| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Horizontal circular track – friction only (no banking) |
| Difficulty | Standard +0.3 Part (a) is direct recall of the centripetal force formula (mv²/r). Part (b) requires resolving forces and using Newton's second law in circular motion, but follows a standard M3 approach: vertical equilibrium gives R cos θ = mg, horizontal centripetal force gives R sin θ = mv²/r, then dividing yields the result. This is a typical textbook exercise in circular motion with 7 total marks, slightly easier than average due to its straightforward structure and guided 'show that' format. |
| Spec | 3.03e Resolve forces: two dimensions6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks |
|---|---|
| (a) Resultant force towards centre \(= \frac{mv^2}{r}\) | B1 |
| (b) Weight \(= mg\), reaction \(R\) at angle \(\theta\) to vertical | B1 B1 B1 |
| \(R \cos \theta = mg\) | |
| Horizontally: \(R \sin \theta = \frac{mv^2}{r}\) | M1 A1 A1 |
| Divide: \(\tan \theta = \frac{v^2}{gr}\) |
**(a)** Resultant force towards centre $= \frac{mv^2}{r}$ | B1 |
**(b)** Weight $= mg$, reaction $R$ at angle $\theta$ to vertical | B1 B1 B1 |
$R \cos \theta = mg$ | |
Horizontally: $R \sin \theta = \frac{mv^2}{r}$ | M1 A1 A1 |
Divide: $\tan \theta = \frac{v^2}{gr}$ | |
**Total: 7 marks**Aliya, whose mass is $m$ kg, is playing rounders. She rounds the first base at a speed of $v$ ms$^{-1}$, making the turn on a horizontal circular path of radius $r$ m.
\begin{enumerate}[label=(\alph*)]
\item Write down, in terms of $m$, $v$ and $r$, the magnitude of the horizontal force acting on her. [1 mark]
\item Show that if she continues on the same circular path, the reaction force exerted on her by the ground must act at an angle $\theta$ to the vertical, where $\tan \theta = \frac{v^2}{gr}$. [6 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q2 [7]}}