Mathematics 9709/43 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.
Two particles P and Q, of masses 0.1 kg and 0.3 kg respectively, are at rest on a smooth horizontal plane. P is projected directly towards Q with speed 4u m s -1. At the same instant, Q is projected directly towards P with speed u m s -1. After P and Q collide, P moves with speed 2 m s -1 and Q moves with speed 4 m s -1.
(a) Find the two possible values of u. [3]
(b) Find the largest possible loss of kinetic energy in the collision. [2]
Mark scheme
1(a) 0.1× 4u − 0.3 × u = 0.1× 2 + 0.3 × 4 M1 For use of conservation of momentum once. Must have
or 0.1× 4u − 0.3 × u = 0.1× ( −2 ) + 0.3 × 4 correct number of terms. Allow g included with all 4
masses and sign errors only.
u = 14 A1 Must be positive.
u = 10 A1 Must be positive.
Allow Max M1 A1 A0 if g included with the masses.
Note: 0.1× 4u − 0.3 × u = 0.1× 2 + 0.3 × ( −4 ) leading to
u = 10 (or –10) scores A0.
Maximum M1 A1 if more than 2 values of u stated.
3
1(b) 1 1 M1 For expression or equivalent difference. Allow sign
± × 0.1× ( 4 × their 14 ) + × 0.3 × ( their 14 )
2 2
errors only. Using their 14 which is the larger of the 2
2 2
values found in part (a). If only one value of u found in
1 1 part (a) then M0. If no value for u substituted, then
− 0.1 × 2 2 − × 0.3 × 4 2
2 2 M0.
± (156.8 + 29.4 − 0.2 − 2.4 )
=
Largest loss = 183.6 J A1 918
Allow −183.6, ± . This mark is dependent on
5
full marks in part (a). Condone negative values for u
and v used. If calculating both KE losses, then largest
must be chosen for this mark. Condone ±184 CWO.
2
A van of mass 4500 kg is towing a trailer of mass 350 kg along a straight horizontal road. The van and trailer are connected by a light rigid tow-bar which is parallel to the road. There are resistance forces of
X N on the van and 120 N on the trailer. The driving force produced by the van’s engine is 2500 N. The tension in the tow-bar is T N, and the acceleration of the van is 0.4 m s -2.
Find the value of X and the value of T. [4]
Mark scheme
2 Attempt at Newton’s Second law on either van, trailer or the system *M1 Must have correct number of terms. Allow sign errors.
Dimensionally correct.
T −120 = 350 × 0.4 A1 Any 2 correct equations.
2500 − X − T = 4500 × 0.4
2500 − X − 120 = ( 4500 + 350 ) × 0.4
Attempt to solve for either T or X DM1 From equation(s) with correct number of
dimensionally correct terms.
Must get T = or X = .
T = 260 X = 440 A1 Both correct.
4
v (m s–1)
5
0 t (s)
0 20 40 50 T
The diagram shows a velocity-time graph which models the motion of a particle. The graph consists of
3 straight line segments. The velocity of the particle at time t s after passing a fixed point O is v m s -1.
The particle leaves O with a velocity of 5 m s -1 and accelerates at 0.75 m s -2 for 20 s. The particle then decelerates for the next 30 s. At t = 40 , the velocity of the particle is zero. After t = 40 , the particle starts to travel back to O, coming to rest at O at time T s.
(a) Find the value of T. [5]
(b) Find the acceleration of the particle from t = 50 to t = T . [2]
Mark scheme
3(a) 3 B1 Allow if seen on diagram.
Velocity at t = 20 is 5 + × 20 = 20 m s–1
4
Speed at t = 50 is 10 m s-1 B1 When v is minimum. Allow if seen on diagram. Allow
−10.
1 1 *M1 Correct method to find the displacement up to t = 40 or
Displacement at t = 40 is ( 5 + their 20 ) × 20 + × 20 × their 20 [ = 450] t = 50. Follow through their 20.
2 2
1 1
OR ( ( their 20 − 5 ) × 20 ) + 20 × 5 + × 20 × their 20 [ = 450]
2 2
OR
Displacement at t = 50 is
1 1
( 5 + their 20 ) × 20 + × ( their 20 − 10 ) × 30 [ = 400]
2 2
1 DM1 For an equation in T (or t) involving their displacement
× (T − 40 ) × their 10 = their 450 at either t = 40 or t = 50 and using their positive 10
2
which must have come from
1 1 0 − their 20
OR × (T − 50 ) × their 10 + × 10 × their 10 = their 450 their 20 + × 30 . Must lead to a value of
2 2 20
T > 0 unless correctly recovered.
1
OR × (T − 50 ) × their 10 = their 400
2
T = 130 A1 Condone t = 130 or 130
5
3(b) their 10 M1 Correct method to find the acceleration using their 10
Acceleration = (or −10 ) and their T – dependent on both M marks in
their 130 − 50
part (a) and must lead to a positive value for the
OR 0 = ( their (−10 )) + a × ( their 130 − 50 ) acceleration.
1 A1
Acceleration = m s−2 or 0.125 m s−2
8
2
A
30°
O a°
B
P Q R
25 kg 20 kg m kg
Three blocks P, Q and R, of masses 25 kg, 20 kg and m kg respectively, are held in equilibrium by three light inextensible strings OP, OQ and OR. The strings OP and OR both pass over small fixed smooth pulleys A and B respectively, with P and R hanging vertically below the pulleys. The block Q hangs vertically below the point O. The angle between OA and the vertical is 30° and the angle BOQ = a°
(see diagram).
Find the value of m and the value of a . [6]
Mark scheme
4 For resolving in either direction to form an equation – diagram for reference: *M1 Correct number of terms. Allow sin/cos mix. Allow
sign errors. Allow g missing. Equation must be in
terms of m and α only (so no marks until tensions
replaced).
For reference: TP = 25 g , TR = mg ,TQ = 20 g ,
TP cos30 = TR cos α + TQ , TR sin α = TP sin30 .
25 g cos30 − 20 g − mg cos α = 0 [ 216.506− 200 − 10m cosα = 0 ] A1 A0 if g missing.
25 g sin 30 − mg sin α = 0 [125 − 10m sin α = 0] A1 A0 if g missing.
25 g sin 30 DM1 For attempt to find α. Must get to 'α =′ Must come
α = tan −1 from equations with the correct number of relevant
25 g cos30 − 20 g
terms.
25 g sin 30
OR α = sin −1
( their m ) g
DM1 For attempt to find m or mg.
mg = (25g sin 30)2 + ( 25g cos30 − 20 g )
2
Must get to ‘m=’ or ‘mg=’
OR finding the tension in string OR (for reference if
25 g sin 30 correct is 126.085…) and then using TR = mg .
OR m =
g sin ( their α ) Must come from equations with the correct number of
relevant terms.
α = 82.5 and m = 12.6 A1 AWRT 82.5, 12.6 ( α = 82.4775 m = 12.608 )
A0 if g missing from original equations.
6
A van of mass 2500 kg travelling at speed v m s -1 experiences a resistance force of kv 2 N . The constant power of the van’s engine is 62.5 kW.
(a) The steady speed that the van could maintain when moving along a straight horizontal road is 50 m s -1.
Show that k = 0.5, and find the acceleration of the van when its speed is 25 m s -1 on this straight horizontal road. [4]
The van begins to ascend a hill inclined at an angle i° to the horizontal. The van travels along a line of greatest slope of the hill. The speed of the van at the start of the hill is 20 m s -1 , and its acceleration is 5a m s -2. Later, on the same hill, the speed of the van is 30 m s -1 , and its acceleration is a m s -2. The power of the van’s engine remains at 62.5 kW, and the resistance force remains at 0.5v 2 N .
(b) Find the value of a and the value of i . [5]
Mark scheme
5(a) 62500 = k × 502 × 50 or 1250 = k × 502 M1 For use of Power = DF × v - allow 62.5 ×103 for
62500, allow 62500 = k × 503 .
k = 0.5 A1 AG – allow a correct equation followed by
k = 0.5 .
62500 M1 For N2L with 3 terms; dimensionally correct but allow
− 0.5 × 252 = 2500a
25 62500
sign errors. If using ( = 1250 ) for the DF then
[ 2500 − 312.5 = 2500a] 50
M0.
Acceleration = 0.875 m s−2 A1 7
Allow .
8
4
5(b) Attempt at Newton’s second law at least once to form an equation *M1 With 4 relevant terms; allow sign errors; Allow sin/cos
mix; condone 30 with 5a, 20 with a, but must be
dimensionally correct.
62500 A2 A1 for either correct equation.
− 0.5 × 302 − 2500 g sin θ = 2500a
30
[ 2083.33− 450 − 25000sin θ = 2500a]
62500
and − 0.5 × 202 − 2500 g sin θ = 2500 × 5a
20
[3125 − 200 − 25000sin θ = 12500a ]
62500 62500 DM1 For attempt to solve for a or θ – from equations with
− 0.5 × 202 − − 0.5 × 302 = 10000a the correct number of relevant terms.
20 30
θ = 3[.00] and a = 0.129 A1 31
Allow a = , 0.129167…
240
Allow 0.130 (0.129973…) from using θ = 3 but not
0.13 unless greater accuracy seen.
5
A
5m
i
B
2.5 m
C
The diagram shows the vertical cross-section ABC of a rough waterslide. The section AB is a straight line of length 5 m inclined at an angle of i to the horizontal, where sin i = 0.8 . The point B is 2.5 m above the level of C. A man of mass 80 kg, modelled as a particle, slides down the waterslide, starting from rest at A. The coefficient of friction between the man and the straight section of the waterslide is 0.1.
(a) Find the speed of the man at B. [5]
It is given that there is no change in the speed of the man when passing through B and that his speed at
C is 11 m s -1.
(b) Find the work done against the resistance force as the man moves from B to C. [4]
Mark scheme
6(a) R = 80 g × 0.6 [ = 480] B1 Allow 80 g × cos53 (or better for θ = 53.1301 )
F = 0.1× 80 g × 0.6 [ = 48] *M1 For use of F = 0.1R with R = 80 g × 0.6 or
R = 80 g × 0.8 or equivalent with cos53 or sin53 or
better.
80 g × 0.8 − F = 80a [ a = 7.4] *M1 For attempt to find an equation for a using N2L with 3
terms; allow sign errors; allow sin/cos mix for the
weight component with cos53 or sin53 or better.
Allow F or their F.
v 2 = ( 0 + ) 2 × ( their a ) × 5 DM1 For attempt to find v 2 or v using their positive a.
Velocity = 8.60 m s−1 A1 Allow 74 but A0 for 8.6 if 3sf or better (8.6023…)
answer not seen.
Alternative for Q6(a) for candidates who use an energy method
R = 80 g × 0.6 [ = 480] B1 Allow 80 g × cos53 (or better for θ = 53.1301 ).
F = 0.1× 80 g × 0.6 [ = 48] *M1 For use of F = 0.1R with R = 80 g × 0.6 or
R = 80 g × 0.8 , or equivalent with cos53 or sin53 or
better.
[Loss in] PE = ±80 g × 5 × 0.8[ = ±3200] B1 Allow cos53 or better for the 0.6 in the WD against
friction term or sin 53 or better for the 0.8 in the PE
OR work done [against] friction = ±0.1× 80 g × 0.6 × 5[ = ±240]
term.
1 DM1 For attempt at work energy equation. 3 relevant terms;
80 g × 5 × 0.8 − 0.1× 80 g × 0.6 × 5 = × 80 × v 2 allow sign errors; allow sin/cos mix (using 53 or
2
better) but must be dimensionally correct, terms that
3200 − 240 = 40v 2
need a component should have a component. M0 if the
distance in the WD against friction term is not 5.
6(a) Velocity = 8.60 m s−1 A1 Allow 74 but A0 for 8.6 if 3sf or better (8.6023…)
answer not seen.
5
6(b) 1 B1FT FT their v 2 from part (a) .
1
(
Change in KE = ± × 80 × 112 − × 80 × their v 2
2 2
)
Change in PE = ± 80 g × 2.5 [ = ±2000] B1 Including PE from A is B0.
M1 For attempt at work energy equation. 4 relevant terms;
1 1
(
80 g × 2.5 − W = × 80 × 112 − × 80 × their v 2
2 2
) allow sign errors but must be dimensionally correct.
M0 if using change in PE from A to C.
( )
2000 − W = 4840 − 40 × their v 2
Work done = 120 J A1 Allow 118(.4) from using 8.6(0) from part (a).
Working must lead to a positive answer for the work
done (so –120 oe is A0).
4
A particle X moves in a straight line. The displacement of X from O at time t s after leaving O is s m, where s = 0.3t 2 + 0.6t for 0 G t G 4 .
(a) Find the velocity of X at t = 4 . [2]
1
For t 2 4 , the acceleration of X at time t s after leaving O is a m s -2, where a = 0.3t 2 . There is no change in the velocity of X at t = 4 . The velocity of X at t = T is 14.2 m s -1.
(b) (i) Find the value of T. [4]
(ii) Find the total distance travelled by X between t = 0 and t = T . [4]
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) v = 0.6t + 0.6 B1 Allow un-simplified.
Velocity at t = 4 is 3 m s−1 B1
2
7(b)(i) 1 3 *M1 3
For attempt at integration of 0.3t 2 to kt 2 ( +c ) Answer must be of the form kt 2 where k ≠ 0 or 0.3
3
Use of v = at or v = k1 ( t − 4 ) 2 + scores M0. Do not
penalise missing c.
3 DM1 Use of their 3 from part (a) and t = 4 to find c (for
Their 3 = 0.2 ( 4 ) 2 + c leading to c = reference if correct then c = 1.4)
3
OR using correct limits of 4 and T in their equation for
v = 0.2t 2 + 1.4
T
3
3
v e.g. 0.2t = 0.2 T 2 − 8 .
2
4
3 DM1 For equating to 14.2. Allow in terms of t
14.2 = 0.2T 2 + their 1.4 Dependent on both previous M marks
3
OR equivalent e.g. 14.2 − their 3 = 0.2 T 2 − 8 .
T = 16 A1 Allow 16 or t = 16.
4
7(b)(ii) Distance travelled in first 4 seconds = 7.2 m B1
5 *B1FT For integrating their v from part (b)(i) correctly which
s = 0.08t 2 + 1.4t 3
must be of the form α t 2 + β with α , β ≠ 0 – allow
un-simplified.
5
5
DM1 Correct use of limits 4 and their 16 ( > 4) must be
0.08 (16 ) 2 + 1.4 × 16 − 0.08 ( 4 ) 2 + 1.4 × 4 equivalent to F(their 16) – F(4).
[ = 81.92 + 22.4 − 2.56 − 5.6 = 104.32 − 8.16 = 96.16]
Distance = 103.36 m A1 2584
Condone 103 or better CWO, . Do not ISW if,
25
for example, 103.36 + 7.2 = 110.56.
If constant of integration c found, then must be correct,
that is c = −0.96 .
If no integration seen than max B1 for 7.2 and B1 for
103 m or better.
4
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