| Exam Board | OCR MEI |
|---|---|
| Module | Further Mechanics Major (Further Mechanics Major) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Find velocity from position |
| Difficulty | Standard +0.3 This is a straightforward Further Maths mechanics question requiring differentiation of vector functions and basic vector operations. Part (a) is routine differentiation, part (b) requires setting up a ratio condition for parallel vectors and solving a quadratic, and part (c) involves second differentiation and solving an equation. All techniques are standard with no novel insight required, making it slightly easier than average even for Further Maths. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | v = ( 4 t − 1 2 ) i + ( 3 t 2 + 6 t − 8 ) j |
| Answer | Marks | Guidance |
|---|---|---|
| (b) | 1(3t2 +6t−8)=−4(4t−12) | M1* |
| Answer | Marks | Guidance |
|---|---|---|
| 3 t 2 + 2 2 t − 5 6 = 0 t = ... | M1dep* | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| t = 2 | A1 | 1.1 |
| v = ( 4 t − 1 2 ) 2 + ( 3 t 2 + 6 t − 8 ) 2 with their positive value of t | M1 | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| v = 1 6 .5 (m s-1) | A1 | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| (c) | a = 4 i + ( 6 t + 6 ) j | B1ft |
| 4 2 + ( 6 t + 6 ) 2 = 2 0 .2 2 | M1 | 1.1 |
| t = 2 .3 | A1 | 1.1 |
Question 6:
6 | (a) | v = ( 4 t − 1 2 ) i + ( 3 t 2 + 6 t − 8 ) j | B1 | 1.1 | Condone lack of brackets and accept column vector
form
[1]
(b) | 1(3t2 +6t−8)=−4(4t−12) | M1* | 3.1b | Setting up a quadratic equation in t only – allow sign
errors (including on the 1 and the 4) and the 1 and the
–4 on the wrong side (or multiples of 1 and –4)
3 t 2 + 2 2 t − 5 6 = 0 t = ... | M1dep* | 1.1 | Re-arrange and solves their three term quadratic
equation in t to obtain a positive value of t
t = 2 | A1 | 1.1 | Ignore t = − 2 83 unless used
v = ( 4 t − 1 2 ) 2 + ( 3 t 2 + 6 t − 8 ) 2 with their positive value of t | M1 | 3.4 | Dependent on both previous M marks – if working
not shown then numerical components of v must be
consistent with their positive value of t
v = 1 6 .5 (m s-1) | A1 | 1.1 | awrt 16.5 (allow exact 4 1 7 ) – for reference:
16.4924225…
[5]
(c) | a = 4 i + ( 6 t + 6 ) j | B1ft | 1.1 | Follow their v from (a)
4 2 + ( 6 t + 6 ) 2 = 2 0 .2 2 | M1 | 1.1 | Setting up an equation in t using 20.2 and their a
t = 2 .3 | A1 | 1.1 | Positive value of t only
[3]At time $t$ seconds, where $t \geq 0$, a particle P has position vector $\mathbf{r}$ metres, where
$$\mathbf{r} = (2t^2 - 12t + 6)\mathbf{i} + (t^3 + 3t^2 - 8t)\mathbf{j}.$$
The velocity of P at time $t$ seconds is $v \text{ m s}^{-1}$.
\begin{enumerate}[label=(\alph*)]
\item Find $v$ in terms of $t$. [1]
\item Determine the speed of P at the instant when it is moving parallel to the vector $\mathbf{i} - 4\mathbf{j}$. [5]
\item Determine the value of $t$ when the magnitude of the acceleration of P is $20.2 \text{ m s}^{-2}$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2023 Q6 [9]}}