AQA A Level Physics 7408/2 Paper 2 Mark Scheme
Actual 2025 with Detailed Rationales | Complete
Exam-Style Questions | Pass Guaranteed – A+ Graded
SECTION 1: Mechanics & Materials (Questions 1–12)
Q1: A car accelerates uniformly from rest to 20 m/s in 5.0 s. What is the distance
travelled during this time?
A. 20 m
B. 50 m [CORRECT]
C. 100 m
D. 200 m
Correct Answer: B
Rationale: Correct because using s = (u + v)t/2 with u = 0, v = 20 m/s, t = 5.0 s gives s
= (0 + 20) × 5. = 50 m; alternatively s = ut + ½at² where a = 4.0 m/s².
Q2: A 5.0 kg object is acted upon by a constant resultant force of 15 N. What is the
acceleration of the object?
A. 0.33 m/s²
B. 3.0 m/s² [CORRECT]
C. 20 m/s²
D. 75 m/s²
Correct Answer: B
Rationale: Correct because applying Newton's second law F = ma gives a = F/m = 15
N / 5.0 kg = 3.0 m/s².
,Q3: A ball of mass 0.40 kg is thrown vertically upwards with an initial velocity of 12
m/s. What is the maximum height reached? (Take g = 9.81 m/s²)
A. 7.3 m [CORRECT]
B. 14.6 m
C. 1.2 m
D. 24.5 m
Correct Answer: A
Rationale: Correct because using v² = u² + 2as with v = 0, u = 12 m/s, a = -9.81 m/s²
gives s = (0 - 144) / (-19.62) = 7.34 m ≈ 7.3 m.
Q4: A spring of original length 0.20 m extends to 0.25 m when a 4.0 N load is
applied. What is the spring constant?
A. 16 N/m
B. 80 N/m [CORRECT]
C. 20 N/m
D. 0.05 N/m
Correct Answer: B
Rationale: Correct because using Hooke's law F = kx with extension x = 0.25 - 0.20 =
0.05 m gives k = F/x = 4.0 N / 0.05 m = 80 N/m.
Q5: A wire of diameter 1.2 mm and original length 2.0 m extends by 3.0 mm when a
tensile force of 80 N is applied. What is the Young modulus of the material?
A. 4.7 × 10⁷ Pa
B. 4.7 × 10¹⁰ Pa [CORRECT]
C. 1.2 × 10¹¹ Pa
D. 2.4 × 10⁸ Pa
Correct Answer: B
Rationale: Correct because E = (F/A)/(e/L) = FL/Ae; cross-sectional area A = π(0.6 ×
10⁻³)² = 1.13 × 10⁻⁶ m², giving E = (80 × 2.0)/(1.13 × 10⁻⁶ × 3.0 × 10⁻³) = 4.7 × 10¹⁰ Pa.
, Q6: A 2.0 kg mass moving at 3.0 m/s collides with a stationary 1.0 kg mass. After
the collision, the 2.0 kg mass moves at 1.5 m/s in the same direction. What is the
velocity of the 1.0 kg mass?
A. 1.5 m/s
B. 3.0 m/s [CORRECT]
C. 4.5 m/s
D. 6.0 m/s
Correct Answer: B
Rationale: Correct because applying conservation of momentum: (2.0 × 3.0) + (1.0 ×
0) = (2.0 × 1.5) + (1.0 × v), giving 6.0 = 3.0 + v, therefore v = 3.0 m/s.
Q7: A car of mass 1200 kg travelling at 15 m/s brakes to a stop. How much work is
done by the braking force?
A. 9.0 × 10³ J
B. 1.35 × 10⁵ J [CORRECT]
C. 1.8 × 10⁴ J
D. 9.0 × 10⁴ J
Correct Answer: B
Rationale: Correct because work done equals change in kinetic energy: W = ½mv² =
0.5 × 1200 × (15)² = 0.5 × 1200 × 225 = 1.35 × 10⁵ J.
Q8: A satellite orbits Earth at a constant speed in a circular orbit. Which statement
correctly describes the forces acting on the satellite?
A. No forces act on the satellite
B. The resultant force is zero
C. A centripetal force directed toward the centre of Earth provides the acceleration
for circular motion [CORRECT]
D. The satellite moves at constant velocity
Correct Answer: C
Actual 2025 with Detailed Rationales | Complete
Exam-Style Questions | Pass Guaranteed – A+ Graded
SECTION 1: Mechanics & Materials (Questions 1–12)
Q1: A car accelerates uniformly from rest to 20 m/s in 5.0 s. What is the distance
travelled during this time?
A. 20 m
B. 50 m [CORRECT]
C. 100 m
D. 200 m
Correct Answer: B
Rationale: Correct because using s = (u + v)t/2 with u = 0, v = 20 m/s, t = 5.0 s gives s
= (0 + 20) × 5. = 50 m; alternatively s = ut + ½at² where a = 4.0 m/s².
Q2: A 5.0 kg object is acted upon by a constant resultant force of 15 N. What is the
acceleration of the object?
A. 0.33 m/s²
B. 3.0 m/s² [CORRECT]
C. 20 m/s²
D. 75 m/s²
Correct Answer: B
Rationale: Correct because applying Newton's second law F = ma gives a = F/m = 15
N / 5.0 kg = 3.0 m/s².
,Q3: A ball of mass 0.40 kg is thrown vertically upwards with an initial velocity of 12
m/s. What is the maximum height reached? (Take g = 9.81 m/s²)
A. 7.3 m [CORRECT]
B. 14.6 m
C. 1.2 m
D. 24.5 m
Correct Answer: A
Rationale: Correct because using v² = u² + 2as with v = 0, u = 12 m/s, a = -9.81 m/s²
gives s = (0 - 144) / (-19.62) = 7.34 m ≈ 7.3 m.
Q4: A spring of original length 0.20 m extends to 0.25 m when a 4.0 N load is
applied. What is the spring constant?
A. 16 N/m
B. 80 N/m [CORRECT]
C. 20 N/m
D. 0.05 N/m
Correct Answer: B
Rationale: Correct because using Hooke's law F = kx with extension x = 0.25 - 0.20 =
0.05 m gives k = F/x = 4.0 N / 0.05 m = 80 N/m.
Q5: A wire of diameter 1.2 mm and original length 2.0 m extends by 3.0 mm when a
tensile force of 80 N is applied. What is the Young modulus of the material?
A. 4.7 × 10⁷ Pa
B. 4.7 × 10¹⁰ Pa [CORRECT]
C. 1.2 × 10¹¹ Pa
D. 2.4 × 10⁸ Pa
Correct Answer: B
Rationale: Correct because E = (F/A)/(e/L) = FL/Ae; cross-sectional area A = π(0.6 ×
10⁻³)² = 1.13 × 10⁻⁶ m², giving E = (80 × 2.0)/(1.13 × 10⁻⁶ × 3.0 × 10⁻³) = 4.7 × 10¹⁰ Pa.
, Q6: A 2.0 kg mass moving at 3.0 m/s collides with a stationary 1.0 kg mass. After
the collision, the 2.0 kg mass moves at 1.5 m/s in the same direction. What is the
velocity of the 1.0 kg mass?
A. 1.5 m/s
B. 3.0 m/s [CORRECT]
C. 4.5 m/s
D. 6.0 m/s
Correct Answer: B
Rationale: Correct because applying conservation of momentum: (2.0 × 3.0) + (1.0 ×
0) = (2.0 × 1.5) + (1.0 × v), giving 6.0 = 3.0 + v, therefore v = 3.0 m/s.
Q7: A car of mass 1200 kg travelling at 15 m/s brakes to a stop. How much work is
done by the braking force?
A. 9.0 × 10³ J
B. 1.35 × 10⁵ J [CORRECT]
C. 1.8 × 10⁴ J
D. 9.0 × 10⁴ J
Correct Answer: B
Rationale: Correct because work done equals change in kinetic energy: W = ½mv² =
0.5 × 1200 × (15)² = 0.5 × 1200 × 225 = 1.35 × 10⁵ J.
Q8: A satellite orbits Earth at a constant speed in a circular orbit. Which statement
correctly describes the forces acting on the satellite?
A. No forces act on the satellite
B. The resultant force is zero
C. A centripetal force directed toward the centre of Earth provides the acceleration
for circular motion [CORRECT]
D. The satellite moves at constant velocity
Correct Answer: C
