AQA AS Level Physics Paper 1, November 2020: Question 3

✅ Correct Answer

v = 0.35 m s⁻¹ (or more than 2 s.f. if correct)

💡 Key Knowledge

• Variables given: initial velocity u = 1.8 m s⁻¹ , acceleration a = -1.2 m s⁻² (negative because it's deceleration), displacement s = 1.3 m .
• Equation to use: v² = u² + 2as

📐 Step-by-Step Calculation

1. State formula: v² = u² + 2as
2. Substitute values: v² = (1.8)² + 2(-1.2)(1.3)
3. Calculate terms: v² = 3.24 - 3.12 = 0.12
4. Square root: v = √0.12 = 0.3464... = 0.35 m s⁻¹

❌ Common Errors

• Forgetting to make acceleration negative ( a = -1.2 ), which leads to a larger final velocity instead of deceleration.

✅ Correct Answer

uᵥ ≈ 6 m s⁻¹ (exact calculation yields 6.2 m s⁻¹ or 6.16 m s⁻¹ )

💡 Key Knowledge

• Trigonometric resolution of vectors: adjacent component is u cos(35) , opposite vertical component is u sin(35) .
• Using ratios: tan(35) = uᵥ / uₕ where uₕ = 8.8 m s⁻¹ .

📐 Step-by-Step Calculation

1. Find total velocity u first using horizontal component: 8.8 = u cos(35) ⇒ u = 8.8 / cos(35) = 10.74 m s⁻¹
2. Calculate vertical component: uᵥ = u sin(35) = 10.74 × sin(35) = 6.16 m s⁻¹
3. Alternative direct method: uᵥ = 8.8 × tan(35) = 6.16 m s⁻¹ (rounds to 6 m s⁻¹ or 6.2 m s⁻¹ ).

✅ Correct Answer

t = 0.63 s (allows ECF from 03.2)

🧠 Exam Technique

Horizontal velocity remains constant ( 8.8 m s⁻¹ ) since weight is the only force acting. Use horizontal displacement or vertical motion parameters depending on available data. Here, vertical motion suvat is most direct if horizontal distance to X isn't given, but wait—let's look at how standard projectile problems link horizontal distance or if time is derived from vertical apex/landing. Note: Examiners accept ECF if consistent with student's previous values.

📐 Step-by-Step Calculation

1. At the highest point X, the vertical velocity v = 0 m s⁻¹ .
2. Use vertical suvat: v = u + at ⇒ 0 = 6.16 - 9.81t
3. Rearrange for time: t = 6.16 / 9.81 = 0.628 s ≈ 0.63 s (using g = 9.8 m s⁻¹ gives 0.63 s ).

✅ Correct Answer

h = 1.9 m (or 1.83 m depending on unrounded intermediate values)

💡 Key Knowledge

Use vertical suvat where final velocity at the peak v = 0 , initial vertical velocity u = 6.16 m s⁻¹ , and acceleration a = -9.81 m s⁻² .

📐 Step-by-Step Calculation

1. Select equation: v² = u² + 2as
2. Substitute values: 0 = (6.16)² + 2(-9.81)h
3. Rearrange for h : h = 37.95 / 19.62 = 1.93 m
4. Format to appropriate sig figs: 1.9 m

✅ Correct Answer

A smooth, parabolic curve starting at the ball's initial position and terminating precisely at point X, featuring a clear maximum turning point (apex).

🧠 Exam Technique

Since the horizontal velocity is smaller while the target destination X remains fixed in space, the ball must take a steeper path to reach the same height and position over a longer time or steeper arc. Ensure your sketch is smooth, clearly curved upwards with a peak, and touches point X within 5 mm.

❌ Common Errors

Drawing straight lines instead of a smooth parabola, or failing to show a clear maximum turning point.

✅ Correct Answer

The golfer must increase the angle of projection to the horizontal. This increases the initial vertical velocity component (making the ball go higher and stay in the air longer), which allows it to cover the same horizontal distance despite having a smaller horizontal velocity.

💡 Key Knowledge

• Horizontal motion equation: Distance = horizontal velocity × time ( sₓ = uₓ t ).
• To keep sₓ constant while decreasing uₓ , time t in the air must increase.
• A larger projection angle achieves this by shifting vector proportions toward a greater vertical component.