| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Find force using F=ma |
| Difficulty | Standard +0.3 This is a straightforward M2 mechanics question requiring differentiation of velocity to find acceleration, then applying F=ma for part (a), and finding when j-component equals zero for parallel motion in part (b). All steps are routine applications of standard techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.03d Newton's second law: 2D vectors |
| Answer | Marks |
|---|---|
| Differentiate v: \(a = (6t - 4)\mathbf{i} + (6t - 8)\mathbf{j}\) | M1 A1 |
| \(\mathbf{F} = m\mathbf{a}\) when \(t = 4\): \(\mathbf{F} = 0.3(20\mathbf{i} + 16\mathbf{j}) = 6\mathbf{i} + 4.8\mathbf{j}\) | M1 |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Motion parallel to i: \(3t^2 - 8t + 4 = 0 = (3t - 2)(t - 2)\) | M1 | |
| \(t = \frac{2}{3}\) or \(t = 2\) | A1 | |
| Integrate v: \(\mathbf{r} = (t^3 - 2t^2(+p))\mathbf{i} + (t^3 - 4t^2 + 4t(+q))\mathbf{j}\) | M1 A1 | |
| Use limits: \(\mathbf{r}_s = (8 - 8(+p))\mathbf{i} + (8 - 16 + 8(+q))\mathbf{j}\) | M1 A1 | |
| \(\mathbf{r}_s = \left(\frac{8}{27} - \frac{8}{9}(+p)\right)\mathbf{i} + \left(\frac{8}{27} - \frac{16}{9} + \frac{8}{3}(+q)\right)\mathbf{j}\) | A1 | |
| \(AB = \pm\left(\frac{16}{27}\mathbf{i} - \frac{32}{27}\mathbf{j}\right)\) | ||
| Pythagoras' theorem: \( | AB | = \frac{16}{27}\sqrt{5} = 1.3\) (or better) (m) |
| (9) | ||
| [12] |
## 5a.
Differentiate v: $a = (6t - 4)\mathbf{i} + (6t - 8)\mathbf{j}$ | M1 A1 |
$\mathbf{F} = m\mathbf{a}$ when $t = 4$: $\mathbf{F} = 0.3(20\mathbf{i} + 16\mathbf{j}) = 6\mathbf{i} + 4.8\mathbf{j}$ | M1 |
(3) | |
## 5b.
Motion parallel to i: $3t^2 - 8t + 4 = 0 = (3t - 2)(t - 2)$ | M1 |
$t = \frac{2}{3}$ or $t = 2$ | A1 |
Integrate v: $\mathbf{r} = (t^3 - 2t^2(+p))\mathbf{i} + (t^3 - 4t^2 + 4t(+q))\mathbf{j}$ | M1 A1 |
Use limits: $\mathbf{r}_s = (8 - 8(+p))\mathbf{i} + (8 - 16 + 8(+q))\mathbf{j}$ | M1 A1 |
$\mathbf{r}_s = \left(\frac{8}{27} - \frac{8}{9}(+p)\right)\mathbf{i} + \left(\frac{8}{27} - \frac{16}{9} + \frac{8}{3}(+q)\right)\mathbf{j}$ | A1 |
$AB = \pm\left(\frac{16}{27}\mathbf{i} - \frac{32}{27}\mathbf{j}\right)$ | |
Pythagoras' theorem: $|AB| = \frac{16}{27}\sqrt{5} = 1.3$ (or better) (m) | DM1 A1 |
(9) | |
[12] | |
**Notes for Q5**
Accept column vectors throughout
5(a)
- First M1 for attempt to differentiate v, at least two powers of $t$ decreasing by one
- A1 for a correct expression. (A0 if i or j omitted)
- Second M1 for multiplying their a by 0.3, substituting $t = 4$ and collecting $\mathbf{i}$'s and $\mathbf{j}$'s. Isw if they find the magnitude.
5(b)
- First M1 for $3t^2 - 8t + 4 = 0$ and attempting to solve. This M mark can be implied by two correct answers but if answer(s) are incorrect, we need to see an explicit attempt at factorising, using the formula or completing the square.
- First A1 for two correct answers, allow 0.67 or better.
- Second M1 for attempt to integrate v, to produce a vector, with at least two powers of $t$ increasing by 1.
- Second A1 for a correct $\mathbf{r}$ (constant not needed).
- Third M1 for substituting both their values of $t$ (which must have come from using a velocity vector) into their $\mathbf{r}$.
- Third A1 for correct unsimplified $\mathbf{r}_1$ ( constant not needed) Allow a point.
- Fourth A1 for correct unsimplified $\mathbf{r}_2$ ( constant not needed).Allow a point.
- Fourth DM1, dependent on previous M mark, for subtracting their velocity vectors (or points) either way and using Pythagoras to find the length.
- Fifth A1 for correct surd answer oe or 1.3 or better.
---5. A particle $P$ of mass 0.3 kg moves under the action of a single force $\mathbf { F }$ newtons. At time $t$ seconds $( t \geqslant 0 ) , P$ has velocity $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$, where
$$\mathbf { v } = \left( 3 t ^ { 2 } - 4 t \right) \mathbf { i } + \left( 3 t ^ { 2 } - 8 t + 4 \right) \mathbf { j }$$
\begin{enumerate}[label=(\alph*)]
\item Find $\mathbf { F }$ when $t = 4$
At the instants when $P$ is at the points $A$ and $B$, particle $P$ is moving parallel to the vector i.
\item Find the distance $A B$.\\
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2018 Q5 [12]}}