| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Newton's second law with vector forces (find acceleration or force) |
| Difficulty | Moderate -0.8 This is a straightforward AS-level mechanics question testing basic vector operations (distance formula), equilibrium of forces (simple addition), and constant acceleration kinematics. All parts are routine applications of standard formulas with no problem-solving insight required—easier than the typical A-level question which would involve more integration of concepts or multi-step reasoning. |
| Spec | 1.10f Distance between points: using position vectors3.02d Constant acceleration: SUVAT formulae3.03b Newton's first law: equilibrium3.03c Newton's second law: F=ma one dimension |
| Answer | Marks |
|---|---|
| 14(a) | Condone omission of units |
| Answer | Marks | Guidance |
|---|---|---|
| Ignore one sign error. | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Obtains correct distance. | 1.1b | A1 |
| Answer | Marks |
|---|---|
| 14(b) | Explains that remains at rest |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | 2.4 | E1 |
| Answer | Marks |
|---|---|
| (c)(i) | Calculates magnitude of given |
| Answer | Marks | Guidance |
|---|---|---|
| obtain a = 6.25i – 15j | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Completes calculation correctly AG | 2.1 | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| (c)(ii) | Uses appropriate suvat equation | |
| with a = 16.25 and their s used | 3.3 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| time to 2sf | 1.1b | A1 |
| Total | 7 | |
| Q | Marking Instructions | AO |
Question 14:
--- 14(a) ---
14(a) | Condone omission of units
throughout this question
Calculates the magnitude of AB
Ignore one sign error. | 1.1a | M1 | Distance =
2 2
�(13−3) +(−22−2)
= 26
Obtains correct distance. | 1.1b | A1
--- 14(b) ---
14(b) | Explains that remains at rest
implies resultant force = 0 and
shows F + F𝐴𝐴 + F = 0
1 2 3
or shows addition of F and F and
1 2
states that F is the opposite
3 | 2.4 | E1 | s(o− F 2𝐢𝐢 i+s t4h𝐣𝐣e) o+pp(o 6 s 𝐢𝐢 i − te 1 = 0 𝐣𝐣 -)4i = +6 4 j 𝐢𝐢 −6𝐣𝐣
3
--- 14
(c)(i) ---
14
(c)(i) | Calculates magnitude of given
force
Or uses Newton’s second law to
obtain a = 6.25i – 15j | 1.1a | M1 | 1|𝑭𝑭3 |== 01.83a
𝑎𝑎 = 16.25
Completes calculation correctly AG | 2.1 | R1
--- 14
(c)(ii) ---
14
(c)(ii) | Uses appropriate suvat equation
with a = 16.25 and their s used | 3.3 | M1 | 2
26 = 0.5×16.2 5×𝑡𝑡
2
𝑡𝑡t == 13.8.2
Solves to find the correct value of
time to 2sf | 1.1b | A1
Total | 7
Q | Marking Instructions | AO | Marks | Typical SolutionTwo particles, $A$ and $B$, lie at rest on a smooth horizontal plane.
$A$ has position vector $\mathbf{r}_A = (13\mathbf{i} - 22\mathbf{j})$ metres
$B$ has position vector $\mathbf{r}_B = (3\mathbf{i} + 2\mathbf{j})$ metres
\begin{enumerate}[label=(\alph*)]
\item Calculate the distance between $A$ and $B$.
[2 marks]
\item Three forces, $\mathbf{F}_1$, $\mathbf{F}_2$ and $\mathbf{F}_3$ are applied to particle $A$, where
$\mathbf{F}_1 = (-2\mathbf{i} + 4\mathbf{j})$ newtons
$\mathbf{F}_2 = (6\mathbf{i} - 10\mathbf{j})$ newtons
Given that $A$ remains at rest, explain why $\mathbf{F}_3 = (-4\mathbf{i} + 6\mathbf{j})$ newtons
[1 mark]
\item A force of $(5\mathbf{i} - 12\mathbf{j})$ newtons, is applied to $B$, so that $B$ moves, from rest, in a straight line towards $A$.
$B$ has a mass of $0.8 \text{kg}$
\begin{enumerate}[label=(\roman*)]
\item Show that the acceleration of $B$ towards $A$ is $16.25 \text{m s}^{-2}$
[2 marks]
\item Hence, find the time taken for $B$ to reach $A$.
Give your answer to two significant figures.
[2 marks]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2019 Q14 [7]}}