| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2008 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Bead on straight wire vector force |
| Difficulty | Standard +0.8 This M5 question requires students to connect work-energy principles with vector mechanics, finding a force parallel to a given vector using the work-energy theorem. It involves multiple steps: calculating displacement, applying the parallel vector condition, using work done equals kinetic energy change, and solving simultaneously. This is more challenging than standard A-level mechanics but typical for Further Maths M5. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(\mathbf{d} = (7\mathbf{i} - 14\mathbf{j}) - (\mathbf{i} - 6\mathbf{j}) = (6\mathbf{i} - 8\mathbf{j})\) | B1 | |
| \((6k\mathbf{i} + k\mathbf{j})\cdot(6\mathbf{i} - 8\mathbf{j}) = \frac{1}{2} \times \frac{1}{2} \times (2\sqrt{7})^2\) | M1 A2 ft | |
| \(28k = 7 \Rightarrow k = \frac{1}{4}\) | DM1 | |
| \(\Rightarrow \mathbf{P} = \frac{3}{2}\mathbf{i} + \frac{1}{4}\mathbf{j}\) | A1 | Total: 6 |
## Question 1:
| Working | Marks | Notes |
|---|---|---|
| $\mathbf{d} = (7\mathbf{i} - 14\mathbf{j}) - (\mathbf{i} - 6\mathbf{j}) = (6\mathbf{i} - 8\mathbf{j})$ | B1 | |
| $(6k\mathbf{i} + k\mathbf{j})\cdot(6\mathbf{i} - 8\mathbf{j}) = \frac{1}{2} \times \frac{1}{2} \times (2\sqrt{7})^2$ | M1 A2 ft | |
| $28k = 7 \Rightarrow k = \frac{1}{4}$ | DM1 | |
| $\Rightarrow \mathbf{P} = \frac{3}{2}\mathbf{i} + \frac{1}{4}\mathbf{j}$ | A1 | **Total: 6** |
---\begin{enumerate}
\item \hspace{0pt} [In this question $\mathbf { i }$ and $\mathbf { j }$ are horizontal unit vectors.]
\end{enumerate}
A small bead of mass 0.5 kg is threaded on a smooth horizontal wire. The bead is initially at rest at the point with position vector $( \mathbf { i } - 6 \mathbf { j } ) \mathrm { m }$. A constant horizontal force $\mathbf { P } \mathrm { N }$ then acts on the bead causing it to move along the wire. The bead passes through the point with position vector ( $7 \mathbf { i } - 14 \mathbf { j }$ ) m with speed $2 \sqrt { 7 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
Given that $\mathbf { P }$ is parallel to ( $6 \mathbf { i } + \mathbf { j }$ ), find $\mathbf { P }$.\\
(6)\\
\hfill \mbox{\textit{Edexcel M5 2008 Q1 [6]}}