| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2004 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Inverse power force - gravitational/escape velocity context |
| Difficulty | Standard +0.8 This M3 question requires understanding inverse square law gravity, setting up and applying the work-energy theorem with variable force integration from R to 3R, and algebraic manipulation. While the proof part (a) is straightforward, part (b) demands integration of a variable force and careful application of energy principles—more sophisticated than typical M1/M2 mechanics but standard for M3 variable force work. |
| Spec | 3.02h Motion under gravity: vector form6.02c Work by variable force: using integration6.02d Mechanical energy: KE and PE concepts |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(F = \frac{k}{x^2}\) | M1 | [k may be seen as \(Gm_1m_2\), for example] |
| Equating \(F\) to \(mg\) at \(x = R\): \(mg = \frac{k}{R^2}\) | M1 | |
| Convincing completion: \([k = mgR^2]\) to give \(F = \frac{mgR^2}{x^2}\) | A1 (3) |
| Answer | Marks |
|---|---|
| (b) Equation of motion: \((m)a = (-)\frac{(m)gR^2}{x^2}\); \((m)v\frac{dv}{dx} = -\frac{(m)gR^2}{x^2}\) | M1;M1 |
| Integrating: \(\frac{1}{2}v^2 = \frac{gR^2}{x} + (c)\) or equivalent | M1A1 |
| Answer | Marks |
|---|---|
| Use of \(v^2 = \frac{3gR}{2}\), \(x = R\) to find \(c\) \([c = -\frac{1}{2}gR]\) or use in def. int. | M1 |
| [Using \(x = 0\) is M0] \(\left[v^2 = \frac{2gR^2}{x} - \frac{gR}{2}\right]\) | |
| Substituting \(x = 3R\) and finding \(V\): \(V = \sqrt{\frac{gR}{6}}\) | M1;A1 (7) |
| Answer | Marks |
|---|---|
| Work/energy: \((-)\int_R^{3R}\frac{mgR^2}{x^2}dx = \frac{1}{2}mv^2 - \frac{1}{2}mu^2\) | M1;M1 |
| Integrating: \(\left[\frac{mgR^2}{x} - \frac{mgR^2}{R}\right] = \frac{1}{2}mv^2 - \frac{1}{2}m\frac{3gR}{2}\) | M1A1M1 |
| Final 2 marks as scheme | M1A1 |
| [Conservation of energy scores 0] |
(a) $F = \frac{k}{x^2}$ | M1 | [k may be seen as $Gm_1m_2$, for example]
Equating $F$ to $mg$ at $x = R$: $mg = \frac{k}{R^2}$ | M1 |
Convincing completion: $[k = mgR^2]$ to give $F = \frac{mgR^2}{x^2}$ | A1 (3) |
*Note: r may be used instead of x throughout, then $r \to x$ at end.*
(b) Equation of motion: $(m)a = (-)\frac{(m)gR^2}{x^2}$; $(m)v\frac{dv}{dx} = -\frac{(m)gR^2}{x^2}$ | M1;M1 |
Integrating: $\frac{1}{2}v^2 = \frac{gR^2}{x} + (c)$ or equivalent | M1A1 |
[S.C: Allow A1 if A0 earlier due to "+" only]
Use of $v^2 = \frac{3gR}{2}$, $x = R$ to find $c$ $[c = -\frac{1}{2}gR]$ or use in def. int. | M1 |
[Using $x = 0$ is M0] $\left[v^2 = \frac{2gR^2}{x} - \frac{gR}{2}\right]$ | |
Substituting $x = 3R$ and finding $V$: $V = \sqrt{\frac{gR}{6}}$ | M1;A1 (7) |
[Using $x = 2R$ is M0]
**Alternative in (b)**
Work/energy: $(-)\int_R^{3R}\frac{mgR^2}{x^2}dx = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$ | M1;M1 |
Integrating: $\left[\frac{mgR^2}{x} - \frac{mgR^2}{R}\right] = \frac{1}{2}mv^2 - \frac{1}{2}m\frac{3gR}{2}$ | M1A1M1 |
Final 2 marks as scheme | M1A1 |
[Conservation of energy scores 0] | |
---3. Above the earth's surface, the magnitude of the force on a particle due to the earth's gravity is inversely proportional to the square of the distance of the particle from the centre of the earth. Assuming that the earth is a sphere of radius $R$, and taking $g$ as the acceleration due to gravity at the surface of the earth,
\begin{enumerate}[label=(\alph*)]
\item prove that the magnitude of the gravitational force on a particle of mass $m$ when it is a distance $x ( x \geq R )$ from the centre of the earth is $\frac { m g R ^ { 2 } } { x ^ { 2 } }$.
A particle is fired vertically upwards from the surface of the earth with initial speed $u$, where $u ^ { 2 } = \frac { 3 } { 2 } g R$. Ignoring air resistance,
\item find, in terms of $g$ and $R$, the speed of the particle when it is at a height $2 R$ above the surface of the earth.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2004 Q3 [10]}}