| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2015 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Work done by vector force displacement |
| Difficulty | Moderate -0.3 This is a straightforward application of the work formula W = F·s requiring vector subtraction to find displacement, then a dot product calculation, followed by solving a quadratic equation. While it involves multiple steps and algebraic manipulation with a parameter, it uses standard M5 techniques without requiring problem-solving insight or novel approaches. |
| Spec | 1.10f Distance between points: using position vectors6.02b Calculate work: constant force, resolved component |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{AB} = (-\mathbf{i} + a\mathbf{j} - \mathbf{k}) - (2\mathbf{i} + 4\mathbf{j} + a\mathbf{k})\) | M1 | Attempt \(\overrightarrow{AB} = B - A\) |
| \(= (-3\mathbf{i} + (a-4)\mathbf{j} + (-1-a)\mathbf{k})\) | A1 | Correct displacement vector |
| Work done \(= \mathbf{F} \cdot \overrightarrow{AB}\) | M1 | Use of \(W = \mathbf{F} \cdot \mathbf{d}\) |
| \(= (2)(-3) + (a)(a-4) + (-3)(-1-a)\) | M1 | Correct dot product attempt |
| \(= -6 + a^2 - 4a + 3 + 3a\) | A1 | Correct expansion |
| \(= a^2 - a - 3 = 3\) | ||
| \(a^2 - a - 6 = 0\) | ||
| \((a-3)(a+2) = 0\) | M1 | Solve quadratic |
| \(a = 3\) (since \(a\) is a positive constant) | A1 | Correct answer with reason for rejecting \(a = -2\) |
## Question 1:
| $\overrightarrow{AB} = (-\mathbf{i} + a\mathbf{j} - \mathbf{k}) - (2\mathbf{i} + 4\mathbf{j} + a\mathbf{k})$ | M1 | Attempt $\overrightarrow{AB} = B - A$ |
|---|---|---|
| $= (-3\mathbf{i} + (a-4)\mathbf{j} + (-1-a)\mathbf{k})$ | A1 | Correct displacement vector |
| Work done $= \mathbf{F} \cdot \overrightarrow{AB}$ | M1 | Use of $W = \mathbf{F} \cdot \mathbf{d}$ |
| $= (2)(-3) + (a)(a-4) + (-3)(-1-a)$ | M1 | Correct dot product attempt |
| $= -6 + a^2 - 4a + 3 + 3a$ | A1 | Correct expansion |
| $= a^2 - a - 3 = 3$ | | |
| $a^2 - a - 6 = 0$ | | |
| $(a-3)(a+2) = 0$ | M1 | Solve quadratic |
| $a = 3$ (since $a$ is a positive constant) | A1 | Correct answer with reason for rejecting $a = -2$ |
---\begin{enumerate}
\item A particle $P$ moves from the point $A$, with position vector ( $2 \mathbf { i } + 4 \mathbf { j } + a \mathbf { k }$ ) m , where $a$ is a positive constant, to the point $B$, with position vector ( $- \mathbf { i } + a \mathbf { j } - \mathbf { k }$ ) m , under the action of a constant force $\mathbf { F } = ( 2 \mathbf { i } + a \mathbf { j } - 3 \mathbf { k } )$ N. The work done by $\mathbf { F }$, as it moves the particle $P$ from $A$ to $B$, is 3 J . Find the value of $a$.\\
(6)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2015 Q1 [6]}}