| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | January |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given velocity function find force |
| Difficulty | Standard +0.3 This is a standard M3 variable force question requiring differentiation of the given velocity function using the chain rule (v dv/dx = a), then applying F=ma. The velocity function is given explicitly, making this a routine application of a well-practiced technique with straightforward algebra, slightly easier than average. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(v = \sqrt{8x^{\frac{3}{2}} - 4}\) | Given expression | |
| \(v^2 = 8x^{\frac{3}{2}} - 4\) | ||
| \(2v\frac{dv}{dx} = 12x^{\frac{1}{2}}\) | M1, A1, A1 | M1: attempting to differentiate \(v^2\) expression - chain rule must be used on lhs; A1: for correct \(x^{\frac{1}{2}}\); A1: for 6 - award both only if work fully correct |
| \(F = 0.5 \times 6x^{\frac{1}{2}} = 3x^{\frac{1}{2}}\) | M1dep | for using NL2 with \(m = 0.5\) to obtain expression for \(F\) in terms of \(x\) |
| \(x = 4 \Rightarrow F = 6\) | A1 | A1cso for \(F = 6\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dv}{dx} = \frac{1}{2}\left(8x^{\frac{3}{2}}-4\right)^{-\frac{1}{2}} \times 12x^{\frac{1}{2}}\) | M1 | Must be complete method to obtain accel in terms of \(x\) |
| \(\frac{dv}{dx} = \frac{1}{2v} \times 12x^{\frac{1}{2}}\), \(v\frac{dv}{dx} = 6x^{\frac{1}{2}}\) | A1 | A1 rhs |
| \(\frac{dv}{dt} = \frac{1}{2}\left(8x^{\frac{3}{2}}-4\right)^{-\frac{1}{2}} \times 12x^{\frac{1}{2}} \times \frac{dx}{dt}\) leading to \(\frac{dv}{dt} = 6x^{\frac{1}{2}}\) | M1A1A1 | Award as above |
## Question 1:
**Working/Answer** | **Marks** | **Guidance**
$v = \sqrt{8x^{\frac{3}{2}} - 4}$ | | Given expression
$v^2 = 8x^{\frac{3}{2}} - 4$ | |
$2v\frac{dv}{dx} = 12x^{\frac{1}{2}}$ | M1, A1, A1 | M1: attempting to differentiate $v^2$ expression - chain rule must be used on lhs; A1: for correct $x^{\frac{1}{2}}$; A1: for 6 - award both only if work fully correct
$F = 0.5 \times 6x^{\frac{1}{2}} = 3x^{\frac{1}{2}}$ | M1dep | for using NL2 with $m = 0.5$ to obtain expression for $F$ in terms of $x$
$x = 4 \Rightarrow F = 6$ | A1 | A1cso for $F = 6$
**Total: 5 marks**
**Alternatives (first 3 marks):**
$\frac{dv}{dx} = \frac{1}{2}\left(8x^{\frac{3}{2}}-4\right)^{-\frac{1}{2}} \times 12x^{\frac{1}{2}}$ | M1 | Must be complete method to obtain accel in terms of $x$
$\frac{dv}{dx} = \frac{1}{2v} \times 12x^{\frac{1}{2}}$, $v\frac{dv}{dx} = 6x^{\frac{1}{2}}$ | A1 | A1 rhs
$\frac{dv}{dt} = \frac{1}{2}\left(8x^{\frac{3}{2}}-4\right)^{-\frac{1}{2}} \times 12x^{\frac{1}{2}} \times \frac{dx}{dt}$ leading to $\frac{dv}{dt} = 6x^{\frac{1}{2}}$ | M1A1A1 | Award as above\begin{enumerate}
\item A particle $P$ of mass 0.5 kg moves along the positive $x$-axis under the action of a single force of magnitude $F$ newtons. The force acts along the $x$-axis in the direction of $x$ increasing. When $P$ is $x$ metres from the origin $O$, it is moving away from $O$ with speed $\sqrt { \left( 8 x ^ { \frac { 3 } { 2 } } - 4 \right) } \mathrm { ms } ^ { - 1 }$.
\end{enumerate}
Find $F$ when $P$ is 4 m from $O$.\\
\hfill \mbox{\textit{Edexcel M3 2014 Q1 [5]}}