A Level Mathematics 9709 Paper 42, May/June 2026

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Question paper

9709/42 · May/June 2026

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1

A car of mass 960 kg is moving on a straight horizontal road. There is a constant force of magnitude 620 N resisting the motion of the car.

a

Calculate the power developed by the engine of the car when it is moving at a constant speed of 25 ms−1.

[2]
b

Given that the power is suddenly increased by 12 kW, find the instantaneous acceleration of the car.

[2]
2

Two particles, A and B, of masses 2.5 kg and 3.5 kg respectively are at rest on a straight smooth horizontal track. Particle B is situated 9 m from a vertical wall which is fixed at right angles to the track. Particle A is projected directly towards B with a speed of 3 ms−1. In the subsequent motion, A collides and coalesces with B to form particle C. Particle C then collides directly with the wall and rebounds. The collision of C with the wall reduces the speed of C by 25%.

a

Find the speed of C after it rebounds from the wall.

[2]
b

Hence find the time from the instant at which A and B collide until C is once again a distance of 9 m from the wall.

[2]
3

Two particles, A and B, of masses 0.5 kg and 0.8 kg respectively are attached to the ends of a light inextensible string. The string passes over a small smooth pulley fixed at the end of a rough horizontal plane and to the top of a smooth inclined plane. Particle A is held on the horizontal plane, while B lies on the inclined plane, which makes an angle of 55∘ with the horizontal. The string is in the same vertical plane as a line of greatest slope of the inclined plane (see diagram). Particle A is released from rest with the string taut. Particle B moves 0.3 m down the inclined plane in 0.4 s.

Figure 1 of 1 for question 3
a

Find the tension in the string.

[4]
b

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

[3]
c

After B has been moving for 0.4 s, the string suddenly breaks. Given that A subsequently comes to rest on the horizontal plane, find the work done by the frictional force in bringing A to rest. You may assume that A does not reach the pulley.

[3]
4

A particle P is projected from a point A with speed 4 ms−1 up a line of greatest slope of a rough plane inclined at an angle θ to the horizontal, where tan⁡θ=43. P comes to instantaneous rest at a point B on the plane (see diagram). The coefficient of friction between P and the plane is 13.

Figure 1 of 1 for question 4
a

Using an energy method, show that the distance AB is 0.8 m.

[4]
b

After coming to instantaneous rest at B, the particle slides down the plane. Find the total time from the instant at which P is projected from A until it returns to A.

[5]
5

The points A and B are at the same vertical height h m above horizontal ground. A particle P is released from rest from A. One second later, a particle Q is projected vertically downwards from B with speed 18 ms−1. Given that P and Q reach the ground at the same time, find the value of h.

[6]
6

A crate of weight W N is at rest on rough horizontal ground. The crate is pulled at a constant acceleration of 1.5 ms−2 along the ground in a straight line by a light rope. The rope is inclined at an angle of θ to the horizontal, where θ=sin⁡−1513. The tension in the rope is 26 N (see diagram). The coefficient of friction between the crate and the ground is 0.25.

Figure 1 of 1 for question 6
a

Find the value of W.

[5]
b

When the crate reaches a point A, the rope is removed. The speed of the crate at A is 8 ms−1. The crate comes to rest at a point B. Find the distance AB.

[3]
7

A particle P moves in a straight line. The velocity v ms−1 of P at time t s, where t⩾0, is given by v=k1(4t+11)12−14(2t+1)2+k2, where k1 and k2 are constants. When t=1.25 the deceleration of P is 0.5 ms−2.

a

Find the value of k1.

[4]
b

Given that P is only at instantaneous rest when t=3.5, find the distance travelled by P in the interval 0⩽t⩽1.

[5]

Mark scheme

9709/42 · May/June 2026

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1a · 2 marks

  • M1

    P25−620=0

  • A1

    15 500 W or 15.5 kW

1b · 2 marks

  • M1

    (their 15500)+1200025−620=960a or 1200025=960a [1100−620=960a] [480=960a]

  • 1 mark

    0.5 ms−2

2a · 2 marks

  • M1

    2.5×3=(2.5+3.5)v

  • A1

    [Speed of C after impact with the wall is] 0.9375 or 1516 m s−1

2b · 2 marks

  • M1

    Times are 9their 1.25 and 9their 0.9375 [=7.2 and 9.6]

  • A1

    16.8 s

3a · 4 marks

  • B1

    0.3=0+12a×0.42 [⇒a=3.75]

  • M1

    Attempt at N2L for B OR Attempt at N2L for the system AND for A

  • 1 mark

    0.8gsin⁡55−T=0.8×(their 3.75) OR 0.8gsin⁡55−FA=(0.8+0.5)×(their 3.75) AND T−FA=0.5×(their 3.75)

  • A1

    3.55 N

3b · 3 marks

  • M1

    (their 3.55)−FA=0.5×(their 3.75) OR 0.8gsin⁡55−FA=(0.8+0.5)×(their 3.75)

  • 1 mark

    (their 3.55)−μ×0.5g=0.5×(their 3.75) [3.55−5μ=1.875] OR 0.8gsin⁡55−μ×0.5g=(0.8+0.5)×(their 3.75) [6.5532...−5μ=4.875]

  • A1

    μ=0.336 or 0.335

3c · 3 marks

  • 1 mark

    Speed when string breaks is v=(their 3.75)×0.4 [v=1.5] OR 0.3=12(0+v)×0.4 OR v2=02+2×0.3×(their 3.75)

  • 1 mark

    Work done by frictional force is ±12×0.5×(their 1.5)2

  • A1

    0.5625 J or 916 J

4a · 4 marks

  • B1

    R=mg×35 [=6m]

  • M1

    F=13×35mg [=2m]

  • 1 mark

    0−12×m×42=−mg×d×45−13×35mg×d [−8m=−8md−2md]

  • A1

    d=[8m8m+2m]=0.8

4b · 5 marks

  • B1

    Time up the plane is 0.4

  • M1

    (m)a=(m)g×45−(m)g×13×35

  • A1

    a=6

  • 1 mark

    0.8=0+12×(their 6)×t2

  • A1

    0.916 s

5 · 6 marks

  • B1

    For P: [h=]±(12gtP2)

  • B1

    For Q: [h=]±(18tQ+12gtQ2)

  • 1 mark

    12g(1+tQ)2=18tQ+12gtQ2 OR 12gtP2=18(tP−1)+12g(tP−1)2

  • 1 mark

    12g(1+2tQ+tQ2)=18tQ+12gtQ2⇒8tQ=5⇒tQ=... OR 12gtP2=18(tP−1)+12g(tP2−2tP+1)⇒8tP=13⇒tP=...

  • A1

    tQ=0.625 or 58 or tP=1.625 or 138

  • A1

    h=13.2

6a · 5 marks

  • M1

    Attempt to apply N2L either horizontally or vertically to form an equation

  • A1

    26×1213−F=Wg×1.5 [24−F=3W20]

  • A1

    26×513+R=W [10+R=W]

  • 1 mark

    26×1213−0.25×(W−26×513)=Wg×1.5 OR 26×1213−0.25×(mg−26×513)=m×1.5 OR 26×1213−0.25R=10+Rg×1.5

  • A1

    W=66.25 or 2654

6b · 3 marks

  • B1

    (Wg)a=−0.25×W OR a=−0.25g

  • 1 mark

    0=82+2×−2.5×AB 0=82+2×(−0.25×g)×AB

  • A1

    12.8

7a · 4 marks

  • M1

    v=k1(4t+11)12−14(2t+1)2+k2⇒[dvdt]=...

  • A1

    [a=]12k1(4t+11)12−1×4−2×14×(2t+1)2−1×2 [=2k1(4t+11)−12−(2t+1)]

  • 1 mark

    2k1(4×1.25+11)−12−(2×1.25+1)=−0.5

  • A1

    k1=6

7b · 5 marks

  • M1

    (their 6)×(4×3.5+11)12−14(2×3.5+1)2+k2=0

  • M1

    Attempt to integrate v=[k1(4t+11)12−14(2t+1)2+k2]

  • 1 mark

    [s=]k1(4t+11)1+124×(1+12)−(2t+1)2+14×2×(2+1)+k2t (+c)=[(4t+11)32−(2t+1)324−14t (+c)]

  • 1 mark

    [s=][(4t+11)32−(2t+1)324−14t]01=[1532−3324−14]−[1132−124]

  • A1

    [42.96975...−36.4412...=] 6.53 m

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