AQA A-Level Mathematics Paper 2, June 2025: Question 14

Mark scheme

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Q Marking instructions AO Marks Typical solution

14 Substitutes

u = 25, a = 9 8 and. s = 2 5.

into 2 2 v2 = 252 + 2 − 9 8. −2 5.

v = u + 2as

k k v2 = 674

PI or

674 674 Speed = 402 + 674

Or 3.3 M1 −1

Speed = 47 7. m s

Substitutes

u = 25, a = −9 8 and. s = −2 5. Speed = 48 m s−1

12 t

into s = ut + at to find

and

substitutes their t into v = u + at

Obtains correct v2 = 674 or

v = 674

If inconsistent signs are used for

u, a and s score A0 1.1b A1

PI AWRT ± 25.96

If vector form is used, both

components must be correct

Finds the magnitude of the 1.1a M1

resultant vector of their vertical

component of velocity and 40

Do not allow 25 for their vertical

component of velocity

Obtains the correct speed of the

arrow.

Accept AWRT 48 1.1b A1

If inconsistent signs are used for

u, a and s score A0

Question 14 Total 4

How to answer it

Projectile Motion: Finding Final Impact Speed

A-Level Mathematics • Mechanics • 4 Marks

What this question tests

This question evaluates your ability to model 2D projectile motion under constant gravity using vector components:

Resolving velocity into independent horizontal ( x ) and vertical ( y ) components.
Applying vertical constant acceleration formulae ( suvat ) with consistent coordinate sign conventions.
Understanding that horizontal velocity remains constant in the absence of air resistance.
Calculating overall speed as the scalar magnitude of the resultant velocity vector using Pythagoras' theorem.

Question 14 (Full Solution & Strategy)

An arrow projected from (0, 2.5) with initial velocity [40, 25]ᵀ m s⁻¹ landing at ground level Q

📐 Step-by-Step Calculation

Step 1: Extract horizontal and vertical components from initial velocity
The arrow is launched with initial velocity vector [40, 25]ᵀ m s⁻¹ :

Horizontal component: u_x = 40 m s⁻¹ (constant throughout motion, as a_x = 0 ).
Vertical component: u_y = +25 m s⁻¹ (taking upwards as positive).

Step 2: Set up vertical motion parameters to ground level Q
Taking upwards as positive:

Initial vertical velocity: u_y = +25 m s⁻¹
Vertical acceleration: a_y = -9.8 m s⁻² (since gravity acts downwards)
Vertical displacement: s_y = -2.5 m (ground level is 2.5 m below release height)

Step 3: Calculate the vertical velocity component at landing ( v_y )
Use the constant acceleration formula relating v, u, a, s :

v_y² = u_y² + 2 a_y s_y
v_y² = (25)² + 2(-9.8)(-2.5)
v_y² = 625 + 49 = 674
v_y = -√674 ≈ -25.9615 m s⁻¹ (downwards)

Marks awarded: [M1] for correct substitution into v² = u² + 2as (or finding t first); [A1] for correctly finding v_y² = 674 or |v_y| = √674 (AWRT ±25.96).

Step 4: Calculate the final speed (magnitude of velocity vector)
At point Q, the velocity components are:
v_x = 40 m s⁻¹ and v_y² = 674 (or v_y ≈ -25.96 m s⁻¹ )

Speed = √(v_x² + v_y²)
Speed = √(40² + 674) = √(1600 + 674) = √2274 ≈ 47.686 m s⁻¹

Marks awarded: [M1] for applying Pythagoras with 40 and their vertical component; [A1] for correct final speed.

✅ Final Answer

Speed at Q:

47.7 m s⁻¹ (3 s.f.)

or

48 m s⁻¹ (2 s.f., appropriate when using g = 9.8 )

Exact value: √2274 m s⁻¹

Accept any value rounding to 48 (AWRT 47.7 or 48).

💡 Key Knowledge

Velocity vs Speed: Velocity is a vector [v_x, v_y]ᵀ ; speed is its scalar magnitude √(v_x² + v_y²) .
Independence of Motion: Horizontal velocity is unchanged ( a_x = 0 ). All acceleration occurs vertically ( a_y = -g ).
Conservation of Energy Alternative:
½mv₁² + mgh = ½mv₂²
v₂ = √(v₁² + 2gh) = √(40² + 25² + 2×9.8×2.5) = √2274
This is an equally valid, ultra-fast method!

🧠 Exam Technique & Insight

Avoid quadratic equations when possible: Using v² = u² + 2as directly eliminates the need to calculate time t using the quadratic formula, saving time and preventing arithmetic slips.
Keep exact values: Store v_y² = 674 directly into Pythagoras: √(40² + 674) . This avoids rounding errors before taking the final square root.
Degree of Accuracy: Since g = 9.8 m s⁻² is given to 2 significant figures, giving your final answer to 2 s.f. (48) or 3 s.f. (47.7) is standard practice in AQA Mechanics.

❌ Common Traps & Pitfalls

Inconsistent Signs: Mixing up signs for u , a , and s . If upwards is positive, s = -2.5 and a = -9.8 . Using s = +2.5 with a = -9.8 gives 625 - 49 = 576 (which implies the arrow landed higher than the release point!). This scores A0.
Confusing Initial Speed with Vertical Velocity: Using the initial vertical component 25 instead of calculating the final vertical component before finding speed.
Stopping at Vertical Velocity: Answering with 26 m s⁻¹ , forgetting that the question asks for the total speed, which includes the horizontal component 40 m s⁻¹ .

Topics

Mechanics · Q: Kinematics