| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Inverse-square gravitational force |
| Difficulty | Standard +0.3 This is a standard M3 gravitational force problem requiring students to (a) equate the given force law to weight at Earth's surface (straightforward substitution), and (b) apply conservation of energy with the inverse-square force law. Part (b) involves integrating F = K/x² to find work done, which is a routine M3 technique. While it requires understanding of variable force and energy conservation, it follows a well-established method taught explicitly in M3 courses. |
| Spec | 6.02c Work by variable force: using integration6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(x = R \Rightarrow F = mg \therefore mg = \frac{K}{R^2}\) | M1 | Setting \(F = mg\) and \(x = R\) |
| \(K = mgR^2\) | A1 | Deducing the given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{mgR^2}{x^2} = -mv\frac{dv}{dx}\) | M1 | Acceleration in form \(v\frac{dv}{dx}\); minus sign may be missing |
| \(g\int\frac{R^2}{x^2}dx = -\int v\,dv\) | ||
| \(-g\frac{R^2}{x} = -\frac{1}{2}v^2\ (+c)\) | DM1A1ft | Attempting integration; correct integration (ft on missing minus sign) |
| \(x = 3R,\, v = V \Rightarrow -g\frac{R^2}{3R} = -\frac{1}{2}V^2 + c\) | M1 | Substituting \(x=3R, v=V\) to obtain equation for \(c\) |
| \(c = -\frac{Rg}{3} + \frac{1}{2}V^2\) | A1 | Correct expression for \(c\) |
| \(x = R \Rightarrow \frac{1}{2}v^2 = -\frac{Rg}{3} + \frac{1}{2}V^2 + \frac{gR^2}{R}\) | M1 | Substituting \(x = R\) and their expression for \(c\) |
| \(v^2 = V^2 + \frac{4Rg}{3}\) | ||
| \(v = \sqrt{V^2 + \frac{4Rg}{3}}\) | A1 cso | Correct expression for \(v\), any equivalent form |
## Question 2(a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $x = R \Rightarrow F = mg \therefore mg = \frac{K}{R^2}$ | M1 | Setting $F = mg$ and $x = R$ |
| $K = mgR^2$ | A1 | Deducing the given answer |
**Total: (2)**
## Question 2(b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{mgR^2}{x^2} = -mv\frac{dv}{dx}$ | M1 | Acceleration in form $v\frac{dv}{dx}$; minus sign may be missing |
| $g\int\frac{R^2}{x^2}dx = -\int v\,dv$ | | |
| $-g\frac{R^2}{x} = -\frac{1}{2}v^2\ (+c)$ | DM1A1ft | Attempting integration; correct integration (ft on missing minus sign) |
| $x = 3R,\, v = V \Rightarrow -g\frac{R^2}{3R} = -\frac{1}{2}V^2 + c$ | M1 | Substituting $x=3R, v=V$ to obtain equation for $c$ |
| $c = -\frac{Rg}{3} + \frac{1}{2}V^2$ | A1 | Correct expression for $c$ |
| $x = R \Rightarrow \frac{1}{2}v^2 = -\frac{Rg}{3} + \frac{1}{2}V^2 + \frac{gR^2}{R}$ | M1 | Substituting $x = R$ and their expression for $c$ |
| $v^2 = V^2 + \frac{4Rg}{3}$ | | |
| $v = \sqrt{V^2 + \frac{4Rg}{3}}$ | A1 cso | Correct expression for $v$, any equivalent form |
**Total: (7) [9]**
---2. A spacecraft $S$ of mass $m$ moves in a straight line towards the centre of the Earth. The Earth is modelled as a sphere of radius $R$ and $S$ is modelled as a particle. When $S$ is at a distance $x , x \geqslant R$, from the centre of the Earth, the force exerted by the Earth on $S$ is directed towards the centre of the Earth. The force has magnitude $\frac { K } { x ^ { 2 } }$, where $K$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $K = m g R ^ { 2 }$\\
(2)
When $S$ is at a distance $3 R$ from the centre of the Earth, the speed of $S$ is $V$. Assuming that air resistance can be ignored,
\item find, in terms of $g , R$ and $V$, the speed of $S$ as it hits the surface of the Earth.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2015 Q2 [9]}}