| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | January |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given acceleration function find velocity |
| Difficulty | Moderate -0.3 This is a straightforward M3 mechanics question requiring the standard technique of using v dv/dx = a, then integrating and applying initial conditions. The integration is simple (polynomial), and the method is a direct application of a core M3 technique with no conceptual complications or multi-step reasoning required. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(v\dfrac{dv}{dx} = 9x\) | M1 | Setting up the differential equation with correct form |
| \(\dfrac{1}{2}v^2 = 9x \quad (+c)\) | A1 | Correct integration of both sides |
| \(v^2 = 9x^2 + c\) | M1dep | Dependent on previous M1 |
| \(x = 2, \; v = 6 \Rightarrow 36 = 9 \times 4 + c \Rightarrow c = 0\) | Substituting initial conditions to find \(c\) | |
| \(v^2 = 9x^2\) | A1 | Correct final answer |
## Question 1:
| Working/Answer | Mark | Guidance |
|---|---|---|
| $v\dfrac{dv}{dx} = 9x$ | M1 | Setting up the differential equation with correct form |
| $\dfrac{1}{2}v^2 = 9x \quad (+c)$ | A1 | Correct integration of both sides |
| $v^2 = 9x^2 + c$ | M1dep | Dependent on previous M1 |
| $x = 2, \; v = 6 \Rightarrow 36 = 9 \times 4 + c \Rightarrow c = 0$ | | Substituting initial conditions to find $c$ |
| $v^2 = 9x^2$ | A1 | Correct final answer |\begin{enumerate}
\item A particle $P$ is moving along the positive $x$-axis. When the displacement of $P$ from the origin is $x$ metres, the velocity of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the acceleration of $P$ is $9 x \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\end{enumerate}
When $x = 2 , v = 6$\\
Show that $v ^ { 2 } = 9 x ^ { 2 }$.\\
(4)\\
\hfill \mbox{\textit{Edexcel M3 2013 Q1 [4]}}