| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Period or time for one revolution |
| Difficulty | Challenging +1.2 This is a structured M3 gravitational mechanics question with clear scaffolding. Parts (a) and (b) are standard derivations of Newton's law of gravitation and orbital period (routine for M3). Parts (c)-(e) require vector addition of three forces and equating to centripetal force, which is more challenging but still follows a predictable M3 pattern with significant guidance from the question structure. The multi-step nature and 16 total marks elevate it slightly above average difficulty. |
| Spec | 6.01d Unknown indices: using dimensions6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(g = \frac{kM(1)}{R^2}\), so \(k = \frac{gR^2}{M}\) | M1 A1 | |
| (b) \(\frac{mv^2}{r} = \frac{kMm}{r^2}\) so \(v^2 = \frac{gR^2}{M} \cdot \frac{M}{r} = \frac{gR^2}{r}\) and \(T = \frac{2\pi r}{v} = 2\pi\sqrt{\frac{r^3}{gR^2}}\) | M1 A1 M1 A1 | |
| (c) Diagram | B1; B2 | |
| (d) Along \(XE\), as \(X\) in circular orbit so central force | ||
| (e) \(\frac{mv^2}{r} = \frac{kR^2}{M}\left(\frac{m}{3^4}\cos 30° + \frac{mM}{r^2} + \frac{m}{3^4}\cos 30°\right)\) | M1 A1 A1 | |
| \(\frac{mv^2}{r} = \frac{kR^2m}{Mr^2}\left(\frac{m\sqrt{3}}{3} \cdot 2 + M\right)\) | M1 A1 | |
| \(T_1 = \frac{2\pi r}{v} = 2\pi r\sqrt{\frac{gR^2(3M + m\sqrt{3})}{gR^2 \cdot 3M}} = 2\pi\sqrt{\frac{r^3}{gR^2}} \cdot \sqrt{\frac{3M + m\sqrt{3}}{3M}} = T\sqrt{\frac{3M + m\sqrt{3}}{3M}}\) | M1 A1 | 16 marks |
**(a)** $g = \frac{kM(1)}{R^2}$, so $k = \frac{gR^2}{M}$ | M1 A1 |
**(b)** $\frac{mv^2}{r} = \frac{kMm}{r^2}$ so $v^2 = \frac{gR^2}{M} \cdot \frac{M}{r} = \frac{gR^2}{r}$ and $T = \frac{2\pi r}{v} = 2\pi\sqrt{\frac{r^3}{gR^2}}$ | M1 A1 M1 A1 |
**(c)** Diagram | B1; B2 |
**(d)** Along $XE$, as $X$ in circular orbit so central force | |
**(e)** $\frac{mv^2}{r} = \frac{kR^2}{M}\left(\frac{m}{3^4}\cos 30° + \frac{mM}{r^2} + \frac{m}{3^4}\cos 30°\right)$ | M1 A1 A1 |
$\frac{mv^2}{r} = \frac{kR^2m}{Mr^2}\left(\frac{m\sqrt{3}}{3} \cdot 2 + M\right)$ | M1 A1 |
$T_1 = \frac{2\pi r}{v} = 2\pi r\sqrt{\frac{gR^2(3M + m\sqrt{3})}{gR^2 \cdot 3M}} = 2\pi\sqrt{\frac{r^3}{gR^2}} \cdot \sqrt{\frac{3M + m\sqrt{3}}{3M}} = T\sqrt{\frac{3M + m\sqrt{3}}{3M}}$ | M1 A1 | 16 marksThe gravitational attraction $F$ N between two point masses $m_1$ kg and $m_2$ kg at a distance $x$ m apart is given by $F = \frac{km_1m_2}{x^2}$, where $k$ is a constant. Given that a small body of mass $1$ kg experiences a force of $g$ N at the surface of the Earth, which has radius $R$ m and mass $M$ kg,
\begin{enumerate}[label=(\alph*)]
\item show that $k = \frac{gR^2}{M}$. [2 marks]
\end{enumerate}
A small communications satellite of mass $m$ kg is put into a circular orbit of radius $r$ m around the Earth. Modelling the Earth as a particle of mass $M$ kg, and using the value of $k$ from (a),
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item prove that the period of rotation, $T$ s, of the satellite is given by $T = \frac{2\pi}{R}\sqrt{\frac{r^3}{g}}$. [4 marks]
\end{enumerate}
To cover transmission to any point on the Earth, three small satellites $X$, $Y$ and $Z$, each of mass $m$ kg, are placed in a common circular orbit of radius $r$ and form an equilateral triangle as shown.
\includegraphics{figure_6}
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Show on a copy of the diagram the direction of the three forces acting on $X$. [1 mark]
\item State, with a reason, the direction of the resultant force on $X$. [2 marks]
\item Show that the period of rotation of $X$ is given by $T\sqrt{\frac{3M}{2M + m\sqrt{3}}}$ s, where $T$ s is the period found in (b). [7 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q6 [16]}}