| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Inverse power force - non-gravitational context |
| Difficulty | Standard +0.8 This M3 variable force question requires integration of F=ma with a rational function force law, then separation of variables for the time calculation. Part (a) is standard work-energy application, but part (c) requires integrating dt = dx/v with a complex expression involving nested square roots and partial fractions, which is technically demanding for A-level though follows established M3 patterns. |
| Spec | 1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks |
|---|---|
| \(F = ma \to \frac{3}{(x+1)^3} = 0.5a = 0.5\frac{dv}{dx}\) | M1A1 |
| \(\int\frac{3}{(x+1)^3}dx = 0.5\int v \, dv\); Separate and \(\int\) | M1 |
| \(-\frac{3}{2(x+1)^2} = \frac{1}{4}v^2(+ c)\) | A1 |
| \(x = 0, v = 0 \Rightarrow c' = -\frac{3}{2}\) \(\therefore v^2 = 6\left(1 - \frac{1}{(x+1)^2}\right)\) | M1A1cso (6) |
| Answer | Marks |
|---|---|
| \(\forall x \quad v^2 < 6 \therefore v < \sqrt{6}\) (\(\because (x+1)^2\) always \(> 0\)) | B1 (1) |
| Answer | Marks |
|---|---|
| \(v = \frac{dx}{dt} = \frac{\sqrt{6}\sqrt{(x+1)^2 - 1}}{x+1}\) | M1 |
| \(\int\frac{x+1}{\sqrt{(x+1)^2 - 1}}dx = \sqrt{6}\int dt\) | M1 |
| \(\sqrt{(x+1)^2 - 1} = \sqrt{6}t + c'\) | M1A1 |
| \(t = 0, x = 0 \Rightarrow c' = 0\) | M1 |
| \(t = 2 \Rightarrow (x+1)^2 - 1 = (2\sqrt{6})^2\) | M1 |
| \((x+1)^2 = 25 \Rightarrow x = 4\) (\(c'\) need not have been found) | A1cao (7) |
**Part (a):**
$F = ma \to \frac{3}{(x+1)^3} = 0.5a = 0.5\frac{dv}{dx}$ | M1A1 |
$\int\frac{3}{(x+1)^3}dx = 0.5\int v \, dv$; Separate and $\int$ | M1 |
$-\frac{3}{2(x+1)^2} = \frac{1}{4}v^2(+ c)$ | A1 |
$x = 0, v = 0 \Rightarrow c' = -\frac{3}{2}$ $\therefore v^2 = 6\left(1 - \frac{1}{(x+1)^2}\right)$ | M1A1cso (6) |
**Part (b):**
$\forall x \quad v^2 < 6 \therefore v < \sqrt{6}$ ($\because (x+1)^2$ always $> 0$) | B1 (1) |
**Part (c):**
$v = \frac{dx}{dt} = \frac{\sqrt{6}\sqrt{(x+1)^2 - 1}}{x+1}$ | M1 |
$\int\frac{x+1}{\sqrt{(x+1)^2 - 1}}dx = \sqrt{6}\int dt$ | M1 |
$\sqrt{(x+1)^2 - 1} = \sqrt{6}t + c'$ | M1A1 |
$t = 0, x = 0 \Rightarrow c' = 0$ | M1 |
$t = 2 \Rightarrow (x+1)^2 - 1 = (2\sqrt{6})^2$ | M1 |
$(x+1)^2 = 25 \Rightarrow x = 4$ ($c'$ need not have been found) | A1cao (7) |
**Total: 14 Marks**\begin{enumerate}
\item A particle $P$ of mass 0.5 kg moves along the positive $x$-axis. It moves away from the origin $O$ under the action of a single force directed away from $O$. When $O P = x$ metres, the magnitude of the force is $\frac { 3 } { ( x + 1 ) ^ { 3 } } \mathrm {~N}$ and the speed of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
Initially $P$ is at rest at $O$.\\
(a) Show that $v ^ { 2 } = 6 \left( 1 - \frac { 1 } { ( x + 1 ) ^ { 2 } } \right)$.\\
(b) Show that the speed of $P$ never reaches $\sqrt { } 6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(c) Find $x$ when $P$ has been moving for 2 seconds.\\
\end{enumerate}
\section*{LL \\
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\hfill \mbox{\textit{Edexcel M3 2008 Q6 [14]}}