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
How the examiner actually awards the marks on this topic.
Gradora's own decode of the marking approach — not the exam board's published mark scheme.
How marks are awarded
For calculation questions, show all steps clearly and include units in your final answer. For explanation questions, provide concise and accurate descriptions using key physics terms.
What the command words demand
- Calculate
- Perform a numerical calculation to find a specific value.
- Determine
- Find the value of a quantity or variable using given data and equations.
- Explain
- Provide a clear and detailed account of why something happens or how it works.
- Describe
- Give a detailed account of the characteristics or features of something.
- Compare
- Identify and explain similarities and differences between two or more concepts.
Model answer
A full-mark response to the question above, worked through step by step.
Timing: Allocate approximately 10 minutes for each question to ensure you have enough time to read the question carefully, plan your response, and check your work.
- Identify the vertical motion: y = v 0y × t + 0.5 × g × t 21 mark
- Since the ball is thrown horizontally, v 0y = 0 . The equation simplifies to: y = 0.5 × g × t 20 marks
- Substitute the values: 20 = 0.5 × 9.81 × t 21 mark
- Solve for t : t 2 = 4.077 → t ≈ 2.02 s1 mark
- Identify the horizontal motion: x = v 0x × t1 mark
- Substitute the values: x = 15 × 2.02 ≈ 30.3 m2 marks
Final answer: Time to hit the ground: 2.02 s, Horizontal distance traveled: 30.3 m
Work through every step correctly and you earn all 6 marks.
Another worked example
A car is traveling at a speed of 30 m/s and experiences a drag force of 1500 N. Calculate the power required to overcome this drag force.
- Use the formula for power: P = F × v1 mark
- Substitute the values: P = 1500 × 30 ≈ 45,000 W2 marks
Final answer: Power required: 45,000 W
Work through every step correctly and you earn all 3 marks.
Common mistakes
Assuming horizontal and vertical motions are dependent on each other.
Why it happens: Students often mix up the independent nature of horizontal and vertical components, leading to incorrect calculations.
Fix: 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.
Why it happens: Students may overlook the impact of air resistance, especially when it is not explicitly mentioned in the problem.
Fix: 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.
Why it happens: Students may confuse the equation for terminal speed with other equations involving drag force.
Fix: 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.
Why it happens: Students may not understand that drag force increases quadratically with speed, leading to incorrect calculations of power or terminal speed.
Fix: 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.
Why it happens: Students may not distinguish between static and dynamic friction, especially when dealing with objects moving through fluids.
Fix: 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.
Why it happens: Students may not break down the initial velocity correctly, leading to incorrect calculations of range or maximum height.
Fix: 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
The question types you’ll meet on this topic and the marks each one carries — so you know what to expect and where to focus.
| Question type | What you’re asked to do | Marks |
|---|---|---|
| Projectile Motion | Find the fall time and horizontal range of a horizontally launched projectile. | 6 |
| Calculate Power | Use the power equation with force and velocity to find the power overcoming drag. | 3 |
| Calculate Terminal Speed | Equate drag and gravitational forces to find the skydiver's terminal velocity with units. | 5 |
| Projectile Motion | Resolve initial velocity and calculate the projectile's maximum height and horizontal range. | 8 |
| Calculate Power | Find the power needed to overcome a drag force at a given speed. | 3 |
| Total across these question types | 25 | |
Question types and mark tariffs are Gradora’s guidance based on how this topic is typically examined — not the board’s official paper structure.