AQA AS Level Physics Paper 2, June 2025: Question 4

📐 Step-by-Step Calculation

1. Identify given data:
   Vertical component: u_v = 8.7 m s⁻¹
   Launch angle: θ = 73° above horizontal
   Time of flight: t = 1.7 s
2. Find horizontal velocity component (u_h):
   tan(73°) = u_v / u_h
   u_h = 8.7 / tan(73°) = 2.660 m s⁻¹
3. Calculate horizontal distance (D):
   Since horizontal acceleration a_h = 0 :
   D = u_h × t = 2.660 × 1.7 = 4.522 m
   D = 4.5 m (to 2 sig figs)

✅ Mark Scheme Breakdown

• Mark 1 (Method): Attempt to determine the horizontal component of velocity ( 8.7 / tan 73° , 8.7 × tan 17° , or 2.66 m s⁻¹ ) OR multiplies their calculated horizontal speed by 1.7 s .
• Mark 2 (Accuracy): Final answer of 4.5 m (accept 4.52 m ).

🧠 Exam Technique

Always draw a small right-angled triangle representing the velocity components:

• Opposite side = u_v = 8.7 m s⁻¹
• Adjacent side = u_h
• Angle with ground = 73°
• Therefore, tan 73° = 8.7 / u_h .

❌ Common Errors

• Assuming 8.7 m s⁻¹ was the hypotenuse (resultant speed) and calculating u_h = 8.7 cos(73°) = 2.54 m s⁻¹ .
• Using an incorrect angle (e.g., using 73° to the vertical instead of the horizontal).

📐 Step-by-Step Calculation

1. Set sign convention (taking upwards as positive):
   u = +8.7 m s⁻¹
   a = -9.81 m s⁻² (or -9.8 m s⁻² )
   t = 1.7 s
2. Calculate net vertical displacement (s):
   s = ut + ½at²
   s = (8.7 × 1.7) + 0.5 × (-9.81) × (1.7)²
   s = 14.79 - 14.175 = +0.615 m
   (The ball ends up 0.615 m higher than its launch point)
3. Relate displacement to height h:
   Final height = 3.10 m
   s = 3.10 - h
   h = 3.10 - 0.615 = 2.485 m
   h = 2.5 m (to 2 sig figs)

✅ Mark Scheme Breakdown

Max 2 from first 3 marks:

• Mark 1: Use of s = ut + ½at² with t = 1.7 and u = 8.7 (allow g = 9.8 or 9.81 ).
• Mark 2: Consistent use of vector signs (e.g. positive u and negative a ).
• Mark 3: Correctly taking into account the 3.10 m basket height ( s = 3.10 - h ).
• Mark 4 (Final Answer): h = 2.5 m (full marks awarded for correct final answer with relevant working).

❌ Common Errors & Traps

• Sign Mismatch: Entering both u and a as positive numbers, resulting in s = 14.79 + 14.18 = 28.97 m .
• Height Confusion: Setting displacement s = h - 3.10 and adding 0.615 to 3.10 to get 3.72 m , forgetting that the launch point is below the basket.

🧠 Alternative Method (Peak Height)

You can also split the motion into two halves:

• Time to top: t₁ = u/g = 8.7 / 9.81 = 0.887 s
• Height gained to top: s₁ = u² / 2g = 3.858 m
• Remaining fall time: t₂ = 1.7 - 0.887 = 0.813 s
• Distance fallen: s₂ = ½ g t₂² = 3.243 m
• Net gain: 3.858 - 3.243 = 0.615 m → h = 3.10 - 0.615 = 2.5 m .

✅ Model Answer

1. Velocity link: A smaller launch angle (at the same initial speed) means a larger horizontal component of velocity (or smaller vertical component).

2. Deduction: Because the horizontal speed is greater across the same distance D , the time of flight is shorter ( t = D / u_h ). Therefore, there are fewer images taken.

🧠 Exam Technique & Mark Scheme Advice

• Mark 1: Mention that lower launch angle means larger horizontal component of velocity (or smaller vertical velocity).
• Mark 2: Conclude that time of flight is shorter, leading to fewer images.
• Examiner Note: Always state the cause-and-effect chain:
  θ decreases → u_h increases → time t decreases → fewer flashes/images .

💡 Key Physics Principles

• Air resistance (drag) always opposes the direction of motion.
• Magnitude of air resistance increases with speed.
• On the way up (s_v increasing): Motion is upward → vertical drag acts downward (same direction as gravity/weight).
• On the way down (s_v decreasing): Motion is downward → vertical drag acts upward (opposite direction to gravity/weight).

✅ Marking Points (Max 3 marks)

• Point A: Air resistance acts in the opposite direction to motion (or magnitude depends on speed).
• Point B (Displacement increasing): a_v is greater than g (greater than 9.81 m s⁻² ) because drag acts downwards, in the same direction as weight: F_net = mg + F_drag .
• Point C (Displacement decreasing): a_v is less than g (less than 9.81 m s⁻² ) because drag acts upwards, opposing weight: F_net = mg - F_drag .
• Point D: As speed decreases going up, a_v decreases towards g ; as speed increases coming down, drag increases so a_v decreases further.

❌ Common Misconceptions

• Saying "acceleration is always 9.81": Gravity is constant, but the question asks for net vertical acceleration a_v !
• Confusing speed with acceleration: Don't just say "the ball slows down". Explicitly state whether the magnitude of acceleration is greater than or less than g .
• Neglecting directions: Top answers clearly state: "Upwards motion: drag is down, so total downward force is mg + drag ."

🧠 Quick Comparison Summary