| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Variable force along axis work-energy |
| Difficulty | Standard +0.8 This M3 question requires setting up work-energy principles with variable force integration, applying F=ma with position-dependent force, and solving for velocity as a function of position. While the integration itself is straightforward (x² integrates to x³/3), students must correctly handle the direction of force, set up the energy equation properly, and work through algebraic manipulation. This is above-average difficulty due to the conceptual setup and multi-step reasoning required, but remains a standard M3 exercise rather than requiring novel insight. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(0.5v\frac{dv}{dx} = -0.375x^2\) | M1 | Equation of motion using \(v\frac{dv}{dx}\) |
| \(\frac{1}{2}v^2 = -0.25x^3 + c\) | M1 A1 | Integration |
| Using \(t=0, v=2, x=8\): \(\frac{1}{2}\times 2^2 = -0.25\times 8^3 + c\), \(c=130\) | ||
| \(\therefore v^2 = -\frac{1}{2}x^3 + 260\) * | A1 | Printed answer (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| \(v=5 \Rightarrow x^3 = 520 - 50\) | M1 | |
| \(x = 7.77\) | A1 | (2) Total: 6 |
# Question 1:
## Part (a)
| Working/Answer | Marks | Notes |
|---|---|---|
| $0.5v\frac{dv}{dx} = -0.375x^2$ | M1 | Equation of motion using $v\frac{dv}{dx}$ |
| $\frac{1}{2}v^2 = -0.25x^3 + c$ | M1 A1 | Integration |
| Using $t=0, v=2, x=8$: $\frac{1}{2}\times 2^2 = -0.25\times 8^3 + c$, $c=130$ | | |
| $\therefore v^2 = -\frac{1}{2}x^3 + 260$ * | A1 | Printed answer **(4)** |
## Part (b)
| Working/Answer | Marks | Notes |
|---|---|---|
| $v=5 \Rightarrow x^3 = 520 - 50$ | M1 | |
| $x = 7.77$ | A1 | **(2) Total: 6** |
---\begin{enumerate}
\item A particle $P$ of mass 0.5 kg moves on the positive $x$-axis under the action of a single force directed towards the origin $O$. At time $t$ seconds the distance of $P$ from $O$ is $x$ metres, the magnitude of the force is $0.375 x ^ { 2 } \mathrm {~N}$ and the speed of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\end{enumerate}
When $t = 0 , O P = 8 \mathrm {~m}$ and $P$ is moving towards $O$ with speed $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(a) Show that $v ^ { 2 } = 260 - \frac { 1 } { 2 } \chi ^ { 3 }$.\\
(b) Find the distance of $P$ from $O$ at the instant when $v = 5$.\\
\hfill \mbox{\textit{Edexcel M3 2011 Q1 [6]}}