| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2009 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Position vector from velocity integration |
| Difficulty | Standard +0.3 This is a straightforward M5 vector mechanics problem requiring application of Newton's second law in vector form, integration to find velocity, then integration to find position, and solving for time. While it involves 3D vectors and multiple steps, the method is standard and algorithmic with no conceptual challenges—slightly easier than average for an M5 question. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10d Vector operations: addition and scalar multiplication3.02d Constant acceleration: SUVAT formulae3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\mathbf{F}_1 + \mathbf{F}_2 = (12\mathbf{i} - 6\mathbf{j} + 12\mathbf{k})\) N | B1 | Adding forces correctly |
| \(\mathbf{a} = \frac{\mathbf{F}}{m} = \frac{1}{3}(12\mathbf{i} - 6\mathbf{j} + 12\mathbf{k})\) | M1 | Use of \(F = ma\) |
| \(\mathbf{a} = (4\mathbf{i} - 2\mathbf{j} + 4\mathbf{k})\) m s\(^{-2}\) | A1 | Correct acceleration vector |
| Displacement \(\overrightarrow{AB} = (8\mathbf{i} - 3\mathbf{j} + 5\mathbf{k}) - (0\mathbf{i} + \mathbf{j} - 3\mathbf{k}) = (8\mathbf{i} - 4\mathbf{j} + 8\mathbf{k})\) m | M1 A1 | Subtracting position vectors |
| Using \(\mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\) with \(\mathbf{u} = 0\): \((8\mathbf{i} - 4\mathbf{j} + 8\mathbf{k}) = \frac{1}{2}(4\mathbf{i} - 2\mathbf{j} + 4\mathbf{k})t^2\) | M1 | Applying constant acceleration equation |
| \(t^2 = 4\), so \(t = 2\) s | A1 | Correct time |
| \(\mathbf{v} = \mathbf{u} + \mathbf{a}t = 0 + (4\mathbf{i} - 2\mathbf{j} + 4\mathbf{k})(2)\) | M1 | |
| \(\mathbf{v} = (8\mathbf{i} - 4\mathbf{j} + 8\mathbf{k})\) m s\(^{-1}\) | A1 | cao |
# Question 1:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\mathbf{F}_1 + \mathbf{F}_2 = (12\mathbf{i} - 6\mathbf{j} + 12\mathbf{k})$ N | B1 | Adding forces correctly |
| $\mathbf{a} = \frac{\mathbf{F}}{m} = \frac{1}{3}(12\mathbf{i} - 6\mathbf{j} + 12\mathbf{k})$ | M1 | Use of $F = ma$ |
| $\mathbf{a} = (4\mathbf{i} - 2\mathbf{j} + 4\mathbf{k})$ m s$^{-2}$ | A1 | Correct acceleration vector |
| Displacement $\overrightarrow{AB} = (8\mathbf{i} - 3\mathbf{j} + 5\mathbf{k}) - (0\mathbf{i} + \mathbf{j} - 3\mathbf{k}) = (8\mathbf{i} - 4\mathbf{j} + 8\mathbf{k})$ m | M1 A1 | Subtracting position vectors |
| Using $\mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2$ with $\mathbf{u} = 0$: $(8\mathbf{i} - 4\mathbf{j} + 8\mathbf{k}) = \frac{1}{2}(4\mathbf{i} - 2\mathbf{j} + 4\mathbf{k})t^2$ | M1 | Applying constant acceleration equation |
| $t^2 = 4$, so $t = 2$ s | A1 | Correct time |
| $\mathbf{v} = \mathbf{u} + \mathbf{a}t = 0 + (4\mathbf{i} - 2\mathbf{j} + 4\mathbf{k})(2)$ | M1 | |
| $\mathbf{v} = (8\mathbf{i} - 4\mathbf{j} + 8\mathbf{k})$ m s$^{-1}$ | A1 | cao |
---\begin{enumerate}
\item At time $t = 0$, a particle $P$ of mass 3 kg is at rest at the point $A$ with position vector $( \mathbf { j } - 3 \mathbf { k } ) \mathrm { m }$. Two constant forces $\mathbf { F } _ { 1 }$ and $\mathbf { F } _ { 2 }$ then act on the particle $P$ and it passes through the point $B$ with position vector $( 8 \mathbf { i } - 3 \mathbf { j } + 5 \mathbf { k } ) \mathrm { m }$.
\end{enumerate}
Given that $\mathbf { F } _ { 1 } = ( 4 \mathbf { i } - 2 \mathbf { j } + 5 \mathbf { k } ) \mathrm { N }$ and $\mathbf { F } _ { 2 } = ( 8 \mathbf { i } - 4 \mathbf { j } + 7 \mathbf { k } ) \mathrm { N }$ and that $\mathbf { F } _ { 1 }$ and $\mathbf { F } _ { 2 }$ are the only two forces acting on $P$, find the velocity of $P$ as it passes through $B$, giving your answer as a vector.\\
\hfill \mbox{\textit{Edexcel M5 2009 Q1 [7]}}