| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Inverse-square gravitational force |
| Difficulty | Standard +0.8 Part (a) is a straightforward 'show that' requiring substitution into the inverse square law. Part (b) requires setting up and integrating a variable gravitational force with respect to distance (not a standard constant acceleration problem), then applying energy conservation between two heights. This goes beyond routine M3 questions by requiring integration of 1/(R+x)² and careful algebraic manipulation, but follows a standard method once recognized. |
| Spec | 6.02c Work by variable force: using integration6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Distance of P from centre of Earth \(= R + x\) | ||
| \(F = \frac{k}{(R+x)^2}\); at \(x=0\), \(F = mg\), so \(k = mg(R)^2\) | M1 A1 | |
| \(F = \frac{mgR^2}{(R+x)^2}\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F = ma\), \(-\frac{gR^2}{(R+x)^2} = v\frac{dv}{dx}\) | M1 A1 | |
| \(\int_V^{\sqrt{\frac{gR}{2}}} v\, dv = \int_R^{2R} -\frac{gR^2}{(R+x)^2} dx\) | M1 A1 | |
| \(\left[\frac{1}{2}v^2\right]_V^{\sqrt{\frac{gR}{2}}} = \left[\frac{gR^2}{R+x}\right]_R^{2R}\) | M1, A1 | |
| \(\frac{1}{2}\times\frac{gR}{2} - \frac{1}{2}V^2 = \frac{gR^2}{3R} - \frac{gR^2}{2R} = -\frac{gR}{6}\) | M1 | |
| \(\frac{V^2}{2} = \frac{gR}{4} + \frac{gR}{6} = \frac{5gR}{12}\), \(V^2 = \frac{5gR}{6}\), \(V = \sqrt{\frac{5gR}{6}}\) | A1, A1 | (9) Total: 12 |
## Question 5:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Distance of P from centre of Earth $= R + x$ | | |
| $F = \frac{k}{(R+x)^2}$; at $x=0$, $F = mg$, so $k = mg(R)^2$ | M1 A1 | |
| $F = \frac{mgR^2}{(R+x)^2}$ | A1 | (3) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F = ma$, $-\frac{gR^2}{(R+x)^2} = v\frac{dv}{dx}$ | M1 A1 | |
| $\int_V^{\sqrt{\frac{gR}{2}}} v\, dv = \int_R^{2R} -\frac{gR^2}{(R+x)^2} dx$ | M1 A1 | |
| $\left[\frac{1}{2}v^2\right]_V^{\sqrt{\frac{gR}{2}}} = \left[\frac{gR^2}{R+x}\right]_R^{2R}$ | M1, A1 | |
| $\frac{1}{2}\times\frac{gR}{2} - \frac{1}{2}V^2 = \frac{gR^2}{3R} - \frac{gR^2}{2R} = -\frac{gR}{6}$ | M1 | |
| $\frac{V^2}{2} = \frac{gR}{4} + \frac{gR}{6} = \frac{5gR}{12}$, $V^2 = \frac{5gR}{6}$, $V = \sqrt{\frac{5gR}{6}}$ | A1, A1 | (9) Total: 12 |
---5. Above the Earth's surface, the magnitude of the gravitational force on a particle due to the Earth is inversely proportional to the square of the distance of the particle from the centre of the Earth. The Earth is modelled as a sphere of radius $R$ and the acceleration due to gravity at the Earth's surface is $g$. A particle $P$ of mass $m$ is at a height $x$ above the surface of the Earth.
\begin{enumerate}[label=(\alph*)]
\item Show that the magnitude of the gravitational force acting on $P$ is
$$\frac { m g R ^ { 2 } } { ( R + x ) ^ { 2 } }$$
A rocket is fired vertically upwards from the surface of the Earth. When the rocket is at height $2 R$ above the surface of the Earth its speed is $\sqrt { } \left( \frac { g R } { 2 } \right)$. You may assume that air resistance can be ignored and that the engine of the rocket is switched off before the rocket reaches height $R$.
Modelling the rocket as a particle,
\item find the speed of the rocket when it was at height $R$ above the surface of the Earth.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2012 Q5 [12]}}