A Level Mathematics 9709 Paper 41, May/June 2026

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9709/41 · May/June 2026

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Coplanar forces of magnitudes $55 N$ , $23 N$ , $20 N$ and $16 N$ act at a point in the directions shown in the diagram. Find the magnitude and direction of the resultant force. [6]

[6]

A car of mass $1200 kg$ is moving on a straight horizontal road. The resistance to the motion of the car has a constant magnitude of $600 N$ and the car's engine is working at a rate of $27 kW$ .

Find the acceleration of the car when its speed is $18 {ms}^{- 1}$ . [3]

[3]

The car now moves up a straight hill which is inclined at an angle $θ$ to the horizontal. The resistance to the motion of the car has now changed to a constant magnitude of $650 N$ . The car moves up the hill at a constant speed of $12 {ms}^{- 1}$ with the car's engine working at a rate of $22 kW$ . Find the value of $θ$ . [3]

[3]

A particle $P$ travels in a straight line, starting from a point $O$ . At time $t$ seconds the displacement of $P$ from $O$ is $s$ metres, where $s = 4 t^{3} - 39 t^{2} + 120 t$ .

Find the values of $t$ for which $P$ is instantaneously at rest. [3]

[3]

Find the total distance travelled by $P$ in the interval $1 ⩽ t ⩽ 4$ . [2]

[2]

Find the magnitude of the acceleration of $P$ when $t = 3$ . [3]

[3]

Three particles $A$ , $B$ and $C$ have masses $3 kg$ , $1.5 kg$ and $m kg$ respectively, where $m > 1.5$ . Particles $A$ and $B$ are attached to the ends of a light inextensible string that passes over a smooth fixed pulley. Particle $C$ is attached to particle $B$ . The system is held at rest with the string taut and the hanging parts of the string vertical (see diagram). The system is released from rest and while the three particles are all attached as described, the tension in the string is $37.5 N$ .

Find the value of $m$ . [4]

[4]

At the instant when $A$ has moved a distance $0.8 m$ upwards from its initial position, $C$ separates from $B$ and falls away. In the subsequent motion, $A$ and $B$ continue to move and $A$ does not reach the pulley. Find the total time from when the system is released until $A$ first comes to instantaneous rest. [6]

[6]

A particle of mass $3.6 kg$ is placed on a rough plane inclined at an angle of $30^{\circ}$ to the horizontal. The particle is held at rest on the plane by a horizontal force of magnitude $24 \sqrt{3} N$ . The force acts in a vertical plane containing a line of greatest slope of the inclined plane (see diagram). The particle is on the point of moving up the plane.

Show that the coefficient of friction between the particle and the plane is $\frac{1}{5} \sqrt{3}$ . [5]

[5]

The force of magnitude $24 \sqrt{3} N$ is now removed, and the particle moves from rest down the plane. Find the speed of the particle after it has travelled $1.5 m$ down the plane. [5]

[5]

Three small smooth spheres $A$ , $B$ and $C$ of equal radii and masses $m kg$ , $2 kg$ and $4 kg$ respectively, lie at rest in that order in a straight line on a smooth horizontal plane. Sphere $A$ is projected directly towards $B$ with speed $2 m {ms}^{- 1}$ and $B$ is projected directly towards $A$ with speed $(m + 2) {ms}^{- 1}$ . After the spheres collide, both $A$ and $B$ move in the same direction as $A$ 's original motion with speeds $2 {ms}^{- 1}$ and $3.25 {ms}^{- 1}$ respectively.

Find the value of $m$ . [4]

[4]

After the collision between $A$ and $B$ , sphere $B$ moves directly towards $C$ . Sphere $B$ now collides with $C$ . After the spheres collide, both $B$ and $C$ move in the same direction and the loss of kinetic energy due to the collision between $B$ and $C$ is $6.875 J$ . Determine if there is a further collision between $A$ and $B$ . [6]

[6]

Mark scheme

9709/41 · May/June 2026

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1 · 6 marks

M1
Resolving either direction
A1
$[Y =] \pm (55 \cos 60 - 23 \sin 15 - 20 \sin 40) [= \pm 8.69140 \dots]$
A1
$[X =] \pm (55 \sin 60 + 23 \cos 15 - 20 \cos 40 - 16) [= \pm 38.5268 \dots]$
1 mark
Attempt to find $θ = \tan^{- 1} (\frac{8.69140 \dots}{38.5268 \dots})$
1 mark
Attempt to find $\sqrt{(38.5268 \dots)^{2} + (8.69140 \dots)^{2}}$
A1
Magnitude 39.5 N and a direction of ${12.7}^{\circ}$ above the positive x-axis Accept equivalent answers for direction e.g. 77.3 to the upward vertical clockwise Or give the bearing as (0)77.3 Or ${12.7}^{\circ}$ above the horizontal to the right. Or ${12.7}^{\circ}$ anticlockwise from the positive x-axis

2a · 3 marks

B1
$D = \frac{27000}{18} [= 1500]$
M1
$\frac{27000}{18} - 600 = 1200 a$
A1
$0.75 {m s}^{- 2}$

2b · 3 marks

B1
[Component of weight parallel to plane =] $1200 g \sin θ$
M1
$\frac{22000}{12} - 650 - 1200 g \sin θ = 0$
A1
$θ = {5.7}^{\circ}$

3a · 3 marks

M1
$s = 4 t^{3} - 39 t^{2} + 120 t \Rightarrow [v =] 12 t^{2} - 78 t + 120$
1 mark
$12 t^{2} - 78 t + 120 = 0$
A1
$t = 2.5$ and $t = 4$

3b · 2 marks

B1
$s (1) = 85, s (2.5) = 118.75, s (4) = 112$
1 mark
Total distance travelled is 40.5 m

3c · 3 marks

M1
$[a =] 24 t - 78$
1 mark
$24 \times 3 - 78 = \dots$
A1
[Magnitude of acceleration is] $6 {m s}^{- 2}$

4a · 4 marks

M1
Attempt at Newton’s 2nd Law
A1
$37.5 - 3 g = 3 a$ $(m + 1.5) g - 37.5 = (m + 1.5) a$ $(m + 1.5) g - 3 g = (m + 1.5 + 3) a$
1 mark
$a = 2.5 \Rightarrow (m + 1.5) g - 37.5 = (m + 1.5) \times 2.5$ OR $\frac{m + 1.5}{3} = \frac{(m + 1.5) g - 37.5}{37.5 - 3 g}$
A1
$m = 3.5$

4b · 6 marks

B1
$v^{2} = 0^{2} + 2 \times 2.5 \times 0.8$
B1
$2 = 0 + 2.5 t_{1}$ or $0.8 = \frac{1}{2} \times 2.5 \times t_{1}^{2}$
M1
Attempt at Newton’s 2nd Law
A1
$T_{1} - 3 g = 3 a$ and $1.5 g - T_{1} = 1.5 a$ or $1.5 g - 3 g = (3 + 1.5) a$
1 mark
$0 = 2 + (- \frac{10}{3}) t_{2}$
A1
Total time $[= 0.6 + 0.8] = 1.4 s$

5a · 5 marks

M1
Resolving perpendicular or parallel to the plane
A1
$R = 3.6 g \cos 30 + 24 \sqrt{3} \sin 30$
A1
$F + 3.6 g \sin 30 = 24 \sqrt{3} \cos 30$
1 mark
$24 \sqrt{3} \cos 30 - 3.6 g \sin 30 = μ (3.6 g \cos 30 + 24 \sqrt{3} \sin 30)$
A1
$24 \sqrt{3} \times \frac{\sqrt{3}}{2} - 3.6 g \times \frac{1}{2} = μ (3.6 g \times \frac{\sqrt{3}}{2} + 24 \sqrt{3} \times \frac{1}{2})$ $\Rightarrow 36 - 18 = μ (18 \sqrt{3} + 12 \sqrt{3})$ so $μ = \frac{18}{30 \sqrt{3}} = \frac{1}{5} \sqrt{3}$

5b · 5 marks

B1
EITHER: $R = 3.6 g \cos 30 [= 18 \sqrt{3}]$ OR: $R = 3.6 g \cos 30 [= 18 \sqrt{3}]$
M1
EITHER: $3.6 g \sin 30 - F_{1} = 3.6 a$ OR: $\pm \frac{\sqrt{3}}{5} (3.6 g \cos 30) (1.5)$ and $\pm 3.6 g \sin 30 (1.5)$
1 mark
EITHER: $3.6 g \sin 30 - \frac{\sqrt{3}}{5} (3.6 g \cos 30) = 3.6 a \Rightarrow a = \dots$ OR: $3.6 g \sin 30 (1.5) - \frac{\sqrt{3}}{5} (3.6 g \cos 30) (1.5) = \frac{1}{2} (3.6) v^{2}$
1 mark
$v^{2} = 0^{2} + 2 \times 2 \times 1.5$
A1
$2.45 {m s}^{- 1}$

6a · 4 marks

M1
Attempt at CLM for the collision between $A$ and $B$
A1
$m \times 2 m - (m + 2) \times 2 = m \times 2 + 2 \times 3.25$
1 mark
$2 m^{2} - 4 m - 10.5 = 0 \Rightarrow m = \dots$
A1
$m = 3.5$ only

6b · 6 marks

M1
$2 \times 3.25 = 2 v_{B} + 4 v_{C} [\Rightarrow v_{B} + 2 v_{C} = 3.25]$
M1
Attempt at change in KE $= \pm 6.875$
A1
$\frac{1}{2} \times 2 \times {3.25}^{2} - \frac{1}{2} \times 2 \times v_{B}^{2} - \frac{1}{2} \times 4 \times v_{C}^{2} = 6.875$
1 mark
$(3.25 - 2 v_{C})^{2} + 2 v_{C}^{2} = 3.6875$ OR $v_{B}^{2} + 2 {(\frac{3.25 - v_{B}}{2})}^{2} = 3.6875$
A1
$6 v_{C}^{2} - 13 v_{C} + 6.875 = 0$ OR $3 v_{B}^{2} - 6.5 v_{B} + 3.1875 = 0$ $\Rightarrow v_{C} = \frac{5}{4}, \frac{11}{12}$ or $v_{B} = \frac{3}{4}, \frac{17}{12}$
A1
$v_{B} = \frac{17}{12}, v_{C} = \frac{11}{12}$ is not possible. So $v_{B} = \frac{3}{4} [0.75], v_{C} = \frac{5}{4}$ and $2 > 0.75$ so yes there is a further collision between $A$ and $B$

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