 and t = 0 DM1 Use of t = 0 and their 1.5 correctly (in their
integrated expression) ONLY;
OR evaluate c ( ≠ 0 ) using t = 0 and substitute t = 1.5 F ( their 1.5) − F ( 0) ;
their 1.5 must have come from an attempt at
differentiation in (a) hence dependent on the first
M1 in (a).
Where their t is such that 0 < t < 3 .
For reference c = −0.2 .
32 9 1 79 A1 CWO.
Distance = − − − 0 = 3.95 or m If no integration seen allow SCB1 for 3.95.
5 4 5 20 For reference c = −0.2 .
4
A particle P of mass 3 kg is projected with a speed of 8 m s -1 up a line of greatest slope of a rough plane inclined at 30° to the horizontal. P is projected from a point A on the plane and comes to instantaneous rest at a point B on the plane. P then slides back down the plane. The coefficient of friction between P
1 and the plane is 12 3.
Using an energy method throughout, find the speed of P at the instant it returns to A. [6]
Additional page
If you use the following page to complete the answer to any question, the question number must be clearly shown.
Mark scheme
7 Apply the work-energy principle for the motion from A to B to form an equation in *M1 Correct number of relevant terms, allow sign
one variable only errors and sin/cos mix.
Dimensionally correct, terms that need a
component should have a component.
1 3 A1 A1 for correct LHS.
× 3 × 82 − 3gd sin 30 = × 3g cos30 × d (where d is the distance AB)
2 12 A1 A1 for correct RHS.
[96 −15d = 3.75d ]
For either d = 5.12 m or work done against friction is 19.2 J A1 Either value stated or clearly implied by later
working.
If using N2L then M0, but SCB1 only (see
below).
1 1 DM1 Either consider the total work done against friction
× 3 × v 2 = × 3 × 82 − 2 × ( their 19.2 ) 1.5v 2 = 96 − 38.4 from A to B and B to A or, consider motion from
2 2
B to A. In both cases must have the correct number
of terms but allow sign errors and cos/sin mix.
OR
Dimensionally correct, terms that need a
1
× 3 × v 2 + ( their 19.2 ) = 3g × ( their 5.12 ) × sin 30 1.5v 2 + 19.2 = 76.8 component should have a component.
2 Dependent on previous M1 or previous SCB1.
OR
1 3
× 3 × v 2 + ( their 5.12 ) × × 3g cos30 = 3g × ( their 5.12 ) × sin 30
2 12
1.5v 2 + 19.2 = 76.8
7 v = 6.20 m s–1 ONLY A1 8 15
, 6.196773354 must be positive.
5
If using N2L then M0, but SCB1 only (see
below).
Allow 6.2 from CWO.
A 3sf answer of 6.19 becoming 6.2 is A0.
This mark is dependent on all previous 5 marks
awarded or on the SCB1 and the previous M1.
Special case for use of N2L for motion up the plane
1 *B1
3a = ± 3g × 3 × cos30 + 3g sin 30 → a = ±6.25
12
and then 0 = 8 + 2 ( −6.25) s → s = 5.12
2 2
Special case for use of N2L for motion down the plane
1 DB1
3a = 3g sin 30 − 3g × 3 × cos30 → a = 3.75
12
and then v2 = 02 + 2 × 3.75 × 5.12 →v = 6.20
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 May/June 2025 mark scheme?+
On this page, free: the mark scheme for AS & A Level Mathematics 9709/42, May/June 2025 (9709_s25_ms_42), and the question paper (9709_s25_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 May/June 2025 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/M/J/25 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.