A Level Mathematics 9709 Paper 42, May/June 2025

Question paper and mark scheme50 marks7 questions

Solve it on ExamPal

Practise

Untimed. Write on the paper and check as you go.

Start

Exam

Timed, with a clock you set, like sitting the real paper.

Start

Opens full screen.

Opening the question paper

Read as textThe questions and the mark scheme as plain text. Easier to read on a phone.

Question paper

9709/42 · May/June 2025

Download PDF
1

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]
2

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 ms−1. At the same instant, Q is projected away from P with speed 2 ms−1. When P collides with Q, the particles coalesce. Find the kinetic energy lost during the collision.

[4]
3

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]
Figure 1 of 1 for question 3
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 ms−1, and then accelerates uniformly at 0.4 ms−2 for 30 seconds. The car then moves at constant speed for 3T seconds, where T<30. The car then decelerates uniformly at 0.2 ms−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]
b

The distance from A to B is 2750 m. Find the value of T.

[6]
c

The car continues its journey from B, decelerating uniformly at 0.5 ms−2 until it comes to rest at a point C on the road. Find the total distance from A to C.

[3]
5

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 ms−2, and neither A nor B reach a pulley.

Figure 1 of 1 for question 5
a

Find the tensions in each of the strings.

[3]
b

Find the coefficient of friction between A and the inclined plane.

[4]
c

When the system has been in motion for 1.5 s, the string attached to A breaks. 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]
6

A particle P moves in a straight line and passes through the point A at time t=0. The velocity v ms−1 of P at time t seconds is given by v=(2t+1)32−2t2, where 0⩽t⩽3.

a

Find the maximum velocity of P in the interval 0⩽t⩽3.

[5]
b

It is given that in the interval 0⩽t⩽3 the velocity of P is always positive. Find the distance of P from A at the instant when P is moving at this maximum velocity.

[4]
7

A particle P of mass 3 kg is projected with a speed of 8 ms−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 and the plane is 1123. Using an energy method throughout, find the speed of P at the instant it returns to A.

[6]

Mark scheme

9709/42 · May/June 2025

Download PDF

1a · 1 mark

  • B1

    1.5 m s−1

1b · 2 marks

  • M1

    Work done =25cos⁡20×12

  • A1

    282 J

1c · 1 mark

  • 1 mark

    35.2 W

2 · 4 marks

  • 1 mark

    0.2×5+0.1×2=(0.2+0.1)v

  • A1

    v=4

  • 1 mark

    Loss in KE =±[12×0.2×52+12×0.1×22−12×(0.2+0.1)×(their 4)2] [=±(2.5+0.2−2.4)]

  • A1

    [KE lost =] 0.3 J

3 · 4 marks

  • M1

    Resolving in any direction to form an equation

  • A1

    T×1213+35×513=mg [0.9230…T+35×0.3846⋯=mg] [0.9230…T+13.461⋯=mg] OR T=mg×1213[T=mg×0.9230… ]

  • A1

    T×513=35×1213 [T×0.3846⋯=35×0.9230… ] [T×0.3846⋯=32.3076… ] OR mg×513=35[mg×0.3846⋯=35]

  • A1

    T=84 and m=9.1

4a · 2 marks

  • B1

    Correct three-line segments for v-t graph

  • B1

    Values/expressions labelled on axes correctly

4b · 6 marks

  • B1

    Speed after 30 seconds =15+0.4×30[=27]

  • 1 mark

    Speed at B is their 27+(−0.2)T

  • 1 mark

    Attempt at distance from A to B and equate to 2750

  • A1

    12(15+27)×30+27×3T+12{(27−0.2T)+27}×T=2750

  • 1 mark

    0.1T2−108T+2120=0⇒T=…

  • A1

    T=20 only

4c · 3 marks

  • 1 mark

    Speed at B is (their 27)+((−0.2)×(their 20))[=23]

  • 1 mark

    Distance from B to C is s where 02=(their 23)2+2×(−0.5)×s and attempt to solve for s

  • A1

    Total distance is [529 + 2750 =] 3279 m

5a · 3 marks

  • M1

    5g−TBC=5×2

  • M1

    TBC−TAB=4×2

  • A1

    TBC=40 N and TAB=32 N

5b · 4 marks

  • 1 mark

    TAB−F−3gsin⁡30=3×2

  • B1

    R=3gcos⁡30

  • 1 mark

    their 32−μ×3gcos⁡30−3gsin⁡30=3×2 and attempt to solve for μ

  • A1

    μ=0.423

5c · 5 marks

  • B1

    Distance travelled by A in first 1.5 seconds is 12×2×1.52[=2.25]

  • B1

    When string breaks A is moving at a speed of 3 (m s−1)

  • 1 mark

    −F−3gsin⁡30=3a[a=−263]

  • 1 mark

    0=32+2×(their a)×s[s=2752]

  • A1

    Total distance travelled by A up the plane is [2.25+2752=]2.77 m

6a · 5 marks

  • 1 mark

    Attempt to differentiate (v=)(2t+1)32−2t2

  • A1

    (dvdt=)32(2t+1)32−1×2−2×2t2−1[=3(2t+1)12−4t]

  • 1 mark

    dvdt=0⇒3(2t+1)12−4t=0⇒9(2t+1)=16t2

  • 1 mark

    16t2−18t−9=0⇒t=…

  • A1

    Maximum velocity is 3.5 m s−1 ONLY

6b · 4 marks

  • 1 mark

    Attempt to integrate (v=)(2t+1)32−2t2

  • A1

    (s=)(2t+1)32+1(32+1)×2−2t2+12+1(+c)[=(2t+1)525−2t33(+c)]

  • 1 mark

    Substitute t=1.5 and t=0 OR evaluate c(≠0) using t=0 and substitute t=1.5

  • A1

    Distance [=(325−94)−(15−0)]=3.95 or 7920 m

7 · 6 marks

  • 1 mark

    Apply the work-energy principle for the motion from A to B to form an equation in one variable only

  • A1

    12×3×82−3gdsin⁡30 [correct LHS]

  • A1

    =312×3gcos⁡30×d (where d is the distance AB) [96−15d=3.75d] [correct RHS]

  • A1

    For either d=5.12 m or work done against friction is 19.2 J

  • 1 mark

    12×3×v2=12×3×82−2×(their 19.2)[1.5v2=96−38.4] OR 12×3×v2+(their 19.2)=3g×(their 5.12)×sin⁡30[1.5v2+19.2=76.8] OR 12×3×v2+(their 5.12)×312×3gcos⁡30=3g×(their 5.12)×sin⁡30[1.5v2+19.2=76.8]

  • A1

    v=6.20 m s−1 ONLY

Questions

  1. 1
    Question 14 marks
    Mark scheme ›
  2. 2
    Question 24 marks
    Mark scheme ›
  3. 3
    Question 34 marks
    Mark scheme ›
  4. 4
    Question 411 marks
    Mark scheme ›
  5. 5
    Question 512 marks
    Mark scheme ›
  6. 6
    Question 69 marks
    Mark scheme ›
  7. 7
    Question 76 marks
    Mark scheme ›

More 9709 papers

Other papers from May/June 2025

Solve it on ExamPalStart