| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2018 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: complete revolution conditions |
| Difficulty | Standard +0.8 This is a challenging M3 vertical circle problem requiring energy methods, circular motion dynamics, and careful analysis of tension conditions at different points. Part (a) requires deriving a general tension formula using both energy conservation and centripetal force. Part (b) needs identifying the critical condition for complete circles (T=0 at top). Part (c) involves simultaneous equations relating tensions at top and bottom positions. The multi-step reasoning and need to connect multiple concepts (energy, forces, critical conditions) places this above average difficulty, though it follows a standard vertical circle framework taught in M3. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{1}{2}mu^2 - \frac{1}{2}mv^2 = mga\sin\theta\) | M1A1A1 | Energy equation from \(A\) to general position; must have difference of two KE terms and gain of PE; \(-1\) each error |
| \(T + mg\sin\theta = m\dfrac{v^2}{a}\) | M1A1A1 | Equation of motion along radius at general position; weight resolved (sin or cos); acceleration in either form; resultant force correct; acceleration correct |
| \(T + mg\sin\theta = \dfrac{mu^2}{a} - 2mg\sin\theta\) | dM1 | Eliminate \(v^2\) between two equations; depends on both previous M marks |
| \(T = \dfrac{m}{a}(u^2 - 3ga\sin\theta)\) ✱ | A1cso (8) | Obtain given expression for \(T\) with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| At top \(T\geq 0 \Rightarrow u^2 \geq 3ag\) | M1 | Use \(\sin\theta=1\) so \(T\geq 0\) at top; allow \(\geq\) or \(>\); or state min \(u\) when \(T=0\) at top |
| \(u_{\min} = \sqrt{3ag}\) | A1 (2) | \(u_{\min}>3ag\) or \(u_{\min}\geq 3ag\) scores A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Least at top: \(T_{\text{least}} = \dfrac{m}{a}(u^2-3ag)\) \((=S)\) | M1A1 | Use result from (a) for tension at top \((\theta=90°)\) |
| Greatest at bottom: \(T_{\text{greatest}} = \dfrac{m}{a}(u^2+3ag)\) \((=4S)\) | A1 | One mark for each correct (M1A1 for either, A1 for second); alt: energy equation from \(A\) to either or both top and bottom |
| \(4\times\dfrac{m}{a}(u^2-3ag) = \dfrac{m}{a}(u^2+3ag)\) | dM1 | Form equation with \(4\times\) their least \(=\) their greatest; both tensions of form \(\frac{m}{a}(u^2\pm kag)\), \(k\neq 0\); depends on previous M mark |
| \(3u^2=15ag,\quad u=\sqrt{5ag}\) | A1 (5) [15] | Correct expression for \(u\) |
# Question 6:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}mu^2 - \frac{1}{2}mv^2 = mga\sin\theta$ | M1A1A1 | Energy equation from $A$ to general position; must have difference of two KE terms and gain of PE; $-1$ each error |
| $T + mg\sin\theta = m\dfrac{v^2}{a}$ | M1A1A1 | Equation of motion along radius at general position; weight resolved (sin or cos); acceleration in either form; resultant force correct; acceleration correct |
| $T + mg\sin\theta = \dfrac{mu^2}{a} - 2mg\sin\theta$ | dM1 | Eliminate $v^2$ between two equations; depends on both previous M marks |
| $T = \dfrac{m}{a}(u^2 - 3ga\sin\theta)$ ✱ | A1cso (8) | Obtain given expression for $T$ with no errors seen |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| At top $T\geq 0 \Rightarrow u^2 \geq 3ag$ | M1 | Use $\sin\theta=1$ so $T\geq 0$ at top; allow $\geq$ or $>$; or state min $u$ when $T=0$ at top |
| $u_{\min} = \sqrt{3ag}$ | A1 (2) | $u_{\min}>3ag$ or $u_{\min}\geq 3ag$ scores A0 |
## Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Least at top: $T_{\text{least}} = \dfrac{m}{a}(u^2-3ag)$ $(=S)$ | M1A1 | Use result from (a) for tension at top $(\theta=90°)$ |
| Greatest at bottom: $T_{\text{greatest}} = \dfrac{m}{a}(u^2+3ag)$ $(=4S)$ | A1 | One mark for each correct (M1A1 for either, A1 for second); alt: energy equation from $A$ to either or both top and bottom |
| $4\times\dfrac{m}{a}(u^2-3ag) = \dfrac{m}{a}(u^2+3ag)$ | dM1 | Form equation with $4\times$ their least $=$ their greatest; both tensions of form $\frac{m}{a}(u^2\pm kag)$, $k\neq 0$; depends on previous M mark |
| $3u^2=15ag,\quad u=\sqrt{5ag}$ | A1 (5) [15] | Correct expression for $u$ |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2cf74ba3-857a-4ce9-ab5b-e6203b279161-18_481_606_246_667}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $a$. The other end of the string is attached to a fixed point $O$. The particle is held at the point $A$, where $O A = a$ and $O A$ is horizontal. The particle is projected vertically upwards with speed $u$, as shown in Figure 2. When the string makes an angle $\theta$ with the horizontal through $O$ and the string is still taut, the tension in the string is $T$.
\begin{enumerate}[label=(\alph*)]
\item Show that $T = \frac { m } { a } \left( u ^ { 2 } - 3 a g \sin \theta \right)$
The particle moves in complete circles.
\item Find, in terms of $a$ and $g$, the minimum value of $u$.
Given that the least tension in the string is $S$ and the greatest tension in the string is $4 S$,
\item find, in terms of $a$ and $g$, an expression for $u$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2018 Q6 [15]}}