| Exam Board | Edexcel |
|---|---|
| Module | FM2 AS (Further Mechanics 2 AS) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| 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 circular motion problem requiring resolution of forces, application of Newton's second law for circular motion, and consideration of tension constraints. While it involves multiple steps (finding geometry, resolving vertically and horizontally, applying tension limits), the techniques are routine for FM2 students and the question provides clear guidance by asking to 'show that' a specific inequality, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| No vertical motion: \(T_A\cos\theta = mg\) | M1 | One equation in \(T_A\) and/or \(T_B\). Dimensionally correct, all relevant terms. Condone sign errors and sin/cos confusion. N.B. If same tension in both parts, score ONLY first M1A1. |
| \(T_A = \dfrac{5mg}{4}\) | A1 | Correct equation (no trig) |
| Circular motion: \(T_B + T_A\sin\theta = m \times \dfrac{v^2}{r}\) | M1 | Form second equation in \(T_A\) and/or \(T_B\). Condone sign errors and sin/cos confusion. Allow \(mr\omega^2\) |
| \(T_B + \dfrac{3}{5}T_A = m\dfrac{v^2}{3a}\) | A1 | Correct equation (no trig) |
| \(T_B > 0 \Rightarrow v^2 > \dfrac{9ag}{4}\) | DM1 | Use model to form one inequality/equation in \(v^2\), \(a\) and \(g\) only, dependent on both M's |
| \(T_B \leq \dfrac{3mg}{2} \Rightarrow m\dfrac{v^2}{3a} - \dfrac{3}{4}mg \leq \dfrac{3}{2}mg, \quad \left(m\dfrac{v^2}{3a} \leq \dfrac{9mg}{4}\right)\) | DM1 | Use model to form second inequality/equation in \(v^2\), \(a\) and \(g\) only, dependent on both M's. Allow use of \(T_B = \dfrac{3mg}{2}\) or \(T_B < \dfrac{3mg}{2}\) |
| \(\Rightarrow \dfrac{9ag}{4} < v^2 \leq \dfrac{27ag}{4}\) * | A1* | Deduce given answer from correct working. Only available if working with inequalities throughout and fully correct. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use \(v^2 = \dfrac{27ag}{4}\) and \(T = \dfrac{2\pi r}{v}\) oe | M1 | Correct method to find \(T\) in terms of \(a\) and \(g\) only. May substitute 9.8 for \(g\). |
| \(T = 4\pi\sqrt{\dfrac{a}{3g}}\) | A1 | Any equivalent form but no fractions within fractions. If 9.8 used for \(g\), numerical part needs to be 2 sf or 3sf i.e. \(2.3\sqrt{a}\) or \(2.32\sqrt{a}\) |
# Question 3:
## Part 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| No vertical motion: $T_A\cos\theta = mg$ | M1 | One equation in $T_A$ and/or $T_B$. Dimensionally correct, all relevant terms. Condone sign errors and sin/cos confusion. N.B. If same tension in both parts, score ONLY first M1A1. |
| $T_A = \dfrac{5mg}{4}$ | A1 | Correct equation (no trig) |
| Circular motion: $T_B + T_A\sin\theta = m \times \dfrac{v^2}{r}$ | M1 | Form second equation in $T_A$ and/or $T_B$. Condone sign errors and sin/cos confusion. Allow $mr\omega^2$ |
| $T_B + \dfrac{3}{5}T_A = m\dfrac{v^2}{3a}$ | A1 | Correct equation (no trig) |
| $T_B > 0 \Rightarrow v^2 > \dfrac{9ag}{4}$ | DM1 | Use model to form one inequality/equation in $v^2$, $a$ and $g$ **only, dependent on both M's** |
| $T_B \leq \dfrac{3mg}{2} \Rightarrow m\dfrac{v^2}{3a} - \dfrac{3}{4}mg \leq \dfrac{3}{2}mg, \quad \left(m\dfrac{v^2}{3a} \leq \dfrac{9mg}{4}\right)$ | DM1 | Use model to form second inequality/equation in $v^2$, $a$ and $g$ **only, dependent on both M's**. Allow use of $T_B = \dfrac{3mg}{2}$ or $T_B < \dfrac{3mg}{2}$ |
| $\Rightarrow \dfrac{9ag}{4} < v^2 \leq \dfrac{27ag}{4}$ * | A1* | Deduce given answer from correct working. Only available if working with inequalities throughout and fully correct. |
**Total: (7)**
## Part 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use $v^2 = \dfrac{27ag}{4}$ and $T = \dfrac{2\pi r}{v}$ oe | M1 | Correct method to find $T$ in terms of $a$ and $g$ only. May substitute 9.8 for $g$. |
| $T = 4\pi\sqrt{\dfrac{a}{3g}}$ | A1 | Any equivalent form but no fractions within fractions. If 9.8 used for $g$, numerical part needs to be 2 sf or 3sf i.e. $2.3\sqrt{a}$ or $2.32\sqrt{a}$ |
**Total: (2)**\begin{enumerate}
\item A light inextensible string has length $8 a$. 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 $A B = 4 a$. A small ball of mass $m$ is attached to a point $P$ on the string, where $A P = 5 a$.
\end{enumerate}
The ball moves in a horizontal circle with constant speed $v$, with both $A P$ and $B P$ taut.\\
The string will break if the tension in it exceeds $\frac { 3 m g } { 2 }$\\
By modelling the ball as a particle and assuming the string does not break,\\
(a) show that $\frac { 9 a g } { 4 } < v ^ { 2 } \leqslant \frac { 27 a g } { 4 }$\\
(b) find the least possible time needed for the ball to make one complete revolution.
\hfill \mbox{\textit{Edexcel FM2 AS 2019 Q3 [9]}}