| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Inverse power force - gravitational/escape velocity context |
| Difficulty | Standard +0.8 Part (a) is a straightforward proof using proportionality and boundary conditions (3 marks of routine algebra). Part (b) requires setting up and integrating a variable force equation F=ma with F(x), then applying limits correctly—this is the core M3 skill but involves careful manipulation over 7 marks. The conceptual demand (variable force, energy methods via integration) and multi-step algebraic complexity place this above average difficulty, though it follows standard M3 patterns rather than requiring novel insight. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(F = (-) \frac{k}{x^2}\) | M1 | |
| \(mg = (-) \frac{k}{R^3}\) | M1 | |
| \(F = \frac{mgR^2}{x^2}\) | A1 | (3 marks total) |
| Answer | Marks | Guidance |
|---|---|---|
| \(m\ddot{x} = -\frac{mgR^2}{x^2}\) | M1 | |
| \(v\frac{dv}{dx} = -\frac{gR^2}{x^2}\) | M1 | |
| \(\frac{1}{2}v^2 = -\int\frac{gR^2}{x^2}dx\) | M1 dep on 1st M mark | |
| \(\frac{1}{2}v^2 = \frac{gR^2}{x} (+c)\) | A1 | |
| At \(x = R, v = 3U\): \(\frac{9U^2}{2} = gR + c\) | M1 dep on 3rd M mark | |
| \(\frac{1}{2}v^2 = \frac{gR^2}{x} + \frac{9U^2}{2} - gR\) | ||
| At \(x = 2R, v = U\): \(\frac{1}{2}U^2 = \frac{gR^2}{2R} + \frac{9U^2}{2} - gR\) | M1 dep on 3rd M mark | |
| \(U^2 = \frac{gR}{8}\) | ||
| \(U = \sqrt{\frac{gR}{8}}\) | A1 | (7 marks total) |
## Part (a)
$F = (-) \frac{k}{x^2}$ | M1 |
$mg = (-) \frac{k}{R^3}$ | M1 |
$F = \frac{mgR^2}{x^2}$ | A1 | (3 marks total)
## Part (b)
$m\ddot{x} = -\frac{mgR^2}{x^2}$ | M1 |
$v\frac{dv}{dx} = -\frac{gR^2}{x^2}$ | M1 |
$\frac{1}{2}v^2 = -\int\frac{gR^2}{x^2}dx$ | M1 dep on 1st M mark |
$\frac{1}{2}v^2 = \frac{gR^2}{x} (+c)$ | A1 |
At $x = R, v = 3U$: $\frac{9U^2}{2} = gR + c$ | M1 dep on 3rd M mark |
$\frac{1}{2}v^2 = \frac{gR^2}{x} + \frac{9U^2}{2} - gR$ | |
At $x = 2R, v = U$: $\frac{1}{2}U^2 = \frac{gR^2}{2R} + \frac{9U^2}{2} - gR$ | M1 dep on 3rd M mark |
$U^2 = \frac{gR}{8}$ | |
$U = \sqrt{\frac{gR}{8}}$ | A1 | (7 marks total)A particle $P$ of mass $m$ is above the surface of the Earth at distance $x$ from the centre of the Earth. The Earth exerts a gravitational force on $P$. The magnitude of this force is inversely proportional to $x^2$.
At the surface of the Earth the acceleration due to gravity is $g$. The Earth is modelled as a sphere of radius $R$.
\begin{enumerate}[label=(\alph*)]
\item Prove that the magnitude of the gravitational force on $P$ is $\frac{mgR^2}{x^2}$. [3]
\end{enumerate}
A particle is fired vertically upwards from the surface of the Earth with initial speed $3U$. At a height $R$ above the surface of the Earth the speed of the particle is $U$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find $U$ in terms of $g$ and $R$. [7]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q2 [10]}}