| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | When moving parallel to given vector |
| Difficulty | Standard +0.3 This is a straightforward M2 vector mechanics question requiring standard techniques: equating components to find when velocity is in direction i+j, differentiating for acceleration, and integrating for displacement. All steps are routine applications of formulas with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10h Vectors in kinematics: uniform acceleration in vector form3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use \(\mathbf{v} = \lambda(\mathbf{i}+\mathbf{j})\): \(6T^2 + 6T = 3T^2 + 24\) | M1 | Form an equation in \(t\), \(T\) or \(\lambda\); \(\lambda^2 - 108\lambda + 2592 = 0\) |
| Solve for \(T\): \(3T^2 + 6T - 24 = 0\) | M1 | Simplify to quadratic in \(t\), \(T\) or \(\lambda\), and solve |
| \((T+4)(T-2) = 0,\ T = 2\) | A1 | \(T = 2\) only |
| (3) | If M1 scored then state \(T=2\), allow 3/3. If guess \(T=2\) and show it works, allow 3/3. If only \(T=2\) with no equation, 0/3 for (a) but full marks available for (b) and (c). |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Differentiate: \(\mathbf{a} = (12t+6)\mathbf{i} + 6t\mathbf{j}\) | M1 | Majority of powers going down, need to be considering both components |
| A1 | Correct in \(t\) or \(T\) | |
| \(= 30\mathbf{i} + 12\mathbf{j}\ (\text{m s}^{-2})\) | A1 | Cao |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Integrate: \(\mathbf{r} = (2t^3 + 3t^2(+A))\mathbf{i} + (t^3 + 24t(+B))\mathbf{j}\) | M1 | Clear evidence of integration. Need to be considering both components. Do not need to see the constant(s). |
| A2 | \(-1\) each error. If integration seen in part (a) it scores no marks at that stage, but if result is used in part (c) then M1A2 available. | |
| \(\mathbf{OA} = 28\mathbf{i} + 56\mathbf{j}\), use their \(T\) | ||
| Distance \(= 28\sqrt{5} = 62.6\) (m) | DM1 | Dependent on previous M1. Use of Pythagoras on their \(\mathbf{OA}\) |
| 63 or better, \(\sqrt{3920}\) | A1 | |
| (5) | NB: Incorrect \(T\) can score 2/3 in (b) and 4/5 in (c) | |
| (11 marks) |
## Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $\mathbf{v} = \lambda(\mathbf{i}+\mathbf{j})$: $6T^2 + 6T = 3T^2 + 24$ | M1 | Form an equation in $t$, $T$ or $\lambda$; $\lambda^2 - 108\lambda + 2592 = 0$ |
| Solve for $T$: $3T^2 + 6T - 24 = 0$ | M1 | Simplify to quadratic in $t$, $T$ or $\lambda$, and solve |
| $(T+4)(T-2) = 0,\ T = 2$ | A1 | $T = 2$ only |
| | **(3)** | If M1 scored then state $T=2$, allow 3/3. If guess $T=2$ and show it works, allow 3/3. If only $T=2$ with no equation, 0/3 for (a) but full marks available for (b) and (c). |
## Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiate: $\mathbf{a} = (12t+6)\mathbf{i} + 6t\mathbf{j}$ | M1 | Majority of powers going down, need to be considering both components |
| | A1 | Correct in $t$ or $T$ |
| $= 30\mathbf{i} + 12\mathbf{j}\ (\text{m s}^{-2})$ | A1 | Cao |
| | **(3)** | |
## Question 3(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Integrate: $\mathbf{r} = (2t^3 + 3t^2(+A))\mathbf{i} + (t^3 + 24t(+B))\mathbf{j}$ | M1 | Clear evidence of integration. Need to be considering both components. Do not need to see the constant(s). |
| | A2 | $-1$ each error. If integration seen in part (a) it scores no marks at that stage, but if result is used in part (c) then M1A2 available. |
| $\mathbf{OA} = 28\mathbf{i} + 56\mathbf{j}$, use their $T$ | | |
| Distance $= 28\sqrt{5} = 62.6$ (m) | DM1 | Dependent on previous M1. Use of Pythagoras on their $\mathbf{OA}$ |
| 63 or better, $\sqrt{3920}$ | A1 | |
| | **(5)** | NB: Incorrect $T$ can score 2/3 in (b) and 4/5 in (c) |
| | **(11 marks)** | |
---3. At time $t$ seconds $( t \geqslant 0 )$ a particle $P$ has velocity $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$, where
$$\mathbf { v } = \left( 6 t ^ { 2 } + 6 t \right) \mathbf { i } + \left( 3 t ^ { 2 } + 24 \right) \mathbf { j }$$
When $t = 0$ the particle $P$ is at the origin $O$. At time $T$ seconds, $P$ is at the point $A$ and $\mathbf { v } = \lambda ( \mathbf { i } + \mathbf { j } )$, where $\lambda$ is a constant.
Find
\begin{enumerate}[label=(\alph*)]
\item the value of $T$,
\item the acceleration of $P$ as it passes through the point $A$,
\item the distance $O A$.\\
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2018 Q3 [11]}}