Mathematics 9709/42 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.
s (m)
x
0
0 4 6 t (s)
The diagram shows the displacement-time graph for the motion of a particle. The particle starts from rest at a point O and travels with constant acceleration a m s -2, taking 4 s to move a distance of x m. The particle then returns to O with constant speed of 40 m s -1, over a period of 2 s.
Find the value of x and the value of a. [3]
* 0000800000004 * DFD
4
Mark scheme
1 Any method to find x , or a or getting an equation relating x and a M1 Condone assumption that speed is 40 at the end of first
For example: phase of motion.
x 1
= 40 Allow e.g. x = × 4 × 40 ,
2 2
or x = 2 × 40 1
x = ( 0 + 40 ) × 4 ,
1 2
or x = 40 × 2 + ( 0 ) 22
2 40 = 02 + 2ax ,
2
x−0 40 − 0
or = −40 a= ,
4−6 4
1 1
or x = × a × 42 x = 40 × 4 − × a × 42 .
2 2
x = 80 A1 CWO.
Must be x =
x−0
M1A0 for those who have = 40 and have
4−6
x = −80 followed by x = 80 without any explanation.
a = 10 A1 CWO.
3
A railway locomotive of mass 240 000 kg is towing a coach of mass 36 000 kg down a hill inclined at an angle of sin -1 0.04 to the horizontal. The driving force produced by the locomotive is 450 000 N and there are resistances to motion of 120 000 N on the locomotive and 15 000 N on the coach. The coupling between the locomotive and the coach is light, rigid and parallel to the hill.
Find the acceleration of the locomotive and the tension in the coupling. [5]
* 0000800000005 * DFD
5
Mark scheme
2 Attempt at Newton’s second law at least once *M1 Correct number of terms, allow sign errors, allow
sin/cos mix, allow g missing.
450000 + 240000 g × 0.04 − 120000 − T = 240000a A2 A1 for any equation correct.
450000 + 96000 − 120000 − T = 240000a A2 for any 2 equations correct.
Allow sin 2.3 or better for 0.04.
426000 − T = 240000a A candidate may be using their a in the either of the
T + 36000 g × 0.04 − 15000 = 36000a first 2 equations if they have used the 3 rd equation to
T + 14400 − 15000 = 36000a find a .
T − 600 = 36000a
450000 + ( 240000 + 36000 ) g × 0.04 − 120000 − 15000 = ( 240000 + 36000 ) a
450000 + 110400 − 120000 − 15000 = 276000a
425400 = 276000a
Solving for a or T, from equation(s) with the correct number of relevant terms DM1 Allow g missing, terms should be components where
necessary.
Must get to a = or T =
Acceleration = 1.54 m s-2 and Tension = 56100 N A1 709
Allow a = , a = 1.54130 , T = 56086.9
460
Allow with use of sin 2.3 .
If using a = 1.54 ,
then T = 56400 from locomotive equation
or T = 56040 from coach equation.
If using a = 1.541 ,
then T = 56160 from locomotive equation
or T = 56076 from coach equation.
5
A car of mass 1400 kg travelling on a straight horizontal road accelerates uniformly from a speed of
15 m s -1 to 20 m s -1 over a distance of 175 m.
(a) Find the time taken to travel the 175 m. [2]
There is a constant resistance force of 800 N to the motion of the car.
(b) (i) Use an energy method throughout to find the average power of the car’s engine as the speed increases from 15 m s -1 to 20 m s -1. [4]
(ii) Find the steady speed that the car could maintain on the horizontal road if the engine is working at the power found in (b)(i). [2]
* 0000800000006 * DFD
6
Mark scheme
3(a) 1 M1 For complete method to get an equation in time only
175 = (15 + 20 ) t using s = 175 , u = 15 and v = 20 .
2
May see a = 0.5 .
Time = 10 s A1
2
3(b)(i) 1 1 B1
Change in KE = ± × 1400 × 202 − × 1400 × 152
2 2
± ( 280000 − 157500) = ±122500
=
OR Work done against resistance = ±800 × 175 [ = ±140000 ]
1 1 *M1 Attempt at work energy equation. Correct number of
Work done by engine = ×1400 × 202 − ×1400 ×152 + 800 ×175 relevant terms; dimensionally correct; allow sign
2 2
errors. May see : Work done by engine = DF ×175 .
1
Using ×1400 × ( 20 − 15) is M0.
2
2
For reference if correct WD = 262500 , DF = 1500 .
1 1 DM1 FT their 10 from 3(a) or if recalculated t in 3(b)(i)
Av power × ( their 10 ) = ×1400 × 202 − × 1400 × 152 + 800 × 175
2 2
Average power = 26 250 W A1FT 262500 1500 × 175
FT their value of t only so or .
their 10 their 10
Allow 26300 from correct working.
Allow 26.25 kW or 26.3 kW with units.
Special case for using N2L Maximum 2 marks
202 = 152 + 2 × a × 175 a = 0.5 ; B1 This is for getting DF = 1500 .
DF − 800 = 1400 × 0.5 DF = 1500
DF × 175 = Average power × ( their 10 ) Average power = 26250 W B1FT 1500 × 175
FT their value of t so
OR P = 1500 ×15 = 22500 and P = 1500 × 20 = 30000 so their 10
1 Allow 26.25 kW with units
Average power = ( 22500 + 30000 ) = 26250 W
2
4
3(b)(ii) their 26250 = 800v M1 For use of P = Fv .
525 A1 OE.
Speed = 32.8125 or m s−1 Allow 32.875 or 32.9 or better from use of 26300.
16
2
A particle P of mass 0.1 kg is projected vertically upwards with speed 30 m s -1 from horizontal ground.
At the same instant a particle Q of mass 0.4 kg is projected vertically upwards with speed 10 m s -1 from a height of 15 m above the ground. P and Q move in the same vertical line.
(a) Find the height above the ground at which P and Q collide. [4]
* 0000800000007 * DFD
7
When P and Q collide, they coalesce.
(b) Find the speed of the combined particle at the instant that it reaches the ground. [5]
* 0000800000008 * DFD
8
Mark scheme
4(a) 1 *M1 1
[Height gained by P ( sP ) =]30t − gt 2 For use of s = ut + at 2 at least once with a = ± g and
2 2
1 u = ±30 or u = ±10 .
[Height lost by Q ( sQ ) =]10t − gt 2 Allow this M1 only if using a found value of t .
2
1 1 DM1 For use of sP = sQ ± 15 with s P and sQ of the correct
Meet when 30t − gt 2 = 10t − gt 2 + 15
2 2 form which would lead to a linear equation in t (so in
1
the expressions for sP and sQ the signs of the gt 2
2
terms must be the same).
3 A1 OE.
t = or 0.75 CWO.
4
315 A1 Allow 19.7 or better.
Height = 19.6875 m or m CWO.
16
DO NOT ISW.
4
4(b) *B1FT For either expression for the speed of P or Q before
[vP =] ± 30 −
3
vQ = ± 10 − g [ = ±2.5]
3
g [ = ±22.5]
4 4 3
impact FT their t = and/or FT their height.
4
OR [vP =] ± 302 − 2 g ×19.6875 [ = ±22.5] 315
= = 19.6875 .
16
vQ = ± 102 − 2 g × (19.6875 − 15) [ = ±2.5]
The 2 M marks in 4(a) must have been awarded.
0.1× 22.5 + 0.4 × 2.5 = ( 0.1 + 0.4) × vPQ DM1 Use of conservation of momentum; correct number of
non-zero terms, allow sign errors.
22.5 and 2.5 coming from a correct method.
Do not allow with vP = ±30 or with vQ = ±10.
If using mg rather than m , then do not allow
subsequent A marks.
vPQ = 6.5 A1 SOI, allow to 2sf.
v 2 = their 6.52 + 2 g × their19.6875 DM1 Dependent on previous M1 and B1.
OR Complete method to get an equation in speed or
(speed)2 .
0 = their 6.52 + 2 ( −g ) s s = 2.1125 , May see t = 2.738061302 .
height above ground = 2.1125 + 19.6875 = 21.8
hence v 2 = 02 + 2 g × 21.8
Speed = 20.9 m s-1 or 2 109 m s-1 A1 20.88061303
Use of g in momentum equation can be awarded
B1M1A0M1A0, 3 marks max.
5
20 N
35°
5 kg
10°
A block of mass 5 kg is being pulled straight down a line of greatest slope of a rough plane by a force of magnitude 20 N. The plane is inclined at an angle of 10° to the horizontal and the 20 N force acts at an angle of 35° above the line of greatest slope of the plane (see diagram). The coefficient of friction between the block and the plane is 0.4. The speed of the block when it passes a point O is 2 m s -1.
Find the speed of the block when it has moved 3 m down the plane from O. [6]
* 0000800000009 * DFD
9
Mark scheme
5 For resolving in any direction to form an equation *M1 Correct number of relevant dimensionally correct
terms; allow sin/cos mix; allow sign errors.
R = 5g cos10 − 20sin35 [ = 37.76885892] A1 OE. For reference F = 15.10754357
20cos35 + 5 g sin10 − F = 5a [ 25.06544977 − F = 5a ] A1 OE, allow with their possibly incorrect F .
Allow F − 20cos35 − 5g sin10 = 5a .
For use of F = 0.4R to get an equation in a only DM1 From both dimensionally correct equations with the
correct number of relevant terms.
For reference a = 1.99 [1.99158124] .
v2 = 22 + 2 × (their 1.99) × 3 DM1 Dependent on both previous M marks.
M0 if their calculated a is negative from using N2L
down the plane.
M0 if their calculated a is positive from using N2L up
the plane.
Their calculated 1.99 and 3 must be same sign.
For use of v 2 = u 2 + 2as or any complete method to get
an equation in v or v 2 , where v 2 > 0 .
Speed = 3.99 m s−1 A1 3.993680939
5 Alternative for Q5 using energy
For resolving perpendicular to the plane to form an equation for normal *M1 Correct number of relevant dimensionally correct
reaction terms; allow sin/cos mix; allow sign errors.
R = 5g cos10 − 20sin35 [ = 37.76885892] A1 OE.
1 1 *M1 Attempt at Work-energy equation. Correct number of
20cos35 × 3 + 5g sin10 × 3 + × 5 × 22 − F × 3 = × 5 × v 2 relevant terms, dimensionally correct, allow sin/cos
2 2
mix, allow sign errors, allow their F .
49.1491+ 26.0472+ 10 − 3F = 2.5v 2
A1 Allow their possibly incorrect F .
Note: 60cos10 comes from 3 × 0.4 ( 5 g cos10 ) .
For use of F = 0.4R to get an equation in v only DM1 Dependent on both previous M marks.
Attempt at Work-energy equation. Correct number of
relevant terms, dimensionally correct.
Speed = 3.99 m s−1 A1 3.993680939
6
A particle starts from rest at a point O. The acceleration of the particle at time t s after leaving O is a m s -2, where
1 a = 2 (t + 1) - 2 - 1 for t H 0 .
Find the distance that the particle travels from O until the time at which its acceleration is zero. [8]
* 0000800000010 * DFD
10
Mark scheme
6 1 1
B1 1
Acceleration = 0 when 2 ( t + 1) 2 − 1 = 0 ( t + 1) 2 = 2 t = 3 CWO. Those who use 2 ( t + 1) 2 or claim that
− −
1 1
2 ( t + 1) 2 − 1 is ( 2t + 2 ) 2 − 1 score 0 marks.
− −
1 *M1 1
For attempt at integration of ( a = ) 2 ( t + 1) 2 − 1 v = at is M0. Of the form a ( t + 1) 2 − t [ +c ] , where
−
a≠2.
2 1 1 A1 Allow un-simplified.
[ v =] 1 ( t + 1)− 2 +1 − t [ +c] = 4 ( t + 1) 2 − t [+c ]
− +1
2
1
DM1 For attempt to find c using v = 0 at t = 0 .
v = 0 at t = 0 c = −4 v = 4 ( t + 1) 2 − t − 4 Must get a value for c . M0 for stating v = 0 at t = 0
c = 0 with no working.
Attempt to integrate their v , which has come from integration DM1 Increase power by 1 and change of coefficient on the
same term from a 3 term expression in t to get an
3
expression of the form p ( t + 1) 2 + qt 2 + rt , where
p ≠ 0 , q ≠ 0 and r ≠ 0 . s = vt is M0.
Dependent on previous 2 M marks.
4 1
1 1+1 8 3
1 A1 Allow un-simplified.
[ s =] 1 ( t + 1) 2 +1 − t − 4t [ + k ] = ( t + 1) 2 − t 2 − 4t [ + k ] Condone using +c again.
+1 1+1 3 2
2
8 3 1 8 DM1 Use of limits 0 and their positive 3 correctly, their 3
= × 4 2 − × 32 − 4 × 3 − − 0 − 0 must have come from considering given a = 0 .
3 2 3
Dependent on all 3 previous M marks.
8
OR evaluate k from using s = 0 when t = 0 AND substitute t = their 3 If evaluating k , then k = − if correct.
3
6 13 A1 SC, If no integrating v seen then SCB1 for 2.17 so
Distance = m or 2.17m
6 B1M1A1M1SCB1 5 marks max
8
NO MISREADS IN THIS QUESTION
30 N
20 N
P 30° P
20°
45° a°
15 N SN
25 N
Fig. 7.1 Fig. 7.2
Four coplanar forces of magnitudes 20 N, 30 N, 15 N and 25 N act at a point P in the directions shown in Fig. 7.1. The forces act in a vertical plane. The resultant of these forces has magnitude S N and acts at an angle a° below the horizontal as shown in Fig. 7.2.
(a) Find the value of S and the value of a . [6]
* 0000800000011 * DFD
11
A small ring of mass 0.6 kg is threaded on a rough straight horizontal wire. The four forces shown in
Fig. 7.1 act on the ring and are in the same vertical plane as the wire. The ring starts from rest and takes
3 s to travel a distance of 2 m along the wire.
(b) Find the coefficient of friction between the ring and the wire. [5]
* 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
7(a) For attempt to resolve in any direction to form an expression or equation *M1 Correct number of terms; allow sign errors and sin/cos
mix ONLY.
[ X =] ± ( 30cos30 + 15cos 45 − 20cos 20 ) A1 OE.
OR [ S cos α =] ± ( 30cos30 + 15cos 45 − 20cos 20 ) ±17.79351142
[Y =] ± ( 25 + 15sin 45 − 20sin 20 − 30sin 30 ) A1 OE.
OR [ S sin α =] ± ( 25 + 15sin 45 − 20sin 20 − 30sin 30 ) ±13.76619885
DM1 For attempt to find force from equations with the
S= ( their 17.7935)2 + ( their 13.76661)2 correct number of relevant terms (only sign errors and
their 17.7935 their 13.76661 sin/cos mix allowed).
OR S = OR S =
cos ( their 37.7278) sin ( their 37.7278)
their 13.7661 DM1 Allow reciprocal in tan −1 only.
α = tan −1
their 17.7935 For attempt to find angle from equations with the
their 17.7935 correct number of relevant terms (only sign errors and
−1 their 13.7661
OR α = cos −1 OR α = sin sin/cos mix allowed).
their 22.4970 their 22.4970
[S =] 22.5 [α = ] 37.7 A1 S = 22.497050406 α = 37.72782933
Condone α = 37.8 .
Allow S = −22.5 becoming 22.5 with no explanation.
Allow α = −37.7 becoming 37.7 with no explanation.
A0 for α = −37.7 ONLY, A0 for S = −22.5 ONLY
6
7(b) ( their 17.7935) − F = 0.6a *M1 Using N2L with correct number of dimensionally
OR 30cos30 + 15cos45 − 20cos20 − F = 0.6a correct terms, allow sign errors, allow sin/cos mix.
their17.7935 must have come from horizontal
component from 7(a) with only sign errors and sin/cos
mix allowed. May be using their F and/or their a
which may be incorrect.
OR ( their S ) cos ( their α ) − F = 0.6a [17.8025 − F = 0.6a ] OR ( their S ) cos ( their α ) − F = 0.6 × a so in this case
dependent on all 3 M marks in part (a) and must be
using cosine.
1 4 *M1 1
2 = 0 + a × 32 a = Use of s = ut + at 2 or other complete method to get
2 9 2
an equation in a , using s = 2, u = 0 and t = 3 .
R = their (13.7661) + 0.6 g [ = 19.7661] *B1FT their 13.7661 must have come from vertical
OR R = 25 + 15sin 45 − 20sin 20 − 30sin30 + 0.6 g component from 7(a) with only sign errors and sin/cos
mix allowed.
OR R = ( their S ) sin ( their α ) + 0.6 g so in this case
OR R = ( their S ) sin ( their α ) + 0.6 g [ R = 13.759 + 0.6 g ]
dependent on all 3 M marks in 7(a) and must be using
sine
DM1 Dependent on all previous marks. Use of F = μ R to
( their 17.7935)− μ ×19.7661 = 0.6 × their
4
get an equation in μ only.
9
μ = 0.887 A1 0.886711
Condone 0.886.
Condone 0.888.
7(b) Alternative scheme for using energy
1 *M1 Work energy equation with correct number of
× 0.6 × v 2 = ( their 17.7935) × 2 − F × 2 dimensionally correct terms, allow sign errors, allow
2
sin/cos mix.
1 their17.7935 must have come from horizontal
OR × 0.6 × v 2 = ( 30cos30 + 15cos 45 − 20cos 20 ) × 2 − F × 2 component from 7(a) with only sign errors and sin/cos
2 mix allowed.
May be using their F and/or their v which may be
incorrect.
1
× 0.6 × v 2 = ( their S ) cos ( their α ) × 2 − F × 2 , so in
1 2
OR × 0.6 × v 2 = ( their S ) cos ( their α ) × 2 − F × 2 this case dependent on all 3 M marks in 7(a) and must
2
be using cosine.
*M1 1
( 0 + v ) × 3 v =
1 4
2= Use of s = ( u + v ) t or other complete method to get
2 3 2
an equation in v , using s = 2 , u = 0 and t = 3 .
R = their (13.7661) + 0.6 g [ = 19.7661] *B1FT their 13.7661 must have come from vertical component
OR R = 25 + 15sin 45 − 20sin 20 − 30sin30 + 0.6 g from 7(a) with only sign errors and sin/cos mix
allowed.
Or R = ( their S ) sin ( their α ) + 0.6 g so in this case
OR R = ( their S ) sin ( their α ) + 0.6 g [ R = 13.759 + 0.6 g ]
dependent on all M marks in 7(a) and must be using
sine.
1 4
2 DM1 Dependent on all previous marks. Use of F = μ R to
× 0.6 × their = ( their 17.7935) × 2 − 19.7661× μ × 2 get an equation in μ only.
2 3
μ = 0.887 A1 0.886711
Condone 0.886.
Condone 0.888.
5
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