| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: resultant and acceleration |
| Difficulty | Moderate -0.3 Part (a) is a routine distance calculation using Pythagoras. Part (b) requires applying SUVAT to find acceleration, then F=ma, but involves straightforward substitution with unit conversion (grams to kg). This is standard AS mechanics with no conceptual challenges, slightly easier than average due to the guided structure and basic calculations. |
| Spec | 1.10f Distance between points: using position vectors3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension |
| Answer | Marks |
|---|---|
| 18(a) | Subtracts given vectors with at |
| Answer | Marks | Guidance |
|---|---|---|
| diagram | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| component differences | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Condone missing units | 1.1b | A1 |
| Subtotal | 3 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 18(b) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| u, t and s | 3.3 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains a = 2 | 1.1b | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Final value for R must be > 0 | 3.4 | B1F |
| Subtotal | 3 | |
| Question 18 Total | 6 | |
| Q | Marking instructions | AO |
Question 18:
--- 18(a) ---
18(a) | Subtracts given vectors with at
least one component correct for
AB or BA
or
Finds difference between i and j
components with at least one
correct, may be seen on a
diagram | 1.1a | M1 | 13 5 8
− =
5 −1 6
8
AB =
6
Distance = 8 2 +6 2
Distance = 10 m
Uses Pythagoras for their i and j
component differences | 1.1a | M1
Shows distance between A and
B is 10 metres AG
Condone missing units | 1.1b | A1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 18(b) ---
18(b) | 1
Selects s =ut+ at2
2
and substitutes given values for
u, t and s | 3.3 | M1 | 1
s =ut+ at2
2
10=6+2a
a = 2
R=0.15×2=0.3
Obtains a = 2 | 1.1b | A1
Obtains value for R using 0.15 ×
their value for a
Final value for R must be > 0 | 3.4 | B1F
Subtotal | 3
Question 18 Total | 6
Q | Marking instructions | AO | Marks | Typical solutionIt is given that two points $A$ and $B$ have position vectors
$$\overrightarrow{OA} = \begin{bmatrix} 5 \\ -1 \end{bmatrix} \text{ metres} \quad \text{and} \quad \overrightarrow{OB} = \begin{bmatrix} 13 \\ 5 \end{bmatrix} \text{ metres.}$$
\begin{enumerate}[label=(\alph*)]
\item Show that the distance from $A$ to $B$ is 10 metres.
[3 marks]
\item A constant resultant force, of magnitude $R$ newtons, acts on a particle so that it moves in a straight line passing through the same two points $A$ and $B$
At $A$, the speed of the particle is 3 m s$^{-1}$ in the direction from $A$ to $B$
The particle takes 2 seconds to travel from $A$ to $B$
The mass of the particle is 150 grams.
Find the value of $R$
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2024 Q18 [6]}}