| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Speed or velocity at specific time |
| Difficulty | Standard +0.3 This is a straightforward mechanics question requiring differentiation of position vectors to find velocity and acceleration, then computing magnitudes. Part (a) is routine calculus with vector magnitude calculation. Part (b) requires finding when |a| = 0 by solving a simple equation, which is standard A-level technique. The question involves multiple steps but no novel insight or challenging problem-solvingβslightly easier than average due to its procedural nature. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07q Product and quotient rules: differentiation1.10d Vector operations: addition and scalar multiplication3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks |
|---|---|
| 14(a) | Differentiates to find expression for |
| Answer | Marks | Guidance |
|---|---|---|
| π―π― | 3.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| ππππππππππ = | π―π― | = οΏ½8 +4 |
| Answer | Marks | Guidance |
|---|---|---|
| has at least one component correct. | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| are non-zero π―π― | 1.1a | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| CSO | 1.1b | A1 |
| Subtotal | 4 | = 4β5 |
| Answer | Marks |
|---|---|
| 14(b) | Differentiates their vector to find |
| Answer | Marks | Guidance |
|---|---|---|
| Do not award foππr a = r | 3.4 | M1 |
| Answer | Marks |
|---|---|
| ππ | = οΏ½(6π‘π‘β 10) +2 |
| Answer | Marks |
|---|---|
| ππ | β₯ne2ver zero π‘π‘ |
| Answer | Marks | Guidance |
|---|---|---|
| always -2 | 2.4 | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Deduction mu | sππt | bβ₯e 2fr oππmππ completely |
| correct reaso | ππn | in>g.0 |
| Subtotal | 3 | |
| Question Total | 7 | |
| Q | Marking Instructions | AO |
Question 14:
--- 14(a) ---
14(a) | Differentiates to find expression for
with at least one component correct
π―π― | 3.4 | M1 | 2
π―π― = οΏ½3π‘π‘ β10π‘π‘οΏ½π’π’+(8β2π‘π‘)π£π£
So when
π‘π‘ = 2
π―π― = β8π’π’+4π£π£
2 2
ππππππππππ = |π―π―|= οΏ½8 +4
m s-1
Substitutes t = 2 into their which
has at least one component correct. | 1.1a | M1
Uses correct process to finπ―π―d the
magnitude of their expression for
provided both of their components
are non-zero π―π― | 1.1a | M1
Obtains correct answer
ISW
ACF 4β5
CSO | 1.1b | A1
Subtotal | 4 | = 4β5
--- 14(b) ---
14(b) | Differentiates their vector to find
expression for with at least one
component correct. ππ
Do not award foππr a = r | 3.4 | M1 | ππ = (6π‘π‘β 10)π’π’β2π£π£
2 2
|ππ|= οΏ½(6π‘π‘β 10) +2
Since for any value of
the magnitude of the acceleration is
|ππ|β₯ne2ver zero π‘π‘
Therefore Bellaβs claim is correct.
Uses a valid explanation as to why
the magnitude must be positive for
all values of t
This could be earned by:
forming a quadratic from acceleration
components, allow sign error
or
Stating that the j component is
always -2 | 2.4 | E1
Deduces that or that Bella
is correct or .
Deduction mu |sππt |bβ₯e 2fr oππmππ completely
correct reaso|ππn|in>g.0 | 2.2a | R1
Subtotal | 3
Question Total | 7
Q | Marking Instructions | AO | Marks | Typical SolutionAt time $t$ seconds a particle, $P$, has position vector $\mathbf{r}$ metres, with respect to a fixed origin, such that
$$\mathbf{r} = (t^3 - 5t^2)\mathbf{i} + (8t - t^2)\mathbf{j}$$
\begin{enumerate}[label=(\alph*)]
\item Find the exact speed of $P$ when $t = 2$
[4 marks]
\item Bella claims that the magnitude of acceleration of $P$ will never be zero.
Determine whether Bella's claim is correct.
Fully justify your answer.
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2020 Q14 [7]}}