| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: complete revolution conditions |
| Difficulty | Standard +0.8 This is a standard M3 vertical circle problem requiring energy conservation and Newton's second law at critical points, but the algebraic manipulation to show u²=6ag from the tension ratio condition requires careful handling of multiple equations. The 'show that' format and the need to work with the 3:1 tension ratio elevates it above routine exercises. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{1}{2}\times 3m\times u^2 - \frac{1}{2}\times 3m\times v^2 = 3mga\) | M1A1 | Energy equation to top; M1 can be to general point; mass can be missing but \(v^2=u^2+2as\) scores M0 |
| \(T + 3mg = 3m\frac{v^2}{a}\) | M1A1 | Equation of motion along radius at top; acceleration in form \(\frac{v^2}{r}\) |
| \(T = 3m\frac{u^2}{a} - 6mg - 3mg\) | DM1 | Eliminate \(v^2\); dependent on both previous M marks |
| \(T \geq 0 \Rightarrow \frac{u^2}{a} \geq 3g\) | M1 | Use \(T \geq 0\) at top to obtain inequality connecting \(a, g\) and \(u\) |
| \(u^2 \geq 3ag\) | A1 cso | Re-arrange to give the given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\frac{1}{2}\times 3m\times V^2 - \frac{1}{2}\times 3m\times u^2 = 3mga\) | M1 | Energy equation to bottom (from \(A\) or from top) |
| \(T_{\max} - 3mg = 3m\frac{V^2}{a}\) | M1 | Equation of motion along radius at bottom |
| \(T_{\max} = 3mg + 6mg + 3m\frac{u^2}{a}\) | A1 | Correct expression for max tension |
| \(T_{\min} = 3m\frac{u^2}{a} - 9mg\) | ||
| \(9mg + 3m\frac{u^2}{a} = 3\!\left(3m\frac{u^2}{a} - 9mg\right)\) | DM1 | Forming equation connecting their tensions; dependent on both previous M marks |
| \(u^2 = 6ag\) | A1 cso | Obtaining the given answer |
## Question 4(a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}\times 3m\times u^2 - \frac{1}{2}\times 3m\times v^2 = 3mga$ | M1A1 | Energy equation to top; M1 can be to general point; mass can be missing but $v^2=u^2+2as$ scores M0 |
| $T + 3mg = 3m\frac{v^2}{a}$ | M1A1 | Equation of motion along radius at top; acceleration in form $\frac{v^2}{r}$ |
| $T = 3m\frac{u^2}{a} - 6mg - 3mg$ | DM1 | Eliminate $v^2$; dependent on both previous M marks |
| $T \geq 0 \Rightarrow \frac{u^2}{a} \geq 3g$ | M1 | Use $T \geq 0$ at top to obtain inequality connecting $a, g$ and $u$ |
| $u^2 \geq 3ag$ | A1 cso | Re-arrange to give the given answer |
**Total: (7)**
## Question 4(b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}\times 3m\times V^2 - \frac{1}{2}\times 3m\times u^2 = 3mga$ | M1 | Energy equation to bottom (from $A$ or from top) |
| $T_{\max} - 3mg = 3m\frac{V^2}{a}$ | M1 | Equation of motion along radius at bottom |
| $T_{\max} = 3mg + 6mg + 3m\frac{u^2}{a}$ | A1 | Correct expression for max tension |
| $T_{\min} = 3m\frac{u^2}{a} - 9mg$ | | |
| $9mg + 3m\frac{u^2}{a} = 3\!\left(3m\frac{u^2}{a} - 9mg\right)$ | DM1 | Forming equation connecting their tensions; dependent on both previous M marks |
| $u^2 = 6ag$ | A1 cso | Obtaining the given answer |
**Total: (5) [12]**
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b7cfcf0a-8f54-4350-8e07-a3b51d94d0f2-07_408_509_246_705}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A particle of mass $3 m$ 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 $O A$ is horizontal and $O A = a$. The particle is projected vertically downwards from $A$ with speed $u$, as shown in Figure 2. The particle moves in complete vertical circles.\\
(a) Show that $u ^ { 2 } \geqslant 3 a g$.
Given that the greatest tension in the string is three times the least tension in the string, (b) show that $u ^ { 2 } = 6 a g$.
\hfill \mbox{\textit{Edexcel M3 2015 Q4 [12]}}