AQA A Level Physics (7408/2) Paper
2Multiple-Choice Assessment: 2026/2027
Specification
Section A: Further Mechanics
Q1: A car travels at a constant speed of 15 m s⁻¹ around a banked curve of radius 50 m. Assuming
no friction is required to prevent skidding, what is the angle of banking of the track?
A. 17°
B. 24°
C. 25° [CORRECT]
D. 65°
Correct Answer: C
Rationale:
For a banked curve with no friction, the horizontal component of the normal force provides the
centripetal force. Resolving forces: Nsinθ=
r
mv
,2
and Ncos θ= mg . Dividing these gives tanθ=
rg
v
2
=
50×9.81
15
2
=
490.5
225
,≈0.459 , so θ=tan
−1
(0.459)≈24.7°≈25° .
Key Principle: Resolution of forces and circular motion dynamics (F=
r
mv
2
).
Distractor Analysis:
● A: Results from sinθ=
● rg
● v
● 2
●
● (incorrect trigonometric ratio).
● B: Arises from tanθ=
● rg
● v
, ●
● (forgetting to square the velocity).
● D: Comes from inverting the relationship: tanθ=
● v
● 2
● rg
●
● .
Q2: A particle executes simple harmonic motion with amplitude 0.10 m and frequency 2.0 Hz. What
is the maximum acceleration of the particle?
A. 0.80 m s⁻²
B. 4.0 m s⁻²
C. 16 m s⁻² [CORRECT]
D. 39 m s⁻²
Correct Answer: C
Rationale:
Maximum acceleration in SHM is given by a
max
2Multiple-Choice Assessment: 2026/2027
Specification
Section A: Further Mechanics
Q1: A car travels at a constant speed of 15 m s⁻¹ around a banked curve of radius 50 m. Assuming
no friction is required to prevent skidding, what is the angle of banking of the track?
A. 17°
B. 24°
C. 25° [CORRECT]
D. 65°
Correct Answer: C
Rationale:
For a banked curve with no friction, the horizontal component of the normal force provides the
centripetal force. Resolving forces: Nsinθ=
r
mv
,2
and Ncos θ= mg . Dividing these gives tanθ=
rg
v
2
=
50×9.81
15
2
=
490.5
225
,≈0.459 , so θ=tan
−1
(0.459)≈24.7°≈25° .
Key Principle: Resolution of forces and circular motion dynamics (F=
r
mv
2
).
Distractor Analysis:
● A: Results from sinθ=
● rg
● v
● 2
●
● (incorrect trigonometric ratio).
● B: Arises from tanθ=
● rg
● v
, ●
● (forgetting to square the velocity).
● D: Comes from inverting the relationship: tanθ=
● v
● 2
● rg
●
● .
Q2: A particle executes simple harmonic motion with amplitude 0.10 m and frequency 2.0 Hz. What
is the maximum acceleration of the particle?
A. 0.80 m s⁻²
B. 4.0 m s⁻²
C. 16 m s⁻² [CORRECT]
D. 39 m s⁻²
Correct Answer: C
Rationale:
Maximum acceleration in SHM is given by a
max
