A-Level · Physics · AQA · Mark scheme decoded
AQA A-Level Physics: Motion in Horizontal and Vertical Directions in a Uniform Gravitational Field — mark scheme explained
The short answer
In this section, we will explore the independent effects of motion in horizontal and vertical directions within a uniform gravitational field. We will also delve into the factors that determine the motion of an object through a fluid, including qualitative treatments of friction, lift, drag forces, and terminal speed.
The question
A ball is thrown horizontally from a height of 20 m with an initial speed of 15 m/s. Calculate the time it takes to hit the ground and the horizontal distance traveled. [Paraphrased for study — not reproduced from any exam paper.]
Mark scheme, decoded
What each mark is really for — in plain English — and the wording trap that loses it.
- S1
Identify the vertical motion: y = v 0y × t + 0.5 × g × t 2
- S2
Since the ball is thrown horizontally, v 0y = 0 . The equation simplifies to: y = 0.5 × g × t 2
- S3
Substitute the values: 20 = 0.5 × 9.81 × t 2
- S4
Solve for t : t 2 = 4.077 → t ≈ 2.02 s
- S5
Identify the horizontal motion: x = v 0x × t
- S6
Substitute the values: x = 15 × 2.02 ≈ 30.3 m
Model answer
Worked through, with each step tagged to the mark it earns.
- S1
Identify the vertical motion: y = v 0y × t + 0.5 × g × t 2
- S2
Since the ball is thrown horizontally, v 0y = 0 . The equation simplifies to: y = 0.5 × g × t 2
- S3
Substitute the values: 20 = 0.5 × 9.81 × t 2
- S4
Solve for t : t 2 = 4.077 → t ≈ 2.02 s
- S5
Identify the horizontal motion: x = v 0x × t
- S6
Substitute the values: x = 15 × 2.02 ≈ 30.3 m
Final answer: Time to hit the ground: 2.02 s, Horizontal distance traveled: 30.3 m
Common mistakes
- Assuming horizontal and vertical motions are dependent on each other. — Treat horizontal and vertical motions separately. Use x = v 0x × t for horizontal motion and y = v 0y × t + 0.5 × g × t 2 for vertical motion.
- Forgetting to include the effect of air resistance in projectile motion problems. — Consider the effects of air resistance on the trajectory and range of a projectile. Air resistance reduces the range and makes the path less symmetrical.
- Using the wrong formula for terminal speed. — Use the correct equation: F drag = F gravity , which simplifies to 0.5 × C d × ρ × A × v 2 = m × g .
- Misinterpreting the relationship between drag force and speed. — Remember that F drag = 0.5 × C d × ρ × A × v 2 . As speed increases, the drag force increases quadratically.
- Confusing static and dynamic friction in problems involving motion through a fluid. — Understand that static friction prevents an object from starting to move, while dynamic friction opposes the motion of an object once it is moving. Dynamic friction is generally less than static friction.
- Failing to resolve initial velocity into horizontal and vertical components in projectile problems. — Always resolve the initial velocity into horizontal ( v 0x ) and vertical ( v 0y ) components using trigonometric functions: v 0x = v 0 × cos(θ) and v 0y = v 0 × sin(θ) .
Where the marks go
- Full worked solution (all marking points)6 marks