Mathematics 9709/41 October/November 2025: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, October/November 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
A car of mass 900 kg is moving along a straight horizontal road against a constant resistance to motion of 350 N. At an instant when the car is moving at 15 m s -1 its acceleration is 0.25 m s -2.
(a) Find the driving force of the car’s engine at this instant. [2]
(b) Find the power of the car’s engine at this instant. [2]
* 0000800000003 * DFD
3
Mark scheme
1(a) D − 350 = 900 × 0.25 M1 Attempt at N2L – correct number of terms. Allow
sign errors but must be dimensionally correct.
D = 575 N A1
2
1(b) P = 575 ×15 M1 Use of P = D × v with their D from 1(a) and v =
15.
P = 8625 W or 8.625 kW A1 Allow 8625 without units,
but 8.625 must have kW.
2
Two particles, P and Q, of masses 3 kg and 5 kg respectively, are at rest on a smooth horizontal plane.
P is projected at a speed of 4 m s -1 directly towards Q. After P and Q collide, P has speed 1 m s -1.
(a) Find the two possible speeds of Q after the collision. [3]
It is given that m J of kinetic energy, where m 2 0 , is lost during the collision.
(b) Find the value of m . [2]
* 0000800000004 * DFD
4
Mark scheme
2(a) 3 × 4 = 3 × ( −1) + 5 ×v M1 Attempt at conservation of linear momentum;
three non-zero terms – allow sign errors. M1 only
if using weight rather than mass.
or 3× 4 = 3×1 + 5v
v = 3 m s −1 A1
v = 1.8 m s−1 A1 If A0 A0 SC B1 for both −3 and −1.8 .
3
2(b) 1 1 1 M1 Attempt at either the total kinetic energy before or
KE = × 3 × 42 ( = 24 ) or KE = × 3 ×12 + × 5 ×1.82 ( = 9.6 ) after the collision.
2 2 2
The total KE (before and after) could be
embedded in a calculation for the KE loss.
1 1
or KE = × 3 ×12 + × 5 × 32 ( = 24 ) M1 be awarded just on sight of 24, 19.6 or 14.4.
2 2
λ = 14.4 ONLY A1 Allow –14.4.
Must have discarded λ = 0 if mentioned.
2
NO MISREADS IN THIS QUESTION
45 N
28 N
35°
50°
70°
35 N
72 N
Coplanar forces of magnitudes 45 N, 28 N, 72 N and 35 N act at a point in the directions shown in the diagram.
Find, in either order, the magnitude and direction of the resultant force. [6]
* 0000800000005 * DFD
5
Mark scheme
3 For attempt to resolve in any direction to form an equation or expression *M1 Correct number of terms; allow sign errors and
sin/cos mix ONLY.
Vertically: (Y = ) ± ( 45 + 28sin 35 − 72sin 50 − 35sin 60) A1 ±24.405948 …
Horizontally: (X =) ± ( 28cos35 + 72cos50 − 35cos60) A1 ±51.716965 …
R = (their 24.4059...)2 + (their 51.7169...)2 DM1 Attempt to solve for R from equations with the
correct number of relevant terms (only sign errors
and sin/cos mix allowed).
their 24.4059... DM1 Attempt to solve for θ or α from equations with
θ = tan −1 = tan 0.4719...
−1
the correct number of relevant terms (only sign
their 51.7169... errors and sin/cos mix allowed).
OR α = tan −1
their 51.7169...
= tan 2.1190...
−1
their 24.4059...
3 R = 57.2 N A1 Both correct.
Direction is 25.3 below the positive x-direction Giving an angle only is insufficient. Direction may
be seen on a diagram, with minimum of arrow on
or bearing 115 resultant OE.
May give direction relative to one of the given
forces e.g. 24.7 anticlockwise from the 72 N
force.
Arrows on both components only is A0 as it
doesn’t show the direction of the resultant.
However the direction is stated, it must be able to
be drawn uniquely.
6
B
3 kg A m kg
0.75 m sin–1 0.6
Two particles, A and B, of masses m kg and 3 kg respectively, are connected by a light inextensible string. Particle B is on a fixed plane which is at an angle of sin -1 0.6 to the horizontal ground. The string passes over a fixed smooth pulley at the top of the plane. Particle A hangs vertically below the pulley and is 0.75 m above the ground (see diagram).
The system is released from rest. In the subsequent motion B moves up a line of greatest slope of the plane and does not reach the pulley. As B moves up the plane there is a constant resistance to its motion of magnitude 10
3 N. The speed of A immediately before it hits the ground is 2 m s .
-1
Use an energy method to find the value of m. [6]
* 0000800000006 * DFD
6
Mark scheme
4 PE lost by A: ±mg × 0.75 B1
PE gained by B: ±3g × 0.75 × 0.6 [ = ±13.5] B1 Allow sin 37 or better for 0.6.
sin −1 0.6 = 36.86989765 .
1 1 B1
KE gained by A and B: ± × 3 × 22 + × m × 22 = ± ( 6 + 2m )
2 2
10 B1
Work done against resistance = ± × 0.75 [ = ±2.5]
3
1 1 M1 Attempt at work-energy equation. Equivalent of 5
× 3 × 22 + × m × 22 = 0.75mg − 13.5 − 2.5 terms, dimensionally correct, allow sign errors, PE
2 2
gained by B must include a component of 3g .
m=4 A1
Special case for use of N2L max 3 marks
8 B1
22 = 02 + 2 × a × 0.75 a =
3
10 M1 Correct number of dimensionally correct terms,
mg − T = ma and T − 3g × 0.6 − = 3a allow sign errors, allow sin/cos mix.
3
Allow sin 37 or better for 0.6.
10 sin −1 0.6 = 36.86989765 .
OR mg − 3g × 0.6 − = ( m + 3) a
3
m=4 A1
6
8mg N
θ°
P
θ°
A particle P of mass m kg is in equilibrium on a rough plane inclined at an angle i° to the horizontal.
The equilibrium of P is maintained by a force of magnitude 8mg N making an angle i° with a line of greatest slope (see diagram). The coefficient of friction between P and the plane is 0.5 and P is on the point of slipping down the plane.
Find the value of i . [6]
* 0000800000007 * DFD
7
Mark scheme
5 Attempt at resolving parallel or perpendicular to the plane to form an equation *M1 Correct number of terms, allow sign errors, allow
sin/cos mix, allow g missing on the weight term
only (so not missing on the 8mg term).
Forces that need resolving should be resolved,
forces that do not need resolving should not be
resolved.
8mg cos θ + mg sin θ = F A1 Allow with their possibly incorrect F .
R = mg cos θ + 8mg sin θ A1
Use of F 0.5R to get an equation in θ [ m and g ] DM1 Where R is a linear combination of a weight (or
mass) component and an 8mg component.
8mg cosθ + mg sin θ = 0.5 ( mg cosθ + 8mg sin θ ) DM1 Re-arranging to tan θ = k where k ≠ 0 (or
equivalent method to reduce to a single trig. term).
– dependent on both previous M1 marks.
tan θ = 2.5
θ = 68.2 A1
6
A particle A of mass 2.5 kg is released from rest from the top of a smooth plane, which makes an angle of sin -1 0.2 with the horizontal. The particle A collides 2 seconds later with a particle B, of mass
3 kg, which is moving up a line of greatest slope of the plane. The speed of B immediately before the collision is 3.5 m s -1. Immediately after the collision, B has a velocity of 0.5 m s -1 down the plane.
Find the distance A moves up the plane after the collision. [6]
* 0000800000008 * DFD
8
Mark scheme
6 Acceleration of A down the plane is g × 0.2 [ = 2] B1 Allow − g × 0.2 .
Allow ± g × sin12 or better.
Allow 2.5a = 2.5 g × 0.2 .
Speed of A before collision is g × 0.2 × 2 [ = 4] *M1 Use of v = u + at with u = 0, t = 2 and their a
which must be equivalent to either ± g × sin12 or
± g × cos12 ONLY.
3× 3.5 + 2.5 × ( −4) = 3× ( −0.5) + 2.5 ×u DM1 Attempt at conservation of linear momentum; four
non-zero terms – allow sign errors. M1A0 only if
using weight rather than mass.
u = 0.8 A1 Correct speed (or velocity) of A after impact.
1 DM1 Use of v 2 = u 2 + 2as (or other complete method
0 = 0.82 + 2 × − g × × s
5 to find s) with v = 0, their u and their −a –
condone sign errors. Dependent on both
previous M1 marks.
s = 0.16 m A1 CWO.
6
A particle P starts from a point O and moves in a straight line. The velocity v m s -1 of P, at time t s after leaving O, is given by v = 13 (2t - 3) (t - 4) .
(a) Find the acceleration of P when t = 2 . [2]
(b) Find the total distance travelled by P in the first 3 seconds of its motion. [5]
* 0000800000009 * DFD
9
(c) Determine whether P returns to O. [2]
* 0000800000010 * DFD
10
Mark scheme
7(a) 1 2 dv 1 M1 Expand to a 3-term quadratic equivalent and
v=
3
( 2t − 11t + 12 ) = ( 4t − 11)
dt 3 attempt to differentiate (oe e.g. product rule),
decrease power by 1 and a change in coefficient in
at least one term; allow un-simplified
v
a= is M0.
t
1 A1 Must be negative.
( 4 × 2 − 11) = −1m s−2
3
2
7(b) Attempt to integrate *M1 Expand to a 3-term quadratic equivalent and
attempt to integrate, increase power by 1 and a
change in coefficient on the same term.
s = vt is M0.
1 2t 3 11t 2 A1 Allow un-simplified.
s= − + 12t ( +c )
3 3 2
[P changes direction when] t = 1.5 *B1 t = 1.5 SOI.
1 2t 3 11t 2 DM1 Using the correct limits correctly.
Attempt to evaluate their − + 12t for t = 0 to t = 1.5 and t = 1.5 to
3 3 2
t=3
Total distance is 2.625 + −1.125 = 3.75 m A1 If no integration seen, then B1 for 3.75 (so 2
marks max.).
5
7(c) Attempt to evaluate their s for t = 0 to t = 4 M1 1.5 4
OE e.g. compare v dt with v dt
0 1.5
or setting their three-term cubic expression in t
(with no constant term) equal to zero and
attempting to solve or attempt discriminant.
4 8 A1 OE e.g. obtaining 4t 2 − 33t + 72 = 0 and showing
v dt =
0 9
> 0 so, no P does not return to O
that this has no real roots (and hence P does not
return to O).
4t 2 − 33t + 72 = 0 gives
b2 − 4ac = ( −33) − 4 × 4 × 72 = −63 .
2
2 2 11
t − t + 4 = 0 gives
9 6
2
11 2 7
b − 4ac = − − 4 × × 4 = − .
2
6 9 36
2
6N
20°
A
15° B
A block A of mass 2 kg and a particle B of mass 0.5 kg are connected by a light inextensible string inclined at 15° to the horizontal. They are pulled across a horizontal surface with acceleration 1.2 m s -2 by a force of magnitude 6 N, applied to A, acting at 20° above the horizontal as shown in the diagram.
The string and the force applied to A are in the same vertical plane.
The contact between B and the surface is smooth and the contact between A and the surface is rough.
(a) Find the tension in the string. [2]
* 0000800000011 * DFD
11
(b) Find the coefficient of friction between A and the surface. [6]
* 0000800000012 * DFD
12
Additional page
If you use the following page to complete the answer to any question, the question number must be clearly shown.
Mark scheme
8(a) T cos15 = 0.5 ×1.2 M1 N2L for B – correct number of terms, allow
sin/cos mix ONLY.
T = 0.621 N A1 3 6 −3 2
Allow .
5
2
8(b) Attempt to resolve vertically for A to form an equation *M1 Correct number of terms, allow sign errors and
sin/cos mix, allow their (possibly incorrect) T
from 8(a).
6sin 20 + RA = 2 g + T sin15 A1 RA = 18.10864
Attempt at N2L for A *M1 Correct number of terms, allow sign errors and
sin/cos mix, allow their (possibly incorrect) T
from 8(a).
6cos 20 − T cos15 − F = 2 ×1.2 A1 F = 2.63815
Use of F = μ R to get an equation in μ only DM1 Dependent on both previous M1 marks.
μ = 0.146 A1 μ = 0.1456848
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 9709/41 October/November 2025 mark scheme?+
On this page, free: the mark scheme for AS & A Level Mathematics 9709/41, October/November 2025 (9709_w25_ms_41), and the question paper (9709_w25_qp_41), to read or download, with each question's marking points written out underneath.
How many marks is 9709/41 out of?+
Mathematics 9709/41 for October/November 2025 is out of 50 marks, with 1 hour 15 minutes to answer it.
Can I get my 9709/41 answers marked?+
Yes. Upload your answers to 9709/41/O/N/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.