| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2010 |
| Session | June |
| 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) requires setting up and proving an inverse square law relationship using proportionality constants, which is moderately challenging. Part (b) involves integrating a variable force (F = mgR²/x²) with respect to distance using work-energy principles, requiring careful setup of limits and algebraic manipulation—this is a multi-step problem requiring solid understanding of variable force integration, which is above average difficulty for M3. |
| Spec | 3.03g Gravitational acceleration6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(F = (-)\dfrac{k}{x^2}\) | M1 | |
| \(mg = (-)\dfrac{k}{R^2}\) | M1 | |
| \(F = \dfrac{mgR^2}{x^2}\) | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(m\ddot{x} = -\dfrac{mgR^2}{x^2}\) | M1 | |
| \(v\dfrac{dv}{dx} = -\dfrac{gR^2}{x^2}\) | M1 | |
| \(\dfrac{1}{2}v^2 = \int\left(-\dfrac{gR^2}{x^2}\right)dx\) | M1 dep | dep on 1st M mark |
| \(\dfrac{1}{2}v^2 = \dfrac{gR^2}{x} \quad (+c)\) | A1 | |
| \(x=R,\ v=3U\): \(\quad\dfrac{9U^2}{2} = gR + c\) | M1 dep | dep on 3rd M mark |
| \(\dfrac{1}{2}v^2 = \dfrac{gR^2}{x} + \dfrac{9U^2}{2} - gR\) | ||
| \(x=2R,\ v=U\): \(\quad\dfrac{1}{2}U^2 = \dfrac{gR^2}{2R} + \dfrac{9U^2}{2} - gR\) | M1 dep | dep on 3rd M mark |
| \(U^2 = \dfrac{gR}{8}\) | ||
| \(U = \sqrt{\dfrac{gR}{8}}\) | A1 | (7) |
# Question 2:
## Part (a)
| Working | Marks | Notes |
|---------|-------|-------|
| $F = (-)\dfrac{k}{x^2}$ | M1 | |
| $mg = (-)\dfrac{k}{R^2}$ | M1 | |
| $F = \dfrac{mgR^2}{x^2}$ | A1 | (3) |
## Part (b)
| Working | Marks | Notes |
|---------|-------|-------|
| $m\ddot{x} = -\dfrac{mgR^2}{x^2}$ | M1 | |
| $v\dfrac{dv}{dx} = -\dfrac{gR^2}{x^2}$ | M1 | |
| $\dfrac{1}{2}v^2 = \int\left(-\dfrac{gR^2}{x^2}\right)dx$ | M1 dep | dep on 1st M mark |
| $\dfrac{1}{2}v^2 = \dfrac{gR^2}{x} \quad (+c)$ | A1 | |
| $x=R,\ v=3U$: $\quad\dfrac{9U^2}{2} = gR + c$ | M1 dep | dep on 3rd M mark |
| $\dfrac{1}{2}v^2 = \dfrac{gR^2}{x} + \dfrac{9U^2}{2} - gR$ | | |
| $x=2R,\ v=U$: $\quad\dfrac{1}{2}U^2 = \dfrac{gR^2}{2R} + \dfrac{9U^2}{2} - gR$ | M1 dep | dep on 3rd M mark |
| $U^2 = \dfrac{gR}{8}$ | | |
| $U = \sqrt{\dfrac{gR}{8}}$ | A1 | (7) |
**Total: [10]**
---2. 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 $\frac { m g R ^ { 2 } } { x ^ { 2 } }$.
A particle is fired vertically upwards from the surface of the Earth with initial speed $3 U$. At a height $R$ above the surface of the Earth the speed of the particle is $U$.
\item Find $U$ in terms of $g$ and $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2010 Q2 [10]}}