Mathematics 9709/42 February/March 2025: question paper and mark scheme
Cambridge AS & A Level Mathematics (9709), Paper 4 Mechanics, February/March 2025. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
30 N
40 N i°
XN
Three coplanar forces of magnitudes 40 N, 30 N and X N act at a point in the directions shown in the diagram.
Given that the forces are in equilibrium, find the values of i and X. [4]
Mark scheme
1 For resolving and forming an equation in any direction *M1 Allow sin/cos mix; correct number of terms.
X cosθ = 40 X sinθ = 30 A1 For correct resolving in two directions.
30 DM1 For attempt to solve for either.
X = 402 + 302 or θ = tan −1
40
X = 50 , θ = 36.9 A1 36.869…
AWRT 50.0.
Alternative for Q1 using triangle of forces
X = 402 + 302 M1 For attempt to solve for X using Pythagoras.
X = 50 A1 AWRT 50.0.
30 M1 For attempt to solve for θ.
tan θ =
40
θ = 36.9 A1 AWRT 36.9.
Alternative for Q1 using Lami’s Theorem
X 30 40 *M1 For any 2 fractions correct.
= =
sin 90 sin (180 − θ ) sin ( 90 + θ )
A1 For all 3 fractions correct.
30 DM1 Solve for θ.
θ = tan −1
40
X = 50 , θ = 36.9 A1 36.869…
AWRT 50.0.
4
A cyclist is travelling along a straight horizontal road at a speed of 4 m s -1 when she passes a point O.
She accelerates at a constant rate for a distance of 42 m, reaching a speed of V m s -1. She maintains the speed of V m s -1 for 50 m and then decelerates at 2 m s –2 before coming to rest. The distance travelled while decelerating is 16 m.
(a) Find the value of V. [2]
(b) Find the total time for which she is in motion from the instant that she passes O. [3]
Mark scheme
2(a) 0 = V 2 − 2 × 2 × 16 B1 For use of constant acceleration to get a correct
equation in V only.
V = 8 only B1
2
2(b) Acceleration section : M1 For attempt to find an equation in t during acceleration
( 4 + ( theirV ) ) t t = 7 or deceleration or constant speed.
42 = [ ] Using their V , s = 16 , a = −2 , u = 4 . Must lead to a
2
Deceleration section: positive t .
0 = ( theirV ) − 2t or 16 = .2t 2 or 16 = ( theirV ) t − .2t 2 [t = 4]
1 1
M1 For attempt to find an equation in t for the other 2
2 2 sections.
[t = 6.25]
50 Using their V , s = 16 , a = −2 , u = 4 . Must lead to a
For constant speed section t =
theirV positive t .
69 A1 AWRT 17.3 from correct work
Total time = s = 17.25 s
4
3
An aeroplane is flying at a constant speed.
(a) The aeroplane is flying horizontally. The aeroplane’s engines are producing a constant power of
5500 kW, and the aeroplane experiences a constant horizontal resistance force of 25 kN.
Find the speed of the aeroplane. [2]
(b) The aeroplane then ascends 300 m in 50 s, while maintaining the same speed. The resistance force is no longer constant, and the work done against the resistance force in ascending the 300 m is
270 000 kJ. The mass of the aeroplane is 60 000 kg.
Find the average power of the aeroplane’s engines. [4]
Mark scheme
3(a) 5500 = 25v OR 5500 000 = 25 000v M1 OE For use of Power = Fv. Allow errors in use of kN
and/or kW.
Speed = 220 m s−1 A1
2
3(b) Change in PE = ±60 × g × 300 [kJ] OR ±60 000 × g × 300 [J] B1 180 000 kJ or 180 000 000 J.
Work done by engines = Power × 50 B1 OE
Power × 50 = 60 × g × 300 + 270 000 [ Power × 50 = 450 000 ] M1 For work energy equation with 3 terms;
Allow with work done by engines instead of
OR Power × 50 = 60 000 × g × 300 + 270 000 000 Power × 50 ;
Allow sign errors; dimensionally correct.
[ Power × 50 = 450 000 000]
Required power = 9000 kW or 9 000 000 W A1
4
0.3 kg 0.1 kg
A B xm
Two particles A and B have masses 0.3 kg and 0.1 kg respectively. The particles are attached to the ends of a light inextensible string. The string passes over a fixed smooth pulley, and the particles hang vertically below the pulley. Both particles are initially at a height of x m above horizontal ground (see diagram). The system is released from rest.
(a) Find the tension in the string and the acceleration of the particles. [4]
During the subsequent motion, B does not reach the pulley. When A reaches the ground, it comes to rest.
(b) Given that the greatest height of B above the ground is 1.2 m, find the value of x. [3]
Mark scheme
4(a) Attempt at Newton’s second law for at least one case *M1 Allow g missing. Correct number of terms. Allow sign
errors.
0.3g − T = 0.3a A1 Any 2 consistent equations, e.g. allow −a for a if
T − 0.1g = 0.1a consistent.
0.3g − 0.1g = ( 0.3 + 0.1) a
Must be same T if individual particle equations.
Attempt to solve for T or a DM1 From equation(s) with correct number of relevant
terms. Allow g missing. Must get to ‘T =’ or ‘a =’.
If no solving seen, must be correct answers for their
equations for this mark.
Acceleration = 5 m s−2 Tension = 1.5 N A1 Allow acceleration = –5 m s−2 .
4
4(b) v2 = 0 + 2 × their a × x *M1 For use of constant acceleration to find v 2 or v in
terms of x . Using their a , a ≠ ± g .
0 = their v2 − 2g × (1.2 − 2x ) DM1 For use of v 2 = u 2 + 2as to get an equation in x only.
Allow a = ± g .
their v 2
OR 0 = their v 2 − 2 gs and 2x + s = 1.2 [leading to 2 x + = 1.2 ]
2g
x = 0.48 A1 OE
3
P 0.6 kg Q 0.4 kg R 0.8 kg
3m 3m
Three particles P, Q and R, of masses 0.6 kg, 0.4 kg and 0.8 kg respectively, are at rest in a straight line on a smooth horizontal plane. The distance from P to Q is 3 m, and the distance from Q to R is also 3 m
(see diagram). P is projected directly towards Q with speed 3 m s -1. After P and Q collide, P continues to move in the same direction with speed 1.5 m s -1.
(a) Find the speed of Q after the collision. [2]
In the subsequent collision between Q and R, these particles coalesce.
(b) Find the speed of the combined particle after this collision. [1]
(c) Find the time that it takes from when P is initially projected until the instant at which P collides with the combined particle. [4]
Mark scheme
5(a) 0.6 × 3 = 0.6 ×1.5 + 0.4v M1 Attempt at conservation of momentum. 3 non-zero
terms. Allow sign errors.
Speed = 2.25 m s−1 A1 OE must be positive.
Allow max M1A0 if g included with the masses.
2
5(b) 0.4 × 2.25 = ( 0.4 + 0.8) × w speed = 0.75 m s−1 B1FT OE condone including g if already penalised in (a).
FT their 2.25.
their 2.25
speed =
3
1
5(c) 3 4 *B1FT
Q takes = s to reach the point at which R was initially.
their 2.25 3
3 *B1FT
× 1.5[ = 2]
their 2.25
3
OR 3 − × 1.5[ = 1]
their 2.25
3
OR × ( their 0.75) [ = 1]
their 2.25
Difference in speeds of P and QR = 1.5 − their 0.75[ = 0 .75] m s−1 DM1 Dependent on both previous B marks.
For attempt to find time.
3
3 − their 2.25 × 1.5
so time = ±
1.5 − their 0.75
3 4
OR ( their 0.75) t ± 3 − × 1.5 = 1.5t → t =
their 2.25 3
3 8
OR ( their 0.75 ) T ± 3 = 1.5T → T =
their 2.25 3
3 3 3 11
OR ( their 0.75 ) T ' ± 3 = 1.5 T ' → T ' =
their 2.25 3 3 3
3 3 3 3 11
OR ( their 0.75) T ' = 1.5 T ′ → T ' =
their 2.25 3 3 1.5 3
5(c) 3 4 4 11 A1 Allow 3.67 s.
Time = + + = s
3 3 3 3
Alternative for Q5(c)
3 4 *B1FT
Q takes = s to reach the point at which R was initially.
their 2.25 3
3 *B1FT
P takes [ = 2] s to reach the point at which R was initially, so combined
1.5
3 3 2
particle has travelled for − = s beyond where R was initially.
1.5 their 2.25 3
3 3
So combined particle is − × ( their 0.75 ) [ = 0.5] m beyond
1.5 their 2.25
where R was initially.
Difference in speeds of P and QR = 1.5 − their 0.75[ = 0 .75] m s−1 M1 Dependent on both previous B marks.
For attempt to find time.
3 3
1.5 − their 2.25 × ( their 0.75 ) 2
so time = ± =
1.5 − their 0.75 3
3 2 11 A1 Allow 3.67 s.
Time = + 2 + = s
3 3 3
4
( )
XN
12 kg
a
A block of mass 12 kg is placed on a rough plane inclined at an angle of a to the horizontal, where a = tan –1 0.5. A force of X N is applied to the block, directly up the plane (see diagram). The coefficient of friction between the block and the plane is n.
(a) It is given that n = 0.15 and X = 20 .
Find the time that it takes for the block to move 2 m down the plane from rest. [6]
(b) It is given instead that n ! 0.15 and that when X = 10 , the block is on the point of moving down the plane.
Find the value of n and the value of X for which the block is on the point of moving up the plane.
[4]
Mark scheme
6(a) 2 B1 2
R = 12 g cos tan −1 0.5 = 12 g × = 12 g × cos 26.565 Allow cos27 or better for . α = 26.56505118 .
5 5
For reference R = 48 5 = 107.3312629 ,
36 5
F= = 16.09968944
5
1 *M1 For use of N2L with 4 terms. Allow sign errors. Allow
12 g × − 20 − F = 12a
5 sin/cos mix. Allow g missing
12 g sin 26.565. − 20 − F = 12a 1
Allow sin 27 or better for .
( )
5
12 g sin tan −1 0.5 − 20 − F = 12a
Allow their possibly incorrect F .
1 2 DM1 For use of F = 0.15R to get an equation in a only,
12 g × − 20 − 0.15 × 12 g × = 12a
5 5 where R is a component of weight or mass.
12 g sin 26.565. − 20 − 0.15 ×12 g × cos 26.565= 12a
( ) ( )
12 g sin tan −1 0.5 − 20 − 0.15 ×12 g × cos tan −1 0.5 = 12a
−25 + 21 5 A1 SOI. Allow AWRT 1.5 a = 1.46382
a = 1.46 or a =
15
1 DM1 Dependent on both M marks.
2 = 0 + × their a t 2 For use of constant acceleration to find t.
2
Allow their a .
t = 1.65s A1 t = 1.65304 Allow 1.66 from using a = 1.46 .
6
6(b) For resolving forces parallel to the slope to form an equation in either case *M1 3 terms; allow sin/cos mix.
1 A1 F = 43.7 [ 43.665] .
10 + F − 12 g × = 0 [10 + F − 12 g × sin 26.565. = 0]
5 1
Allow sin 27 or better for .
(
10 + F − 12 g × sin tan 0.5 = 0
−1
) 5
AND 2
Allow cos27 or better for .
1 5
X − F − 12 g × =0
5
[ X − F − 12 g × sin 26.565. = 0]
( )
X − F − 12 g × sin tan −1 0.5 = 0
Solve for X or μ DM1 Solving for μ must be using R as a component of
weight.
From equation(s) with the correct number of relevant
terms and no sign errors.
X = 97.3 and A1 X = −10 + 48 5 .
μ = 0.407
12 − 5
μ= .
24
Allow X = 97.4 or 97.5 from correct work.
Allow μ = 0.408 from correct work.
4
A particle moves in a straight line. The velocity v m s -1 of the particle t s after leaving a fixed point O is given by v = k (20 + pt - 6t 2) , where k and p are constants. The acceleration of the particle at t = 1 is
42 m s -2 , and the displacement of the particle from O at t = 1 is 93 m.
(a) Show that k = 3 and p = 26 . [6]
(b) Find the distance moved by the particle between the time at which its acceleration is zero and the time at which its velocity is zero. [5]
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 differentiate v *M1 Decrease power by 1 and a change in coefficient in at
least one term (which must be the same term);
v
a = is M0
t
Substitute a = 42 and t = 1 to get A1 OE; Allow unsimplified.
( )
k ( p − 12 ×1) = kp − 12k ×1
42 = k p − 2 × 6 ×1 = 1
For attempt to integrate v *M1 Increase power by 1 and a change in coefficient in at
least one term (which must be the same term)
s = vt is M0
Substitute s = 93 and t = 1 to get A1 OE; Allow unsimplified.
20 1 6 1
93 = k × 11 + p × 11+1 − × 12+1 = k 20 × 1 + p × 12 − 2 × 13
1 1+1 2 +1 2
solving simultaneously for p or k DM1 Dependent on both previous M marks.
Allow sign errors only in solving.
1
42 = k ( p − 12 ) and93 = k 20 + 2 p − 2 Must be solving the correct equations.
Must have c = 0 if evaluated.
Must get to ‘p =’ or ‘k =’or attempt to verify for both
equations.
Working must be seen for this mark.
Must see at least one line of working once either p or
k have been eliminated.
p = 26 k = 3 A1 AG Any error seen is A0
6
7(b) 13 *M1 Using their 2 term linear a that has come from
a = 0 3 ( 26 − 12t ) = 0 t = differentiation to solve for t , which must be positive,
6
using correct p .
26
OE e.g. .
12
(
v = 0 3 20 + 26t − 6t 2 = 0 ) *M1 Attempt to solve given quadratic expression equated to
0 using correct p and k .
Must get at least 1 t value.
2 A1 If 2 values given, they must be both correct.
t = 5 or t = −
3
( ) DM1 Dependent on previous 2 M marks.
5
Distance = 3 20t + 13t 2 − 2t 3 13
For using their positive limits correctly in their s
6
which has come from integration.
May be implied by correct answer.
4537 A1 4913
Distance = 525 − = 525 − 252.05555 273 m Allow .
18 18
272.944
SCB1 for the last 2 marks for answer without seeing
( )
3 20t + 13t 2 − 2t 3 .
5
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Mathematics 9709/42 for February/March 2025 is out of 50 marks, with 1 hour 15 minutes to answer it.
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