| Exam Board | Edexcel |
|---|---|
| Module | FM2 AS (Further Mechanics 2 AS) |
| Session | Specimen |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Two strings, two fixed points |
| Difficulty | Challenging +1.2 This is a standard Further Mechanics conical pendulum problem with two strings, requiring resolution of forces in vertical and horizontal directions, geometry to find angles, and algebraic manipulation. The 'show that' format guides students through the solution, and while it involves multiple steps and careful bookkeeping, it follows a well-established method taught in FM2 with no novel insight required. Slightly above average difficulty due to the two-string setup and the inequality derivation in part (b). |
| Spec | 1.05b Sine and cosine rules: including ambiguous case3.03d Newton's second law: 2D vectors6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos\alpha = \frac{4}{5}\) or \(\sin\alpha = \frac{3}{5}\) | B1 | For correct trig. ratio seen |
| \(r = 4a\sin\alpha\) | B1 | For a correct radius expression seen |
| Resolving vertically | M1 | For resolving vertically with correct no. of terms and tensions resolved |
| \(T_1\cos\alpha - T_2\sin\alpha = mg\) | A1 | For a correct equation |
| Resolving horizontally | M1 | For resolving horizontally with correct no. of terms and tensions resolved |
| \(T_1\sin\alpha + T_2\cos\alpha = mr\omega^2\) | A1A1 | For a correct equation |
| Solving for either tension | M1 | For solving their two equations to find either tension |
| \(T_1 = \frac{4m}{25}(9a\omega^2 + 5g)\) | A1* | For the given answer |
| \(T_2 = \frac{3m}{25}(16a\omega^2 - 5g)\) | A1* | For the given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{4m}{25}(9a\omega^2 + 5g) < 4mg\) | M1 | For use of \(T_1 < 4mg\) |
| \(\frac{3m}{25}(16a\omega^2 - 5g) > 0\) | M1 | For using \(T_2 > 0\) |
| \(\omega > \sqrt{\frac{5g}{16a}}\) or \(\omega < \sqrt{\frac{20g}{9a}}\) | A1 | For a correct inequality (either) for \(\omega\) |
| \(S = \frac{2\pi}{\omega}\) | M1 | For use of \(S = \frac{2\pi}{\omega}\) with either critical value |
| \(3\pi\sqrt{\frac{a}{5g}} < S < 8\pi\sqrt{\frac{a}{5g}}\) | A1* | For given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| String being light implies that the tension is constant in both portions of the string | B1 | For a clear explanation |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos\alpha = \frac{4}{5}$ or $\sin\alpha = \frac{3}{5}$ | B1 | For correct trig. ratio seen |
| $r = 4a\sin\alpha$ | B1 | For a correct radius expression seen |
| Resolving vertically | M1 | For resolving vertically with correct no. of terms and tensions resolved |
| $T_1\cos\alpha - T_2\sin\alpha = mg$ | A1 | For a correct equation |
| Resolving horizontally | M1 | For resolving horizontally with correct no. of terms and tensions resolved |
| $T_1\sin\alpha + T_2\cos\alpha = mr\omega^2$ | A1A1 | For a correct equation |
| Solving for either tension | M1 | For solving their two equations to find either tension |
| $T_1 = \frac{4m}{25}(9a\omega^2 + 5g)$ | A1* | For the given answer |
| $T_2 = \frac{3m}{25}(16a\omega^2 - 5g)$ | A1* | For the given answer |
**Total: 10 marks**
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{4m}{25}(9a\omega^2 + 5g) < 4mg$ | M1 | For use of $T_1 < 4mg$ |
| $\frac{3m}{25}(16a\omega^2 - 5g) > 0$ | M1 | For using $T_2 > 0$ |
| $\omega > \sqrt{\frac{5g}{16a}}$ or $\omega < \sqrt{\frac{20g}{9a}}$ | A1 | For a correct inequality (either) for $\omega$ |
| $S = \frac{2\pi}{\omega}$ | M1 | For use of $S = \frac{2\pi}{\omega}$ with either critical value |
| $3\pi\sqrt{\frac{a}{5g}} < S < 8\pi\sqrt{\frac{a}{5g}}$ | A1* | For given answer |
**Total: 5 marks**
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| String being light implies that the tension is constant in both portions of the string | B1 | For a clear explanation |
**Total: 1 mark**
---\begin{enumerate}
\item A light inextensible string has length 7a. One end of the string is attached to a fixed point $A$ and the other end of the string is attached to a fixed point $B$, with $A$ vertically above $B$ and $A B = 5 a$. A particle of mass $m$ is attached to a point $P$ on the string where $A P = 4 a$. The particle moves in a horizontal circle with constant angular speed $\omega$, with both $A P$ and $B P$ taut.\\
(a) Show that\\
(i) the tension in $A P$ is $\frac { 4 m } { 25 } \left( 9 a \omega ^ { 2 } + 5 g \right)$\\
(ii) the tension in $B P$ is $\frac { 3 m } { 25 } \left( 16 a \omega ^ { 2 } - 5 g \right)$.
\end{enumerate}
The string will break if the tension in it reaches a magnitude of $4 m g$.\\
The time for the particle to make one revolution is $S$.\\
(b) Show that
$$3 \pi \sqrt { \frac { a } { 5 g } } < S < 8 \pi \sqrt { \frac { a } { 5 g } }$$
(c) State how in your calculations you have used the assumption that the string is light.
\hfill \mbox{\textit{Edexcel FM2 AS Q2 [16]}}