| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2021 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: complete revolution conditions |
| Difficulty | Challenging +1.2 This is a standard Further Mechanics vertical circle problem with three parts: (a) finding minimum speed for complete circles using energy conservation and tension condition at the top (routine), (b) finding angle when string breaks requiring energy methods and solving for position, and (c) finding acceleration components. While it requires multiple techniques (energy conservation, circular motion dynamics, resolving forces), these are well-practiced FM2 procedures with no novel insight needed. The multi-part structure and calculation complexity place it slightly above average difficulty. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equation of motion along the string at the top of the circle | M1 | Correct number of terms |
| \(T + mg = \frac{mv^2}{a}\) | A1 | Correct equation |
| Conservation of energy | M1 | All terms needed and dimensionally correct |
| \(\frac{1}{2}mU^2 - \frac{1}{2}mv^2 = mga\) | A1 | Correct equation |
| Solve for \(T\) and use \(T = 0\) | M1 | Solve for \(T\) and use \(T = 0\) (allow \(T \geq 0\)) |
| \(U = \sqrt{3ga}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equation of motion along the string at instant string breaks | M1 | Correct no. of terms with \(mg\) resolved and correct acceleration component |
| \(\frac{11mg}{2} - mg\cos\alpha = \frac{mv^2}{a}\) | A1 | Correct equation |
| Conservation of energy | M1 | All terms needed and dimensionally correct |
| \(\frac{1}{2}mv^2 - \frac{1}{2}m.4ag = mga\cos\alpha\) | A1 | Correct equation |
| Solve for \(\cos\alpha\) \(\left(= \frac{1}{2}\right)\) | M1 | Solve for \(\cos\alpha\) |
| Angle turned through is \(210°\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Find radial component of acceleration: \(\frac{v^2}{a}\) \((= 5g)\) | M1 | Uses their value of \(v\) from part (b) |
| Find tangential component of acceleration: \(g\sin\alpha\) \(\left(= \frac{\sqrt{3}}{2}g\right)\) | M1 | Equation of motion along the tangent |
| Square, add and square root | M1 | Find the magnitude of the resultant acceleration |
| \(\frac{\sqrt{103}}{2}g\) or \(49.7\ (\text{m s}^{-2})\) or \(50\ (\text{m s}^{-2})\) | A1 | cao |
## Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion along the string at the top of the circle | M1 | Correct number of terms |
| $T + mg = \frac{mv^2}{a}$ | A1 | Correct equation |
| Conservation of energy | M1 | All terms needed and dimensionally correct |
| $\frac{1}{2}mU^2 - \frac{1}{2}mv^2 = mga$ | A1 | Correct equation |
| Solve for $T$ and use $T = 0$ | M1 | Solve for $T$ and use $T = 0$ (allow $T \geq 0$) |
| $U = \sqrt{3ga}$ | A1 | cao |
## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion along the string at instant string breaks | M1 | Correct no. of terms with $mg$ resolved and correct acceleration component |
| $\frac{11mg}{2} - mg\cos\alpha = \frac{mv^2}{a}$ | A1 | Correct equation |
| Conservation of energy | M1 | All terms needed and dimensionally correct |
| $\frac{1}{2}mv^2 - \frac{1}{2}m.4ag = mga\cos\alpha$ | A1 | Correct equation |
| Solve for $\cos\alpha$ $\left(= \frac{1}{2}\right)$ | M1 | Solve for $\cos\alpha$ |
| Angle turned through is $210°$ | A1 | cao |
## Question 5(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Find radial component of acceleration: $\frac{v^2}{a}$ $(= 5g)$ | M1 | Uses their value of $v$ from part (b) |
| Find tangential component of acceleration: $g\sin\alpha$ $\left(= \frac{\sqrt{3}}{2}g\right)$ | M1 | Equation of motion along the tangent |
| Square, add and square root | M1 | Find the magnitude of the resultant acceleration |
| $\frac{\sqrt{103}}{2}g$ or $49.7\ (\text{m s}^{-2})$ or $50\ (\text{m s}^{-2})$ | A1 | cao |\begin{enumerate}
\item A light inextensible string of length $a$ has one end attached to a fixed point $O$. The other end of the string is attached to a small stone of mass $m$. The stone is held with the string taut and horizontal. The stone is then projected vertically upwards with speed $U$.
\end{enumerate}
The stone is modelled as a particle and air resistance is modelled as being negligible.\\
Assuming that the string does not break, use the model to\\
(a) find the least value of $U$ so that the stone will move in complete vertical circles.
The string will break if the tension in it is equal to $\frac { 11 m g } { 2 }$\\
Given that $U = 2 \sqrt { a g }$, use the model to\\
(b) find the total angle that the string has turned through, from when the stone is projected vertically upwards, to when the string breaks,\\
(c) find the magnitude of the acceleration of the stone at the instant just before the string breaks.
\hfill \mbox{\textit{Edexcel FM2 2021 Q5 [16]}}