| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Inverse power force - gravitational/escape velocity context |
| Difficulty | Standard +0.3 This is a standard M3 variable force question requiring application of F=ma with v dv/dx and integration. Part (a) is a 'show that' guiding students through the derivation, and part (b) involves straightforward substitution of the given energy condition. The technique is routine for M3 students who have practiced inverse square law problems. |
| Spec | 6.02c Work by variable force: using integration6.02d Mechanical energy: KE and PE concepts6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| N2L: \(ma = -\dfrac{cm}{x^2}\) | B1 | |
| \(\dfrac{d}{dx}\left(\dfrac{1}{2}v^2\right) = -\dfrac{c}{x^2} \Rightarrow \dfrac{1}{2}v^2 = A + \dfrac{c}{mx}\) | M1 A1 | ignore \(A\) |
| \(v^2 = B + \dfrac{2c}{mx}\) | ||
| \(x = R, v = U \Rightarrow B = U^2 - \dfrac{2c}{mR}\) | M1 | |
| Leading to \(v^2 = U^2 + 2c\left(\dfrac{1}{x} - \dfrac{1}{R}\right)\) ✱ | A1 | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\dfrac{1}{2}\left[\dfrac{1}{2}mU^2\right] = \dfrac{1}{2}m\left[U^2 + 2c\left(\dfrac{1}{2R} - \dfrac{1}{R}\right)\right]\) | M1 A1 | |
| Leading to \(c = \dfrac{1}{2}RU^2\) | A1 |
## Question 3:
**Part (a)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| N2L: $ma = -\dfrac{cm}{x^2}$ | B1 | |
| $\dfrac{d}{dx}\left(\dfrac{1}{2}v^2\right) = -\dfrac{c}{x^2} \Rightarrow \dfrac{1}{2}v^2 = A + \dfrac{c}{mx}$ | M1 A1 | ignore $A$ |
| $v^2 = B + \dfrac{2c}{mx}$ | | |
| $x = R, v = U \Rightarrow B = U^2 - \dfrac{2c}{mR}$ | M1 | |
| Leading to $v^2 = U^2 + 2c\left(\dfrac{1}{x} - \dfrac{1}{R}\right)$ ✱ | A1 | cso |
**(5 marks)**
**Part (b)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\dfrac{1}{2}\left[\dfrac{1}{2}mU^2\right] = \dfrac{1}{2}m\left[U^2 + 2c\left(\dfrac{1}{2R} - \dfrac{1}{R}\right)\right]$ | M1 A1 | |
| Leading to $c = \dfrac{1}{2}RU^2$ | A1 | |
**(3 marks) — Total 8 marks**
---3. A rocket is fired vertically upwards with speed $U$ from a point on the Earth's surface. The rocket is modelled as a particle $P$ of constant mass $m$, and the Earth as a fixed sphere of radius $R$. At a distance $x$ from the centre of the Earth, the speed of $P$ is $v$. The only force acting on $P$ is directed towards the centre of the Earth and has magnitude $\frac { c m } { x ^ { 2 } }$, where $c$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Show that $v ^ { 2 } = U ^ { 2 } + 2 c \left( \frac { 1 } { x } - \frac { 1 } { R } \right)$.
The kinetic energy of $P$ at $x = 2 R$ is half of its kinetic energy at $x = R$.
\item Find $c$ in terms of $U$ and $R$.\\
(3)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2006 Q3 [8]}}