| Exam Board | AQA |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2021 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Particle motion: 2D constant acceleration |
| Difficulty | Easy -1.2 This is a straightforward mechanics question requiring only basic vector magnitude calculation, constant acceleration formula (v=u+at), and Newton's second law (F=ma). All three parts are direct applications of standard formulas with no problem-solving insight needed, making it easier than average for A-level. |
| Spec | 1.10c Magnitude and direction: of vectors3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension |
| Answer | Marks | Guidance |
|---|---|---|
| 12(a) | Uses Pythagoras to show given | |
| magnitude AG | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 12(b) | Uses appropriate suvat equation |
| Answer | Marks | Guidance |
|---|---|---|
| units | 1.1b | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Subtotal | 1 | |
| Q | Marking instructions | AO |
| Answer | Marks |
|---|---|
| 12(c) | Uses F = ma, using magnitude |
| Answer | Marks | Guidance |
|---|---|---|
| PI by correct answer | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| missing units | 1.1b | A1 |
| Subtotal | 2 | |
| Question Total | 4 | |
| Q | Marking instructions | AO |
Question 12:
--- 12(a) ---
12(a) | Uses Pythagoras to show given
magnitude AG | 1.1b | B1 | 2 2
�8 +6 = 10
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 12(b) ---
12(b) | Uses appropriate suvat equation
with given values to find correct
speed value condone missing
units | 1.1b | B1 | ,
so
𝑢𝑢 = 0 𝑎𝑎 = 10, 𝑡𝑡 = m3 s–1
𝑣𝑣 = 0+(3)(10)= 30
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 12(c) ---
12(c) | Uses F = ma, using magnitude
of force with magnitude of
acceleration or in vector form
PI by correct answer | 3.4 | M1 | 2 2
�22.5+ =1 .5 ×= 102 .5
0𝑚𝑚.25 kg
𝑚𝑚 =
Finds correct mass condone
missing units | 1.1b | A1
Subtotal | 2
Question Total | 4
Q | Marking instructions | AO | Marks | Typical solutionA particle P lies at rest on a smooth horizontal table.
A constant resultant force, F newtons, is then applied to P.
As a result P moves in a straight line with constant acceleration $\begin{bmatrix}8\\6\end{bmatrix}$ m s⁻²
\begin{enumerate}[label=(\alph*)]
\item Show that the magnitude of the acceleration of P is 10 m s⁻²
[1 mark]
\item Find the speed of P after 3 seconds.
[1 mark]
\item Given that $\mathbf{F} = \begin{bmatrix}2\\1.5\end{bmatrix}$ N, find the mass of P.
[2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA AS Paper 1 2021 Q12 [4]}}