AQA AS Level Mathematics Paper 1, June 2025: Question 18

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Q Marking instructions AO Marks Typical solution

18 Selects an appropriate equation

of constant acceleration u = 12, a = –10 v = 0

and substitutes u = ±12, a = ±10 3.1b M1

v = 0 Using v2 = u2 + 2as

Condone use of g = 9.8 or 9.81

Substitutes consistent correct 02 = 122 + 2(–10)s

values to obtain s or h 1.1b A1

Condone h + 1.8 for s s = 7.2

Obtains 9 3.2a B1 h = 7.2 + 1.8 = 9

Question 18 Total 3

📋 What This Question Tests

Equations of Constant Acceleration (SUVAT): Selecting and correctly manipulating the equation linking initial velocity, final velocity, acceleration, and displacement ( v² = u² + 2as ).
Physical Conditions at Turning Points: Recognising that at the greatest height reached, instantaneous vertical velocity is zero ( v = 0 ).
Modelling with Gravity & Specified Constants: Correctly adopting a sign convention for upward and downward vectors, and strictly following the given value of g = 10 m s⁻² .
Reference Frames / Initial Position: Accounting for displacement from the point of release versus the total height above ground level.

Establish a sign convention:
Take vertically upwards as the positive direction ( ↑ +ve ).
Identify the variables for the upward journey:
• Initial velocity, u = +12 m s⁻¹
• Velocity at maximum height, v = 0 m s⁻¹
• Acceleration, a = -g = -10 m s⁻²
• Vertical displacement from launch, s
Apply the appropriate SUVAT equation:
v² = u² + 2as
0² = 12² + 2(-10)s
0 = 144 - 20s
20s = 144
s = 7.2 m
Calculate total height above the ground ( h ):
The ball was projected from an initial height of 1.8 m .
h = 1.8 + s
h = 1.8 + 7.2 = 9 m

Value of h: h = 9 (or 9 m )

Mark Scheme Breakdown:

[M1] (AO 3.1b): Selects v² = u² + 2as and substitutes u = ±12 , a = ±10 , v = 0 . (Condones g = 9.8 or 9.81 ).
[A1] (AO 1.1b): Substitutes consistent correct signs to obtain displacement s = 7.2 (or equates directly with s = h - 1.8 ).
[B1] (AO 3.2a): Obtains final answer 9 by adding the launch height of 1.8 m .

Apex Condition: Whenever an object is projected vertically under gravity, it stops momentarily at its highest point; hence v = 0 m s⁻¹ .
Vector Consistency: Direction matters! If upwards is positive, then upward velocity is positive ( +12 ) and downward acceleration due to gravity is negative ( -10 ).
Displacement vs Height: SUVAT gives the displacement s relative to the point of projection, not automatically the height above the ground.

Read the Header Instruction: The paper explicitly states: "In this question use g = 10 m s⁻²". Always scan the question stem carefully so you do not default automatically to 9.8 .
Annotate a Quick Sketch: Draw a ground line, mark 1.8 m to the ball's release position, draw an upward arrow with s , and label the total height h . This eliminates omission errors.

The "Premature Stop": Writing h = 7.2 as the final answer, completely forgetting to add the release height ( 1.8 m ) from the ground.
Sign Mismatch: Writing 0 = 12² + 2(10)s leading to s = -7.2 or an invalid mathematical step (e.g. ignoring the minus sign).
Defaulting to Standard g: Calculating with g = 9.8 m s⁻² gives s ≈ 7.35 m and h ≈ 9.15 m . Although M1 is condoned, students forfeit the final accuracy mark if they fail to follow the explicit instruction g = 10 .

Mechanics · Q: Kinematics