A-Level · Mathematics · AQA · Mark scheme decoded
AQA A-Level Mathematics: Addition of Forces and Resultant Forces in Dynamics — mark scheme explained
The short answer
In AQA A-Level Mathematics, understanding the addition of forces and resultant forces is crucial for solving dynamics problems involving motion in a plane. This topic covers how to combine multiple forces acting on an object to determine the overall effect they have on its motion.
The question
A block of mass 2 kg is on a frictionless surface. Two forces act on it: one of 6 N at 30° to the horizontal and another of 8 N at 120° to the horizontal. Find the resultant force and the acceleration of the block. [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
Resolve each force into its horizontal and vertical components:
- S2
F x1 = 6 cos 30° ≈ 5.2 N, F y1 = 6 sin 30° = 3 N.
- S3
F x2 = 8 cos 120° = -4 N, F y2 = 8 sin 120° ≈ 6.93 N.
- S4
Add the horizontal and vertical components:
- S5
F x = 5.2 - 4 = 1.2 N, F y = 3 + 6.93 ≈ 9.93 N.
- S6
Find the magnitude of the resultant force using Pythagoras' theorem:
- S7
R = √(1.2 2 + 9.93 2 ) ≈ 9.98 N.
- S8
Find the direction of the resultant force using the tangent function:
- S9
tan θ = F y / F x ≈ 9.93 / 1.2 ≈ 8.275, θ ≈ tan -1 (8.275) ≈ 83.2° to the horizontal.
- S10
Apply Newton's second law to find the acceleration:
- S11
a = F / m = 9.98 N / 2 kg ≈ 4.99 m/s 2 .
Model answer
Worked through, with each step tagged to the mark it earns.
- S1
Resolve each force into its horizontal and vertical components:
- S2
F x1 = 6 cos 30° ≈ 5.2 N, F y1 = 6 sin 30° = 3 N.
- S3
F x2 = 8 cos 120° = -4 N, F y2 = 8 sin 120° ≈ 6.93 N.
- S4
Add the horizontal and vertical components:
- S5
F x = 5.2 - 4 = 1.2 N, F y = 3 + 6.93 ≈ 9.93 N.
- S6
Find the magnitude of the resultant force using Pythagoras' theorem:
- S7
R = √(1.2 2 + 9.93 2 ) ≈ 9.98 N.
- S8
Find the direction of the resultant force using the tangent function:
- S9
tan θ = F y / F x ≈ 9.93 / 1.2 ≈ 8.275, θ ≈ tan -1 (8.275) ≈ 83.2° to the horizontal.
- S10
Apply Newton's second law to find the acceleration:
- S11
a = F / m = 9.98 N / 2 kg ≈ 4.99 m/s 2 .
Final answer: Resultant force: 9.98 N at 83.2° to the horizontal; Acceleration: 4.99 m/s 2
Common mistakes
- Forgetting to resolve forces into components. — Always resolve forces into their horizontal and vertical components before adding them.
- Incorrectly using trigonometric functions for resolving forces. — Remember that the horizontal component is given by F cos θ and the vertical component by F sin θ .
- Forgetting to apply Newton's second law separately for each direction. — Apply F = ma separately for the horizontal and vertical directions.
- Incorrectly calculating the magnitude of the resultant force. — Use R = √( F x 2 + F y 2 ) to find the magnitude of the resultant force accurately.
- Incorrectly calculating the direction of the resultant force. — Use tan θ = F y / F x and ensure you interpret the angle correctly based on the quadrant.
- Forgetting to consider all forces acting on an object. — Always list and consider all forces acting on the object before performing any calculations.
- Incorrectly applying vector addition methods. — Practice and understand both methods thoroughly. Ensure you draw vectors accurately and follow the steps correctly.
- Forgetting to convert angles from degrees to radians when using trigonometric functions. — Ensure you are using the correct mode (degrees or radians) on your calculator and convert as necessary.
Where the marks go
- Full worked solution (all marking points)6 marks