| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2005 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Work done by vector force displacement |
| Difficulty | Standard +0.3 This is a straightforward application of work = force ยท displacement in 3D vectors. Students must convert direction vectors to unit vectors, scale by magnitudes, find displacement, then compute two dot products and sum. All steps are routine M5 techniques with no conceptual challenges or novel problem-solving required, making it slightly easier than average. |
| 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 vectors1.10d Vector operations: addition and scalar multiplication6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Unit vector in direction \(2\mathbf{i} + \mathbf{j} + 2\mathbf{k}\): magnitude \(= \sqrt{4+1+4} = 3\) | M1 | Finding unit vector for either force |
| \(\mathbf{F}_1 = 9 \times \frac{1}{3}(2\mathbf{i} + \mathbf{j} + 2\mathbf{k}) = 3(2\mathbf{i} + \mathbf{j} + 2\mathbf{k}) = 6\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}\) | A1 | Correct \(\mathbf{F}_1\) |
| Unit vector in direction \(\mathbf{i} + 8\mathbf{j} - 4\mathbf{k}\): magnitude \(= \sqrt{1+64+16} = 9\) | M1 | |
| \(\mathbf{F}_2 = 18 \times \frac{1}{9}(\mathbf{i} + 8\mathbf{j} - 4\mathbf{k}) = 2\mathbf{i} + 16\mathbf{j} - 8\mathbf{k}\) | A1 | Correct \(\mathbf{F}_2\) |
| Displacement \(= (3\mathbf{i} + 2\mathbf{j} - \mathbf{k}) - (\mathbf{i} + \mathbf{j} + \mathbf{k}) = 2\mathbf{i} + \mathbf{j} - 2\mathbf{k}\) | B1 | Correct displacement vector |
| Total force \(= 8\mathbf{i} + 19\mathbf{j} - 2\mathbf{k}\) | ||
| Work done \(= (8\mathbf{i} + 19\mathbf{j} - 2\mathbf{k}) \cdot (2\mathbf{i} + \mathbf{j} - 2\mathbf{k}) = 16 + 19 + 4 = 39\) J | M1 A1 | Dot product of total force with displacement; 39 J |
# Question 1:
| Working/Answer | Mark | Guidance |
|---|---|---|
| Unit vector in direction $2\mathbf{i} + \mathbf{j} + 2\mathbf{k}$: magnitude $= \sqrt{4+1+4} = 3$ | M1 | Finding unit vector for either force |
| $\mathbf{F}_1 = 9 \times \frac{1}{3}(2\mathbf{i} + \mathbf{j} + 2\mathbf{k}) = 3(2\mathbf{i} + \mathbf{j} + 2\mathbf{k}) = 6\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}$ | A1 | Correct $\mathbf{F}_1$ |
| Unit vector in direction $\mathbf{i} + 8\mathbf{j} - 4\mathbf{k}$: magnitude $= \sqrt{1+64+16} = 9$ | M1 | |
| $\mathbf{F}_2 = 18 \times \frac{1}{9}(\mathbf{i} + 8\mathbf{j} - 4\mathbf{k}) = 2\mathbf{i} + 16\mathbf{j} - 8\mathbf{k}$ | A1 | Correct $\mathbf{F}_2$ |
| Displacement $= (3\mathbf{i} + 2\mathbf{j} - \mathbf{k}) - (\mathbf{i} + \mathbf{j} + \mathbf{k}) = 2\mathbf{i} + \mathbf{j} - 2\mathbf{k}$ | B1 | Correct displacement vector |
| Total force $= 8\mathbf{i} + 19\mathbf{j} - 2\mathbf{k}$ | | |
| Work done $= (8\mathbf{i} + 19\mathbf{j} - 2\mathbf{k}) \cdot (2\mathbf{i} + \mathbf{j} - 2\mathbf{k}) = 16 + 19 + 4 = 39$ J | M1 A1 | Dot product of total force with displacement; 39 J |
---\begin{enumerate}
\item Two constant forces $\mathbf { F } _ { 1 }$ and $\mathbf { F } _ { 2 }$ are the only forces acting on a particle. $\mathbf { F } _ { 1 }$ has magnitude 9 N and acts in the direction of $2 \mathbf { i } + \mathbf { j } + 2 \mathbf { k } . \mathbf { F } _ { 2 }$ has magnitude 18 N and acts in the direction of $\mathbf { i } + 8 \mathbf { j } - 4 \mathbf { k }$.
\end{enumerate}
Find the total work done by the two forces in moving the particle from the point with position vector $( \mathbf { i } + \mathbf { j } + \mathbf { k } ) \mathrm { m }$ to the point with position vector $( 3 \mathbf { i } + 2 \mathbf { j } - \mathbf { k } ) \mathrm { m }$.\\
(Total 6 marks)\\
\hfill \mbox{\textit{Edexcel M5 2005 Q1 [6]}}