AQA A-Level Physics Paper 1, June 2025: Question 6

8 marks · Medium difficulty · Short Answer

Calculate the kinetic energy transferred during a football kick, determine the force scale on a force–time graph from impulse, and find the flight time of a projectile kicked at an angle.

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Question

Question text

06 A person kicks a stationary football horizontally.

During the kick, the person gives an impulse of 6.55 N s to the ball during a contact

time of 12.0 ms.

The mass of the football is 410 g.

Assume that friction forces are negligible.

06.1 Calculate the kinetic energy transferred to the ball during the kick.

[2 marks]

kinetic energy = J

06.2 Figure 8 shows how the force F acting on the ball varies with time t.

Figure 8

Determine the scale for the F / N axis and add your scale to13 Figure 8.

[3 marks]

06.3 A player jumps up to head a football when it is 2.0 m above the ground.

The initial velocity of the ball is 18 m s−1 at an angle of 48° to the horizontal.

Figure 9 shows the path of the ball above horizontal ground.

Figure 9

Determine the time that the ball takes to reach the ground from the instant it leaves

the player’s head.

Assume that the frictional forces acting on the ball are negligible.

[3 marks]

time = s

Mark scheme

Show the mark scheme

Question Answers Additional comments/Guidance Mark AO

06.1 Max 1 from: For 1 expect to see 2 1 × AO1

Calculation of Δv Ft = mΔv and u = 0 1 × AO2

Calculation of E using their Δv 1 6.55

v = 0.410 = 15.9756 m s−1

Condone POT error in 1

Ek = pv = × 6.55 × 15.9756

52(.3) J 2 2 2

= 52.320 (J)

Do not allow rounding error.

If no other mark given, award max 1 for a

combination of ke and momentum equations

1 (mv) 2 seen in numbers or symbols

Ek =

2 m

06.2 Area under graph = 8.5 large squares Allow 8 to 9 or (small ) 200 to 225 3 1 × AO2

OR Accept attempt to calculate area using y-axis 2 × AO3

interval as unknown quantity

212 small squares 1

18 Area = 6.55 N s on its own is insufficent for

uses area of graph = impulse 2

Allow evaluation of impulse ÷ their number of

squares

One square = F × 2 × 10−3 =

6.55

their number of squares

And F = their area of one square ÷ 2 × 10−3 s

judges that the interval is 400 N

AND

annotates y-axis with 400 N intervals 3

3 : minimum 3, non-zero, major grid lines

labelled

06.3 Any two from: 1 2 18sin48 seen or inferred from use of 3 1 × AO1

−1 (−)13.37(7) 2 × AO2

• resolve 18 m s vertically (to obtain u)

12 use of v = u + at with v = 0 and s = ut + at

• use of SUVAT with their u eg s = (their u)× t + at 2

with u = 0

• evidence of consistent sign for s, u and a

−2.0 = (18sin48)t + (−9.8)t

2.0 = (−18sin48)t + (9.8)t

Do not award MP3 if negative solution

t = 2.87 (s) 3 quoted.

Allow 2.86 or 2.9

Allow any correct alternative route

Total 8

How to answer it

Impulse, Force-Time Graphs & Projectile Motion

📋 What this question tests

This question assesses core concepts across linear mechanics, graphical analysis, and 2D kinematics:

Impulse and Momentum: Relating impulse (Δp = FΔt) to velocity change and deriving kinetic energy (E_k = ½mv² or E_k = p² / 2m).
Graphical Calculus: Understanding that the area under a force–time (F–t) graph represents impulse, and using square-counting to scale axes.
2D Projectile Motion: Resolving initial velocity vectors into vertical components and applying constant acceleration equations (SUVAT) with consistent directional signs.

Question 06.1 • 2 Marks

Kinetic Energy Transferred During Kick

Impulse, momentum, and work done on a stationary object

📐 Step-by-Step Calculation

1. Identify given quantities:

Impulse, Δp = 6.55 N s
Mass, m = 410 g = 0.410 kg
Initial velocity, u = 0 m s⁻¹

2. Calculate final velocity (v):
Impulse = Δp = m(v - u) = mv
v = 6.55 / 0.410 = 15.9756 m s⁻¹

3. Calculate Kinetic Energy (E_k):
E_k = ½ m v²
E_k = 0.5 × 0.410 × (15.9756)² = 52.32 J
Alternatively: E_k = p² / (2m) = (6.55)² / (2 × 0.410) = 52.32 J

✅ Correct Answer & Awarding

Final Answer: 52 J (or 52.3 J )

Mark Scheme Breakdown:
• Mark 1: Calculation of Δv ( 15.98 m s⁻¹ ) OR valid combination of E_k and momentum equations ( E_k = ½pv or p² / 2m ).
• Mark 2: Correct final value of 52 J or 52.3 J .

🧠 Exam Technique: Shortcut Formula

Instead of calculating velocity first, use the direct relationship between momentum and kinetic energy: E_k = p² / 2m . It is faster and eliminates rounding errors intermediate steps.

❌ Common Errors

Unit Conversion Trap: Forgetting to convert 410 g to 0.410 kg .
Redundant Data Trap: Using contact time ( 12.0 ms ) unnecessarily to find average force. While valid, it adds needless steps and risks compounding rounding errors.

Question 06.2 • 3 Marks

Determining the Force Axis Scale

Interpreting the area under an F–t curve

💡 Key Knowledge

Area under F–t curve = Impulse (Δp) = 6.55 N s.
Grid dimensions: The base spans from 0 to 12 ms (divided into 6 large squares, each representing 2 ms = 2 × 10⁻³ s horizontally).
Total area under the curve is estimated by square counting.

📐 Step-by-Step Calibration

1. Count the area under the curve:
Number of large squares ≈ 8.5 (Acceptable range: 8 to 9 large squares, or 200 to 225 small squares).

2. Determine the value of one large square:
Area per large square = 6.55 N s / 8.5 squares ≈ 0.7706 N s

3. Calculate the vertical interval per large square (F_grid):
Each large square has width Δt = 2.0 ms = 2.0 × 10⁻³ s .
F_grid × (2.0 × 10⁻³ s) = 0.7706 N s
F_grid = 0.7706 / (2.0 × 10⁻³) ≈ 385 N ≈ 400 N

4. Annotate the y-axis:
Major grid line intervals should be marked at increments of 400 N : 400, 800, 1200, 1600.

✅ Correct Annotations & Marking Points

Mark 1: Determining area under graph = 8.5 large squares (or 8 to 9 large squares / 200 to 225 small squares).
Mark 2: Equating graph area to the total impulse ( 6.55 N s ) to find value of 1 square.
Mark 3: Concluding that each major grid interval is 400 N , and annotating at least 3 non-zero major grid lines ( 400, 800, 1200 ).

❌ Common Errors & Examiner Commentary

Assuming Triangle Approximation: Approximating the peak as a simple triangle (½ × base × height) yields F_max ≈ 1090 N , which incorrectly gives ~360 N per line and fails to count actual grid squares.
Unit Error on Time: Forgetting that the horizontal axis is in milliseconds ( ms ), leading to an answer out by a factor of 1000.
Incomplete Axis: Labelling only one or two numbers; the mark scheme requires a minimum of 3 non-zero major grid lines labelled.

Question 06.3 • 3 Marks

Time of Flight for Projectile Motion

Resolving velocity vectors and vertical SUVAT with displaced landing point

📐 Step-by-Step Calculation

1. Resolve initial vertical velocity (u_y):
Taking upwards as positive (+):
u_y = 18 × sin(48°) = +13.376 m s⁻¹

2. Define vertical SUVAT parameters:
Because the ball lands on the ground 2.0 m below the release point:
s = -2.0 m
u = +13.376 m s⁻¹
a = -9.81 m s⁻² (or -9.8 m s⁻² )
t = ?

3. Set up and solve the quadratic equation:
s = ut + ½at²
-2.0 = 13.376t - 4.905t²
4.905t² - 13.376t - 2.0 = 0

Applying the quadratic formula: t = [-b ± √(b² - 4ac)] / 2a
t = [13.376 ± √(178.92 + 39.24)] / 9.81
t = [13.376 + 14.77] / 9.81 = 2.869 s

✅ Correct Answer & Alternative Routes

Final Answer: t = 2.87 s (accept 2.86 s to 2.9 s )

Mark Breakdown:
• Marks 1 & 2 (Any two from):
1. Resolve vertically: 18 sin 48° ≈ 13.38 m s⁻¹
2. Valid SUVAT equation for time with their u_y .
3. Evidence of consistent signs for s , u , and a .
• Mark 3: Final time 2.87 s (negative root must be discarded).

Alternative Two-Stage Route:

1. Time to peak: t₁ = u_y / g = 13.38 / 9.81 = 1.364 s
2. Peak height above release: h = u_y² / 2g = 9.12 m
3. Total fall height: H = 9.12 + 2.0 = 11.12 m
4. Fall time: t₂ = √(2H / g) = √(22.24 / 9.81) = 1.506 s
5. Total time: t = t₁ + t₂ = 1.364 + 1.506 = 2.87 s

❌ Major Pitfalls & Lost Marks

Sign Inconsistency: Writing 2.0 = 13.38t - 4.9t² (treating downward displacement as positive while velocity is upward). This produces no real solutions or completely incorrect values.
Using Cosine Instead of Sine: Using 18 cos 48° resolves horizontally instead of vertically. Always check: vertical is opposite the angle to the horizontal, so use sin.
Ignoring the 2.0 m Launch Height: Calculating time for a projectile returning to its original launch height ( s = 0 , which gives t = 2.73 s ). The ball travels all the way down to the ground.

🧠 Exam Tip: Choosing Your Approach

Using a single quadratic equation ( s = ut + ½at² ) is much faster than splitting the flight into upwards and downwards stages. Ensure you have your calculator set up to solve polynomials quickly to check your working during the exam!

Topics

Physics · Practical skills · 3.4 Mechanics and materials · Data analysis

Question and mark scheme from the AQA A-Level Physics examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.