| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2018 |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: complete revolution conditions |
| Difficulty | Standard +0.3 This is a standard M3 vertical circle problem requiring energy conservation and Newton's second law at critical points. Part (a) uses the standard condition for complete circles (minimum speed at top), and part (b) applies tension conditions with straightforward algebra. Both are textbook exercises with well-rehearsed techniques, making this slightly easier than average. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods |
| VIIIV SIHI NI JIIYM ION OC | VIIVV SIHI NI JIIIAM ION OO | VEYV SIHIL NI JIIIM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Energy to top: \(\frac{1}{2}\times3m\times u^2 - \frac{1}{2}\times3mv^2 = 3mga\) | M1, A1 | Attempting energy equation; can be to general point. Mass may be missing but use of \(v^2=u^2+2as\) scores M1. Correct equation from \(A\) to top |
| NL2 at top: \(T + 3mg = 3m\frac{v^2}{a}\) | M1, A1 | Attempting equation of motion along radius at top; correct equation with acceleration in form \(\frac{v^2}{r}\) |
| \(T = 3m\frac{u^2}{a} - 6mg - 3mg\) | dM1 | Eliminating \(v^2\) to get expression for \(T\) dependent on both previous M marks |
| \(T \geqslant 0 \Rightarrow \frac{u^2}{a} \geqslant 3g\) | M1 | Using \(T \geqslant 0\) at top to obtain inequality connecting \(a\), \(g\) and \(u\) |
| \(u^2 \geqslant 3ag\) | A1 cso | Re-arranging to obtain the given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{1}{2}\times3m\times V^2 - \frac{1}{2}\times3mu^2 = 3mga\) | M1 | Attempting energy equation to the bottom, maybe from \(A\) or from the top |
| \(T_{\max} - 3mg = 3m\frac{V^2}{a}\) | M1 | Attempting equation of motion along radius at the bottom |
| \(T_{\max} = 3mg + 6mg + 3m\frac{u^2}{a}\) | A1 | Correct expression for the max tension |
| \(T_{\min} = 3m\frac{u^2}{a} - 9mg\) | — | (Expression for min tension at top) |
| \(9mg + 3m\frac{u^2}{a} = 3\!\left(3m\frac{u^2}{a} - 9mg\right)\) | dM1 | Forming equation connecting their tension at top with their tension at bottom; if 3 is multiplying wrong tension this mark can still be gained. Dependent on both previous M marks |
| \(u^2 = 6ag\) | A1 cso | Obtaining the given answer |
## Question 4:
### Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Energy to top: $\frac{1}{2}\times3m\times u^2 - \frac{1}{2}\times3mv^2 = 3mga$ | M1, A1 | Attempting energy equation; can be to general point. Mass may be missing but use of $v^2=u^2+2as$ scores M1. Correct equation from $A$ to top |
| NL2 at top: $T + 3mg = 3m\frac{v^2}{a}$ | M1, A1 | Attempting equation of motion along radius at top; correct equation with acceleration in form $\frac{v^2}{r}$ |
| $T = 3m\frac{u^2}{a} - 6mg - 3mg$ | dM1 | Eliminating $v^2$ to get expression for $T$ dependent on both previous M marks |
| $T \geqslant 0 \Rightarrow \frac{u^2}{a} \geqslant 3g$ | M1 | Using $T \geqslant 0$ at top to obtain inequality connecting $a$, $g$ and $u$ |
| $u^2 \geqslant 3ag$ | A1 cso | Re-arranging to obtain the given answer |
**(7 marks)**
### Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}\times3m\times V^2 - \frac{1}{2}\times3mu^2 = 3mga$ | M1 | Attempting energy equation to the bottom, maybe from $A$ or from the top |
| $T_{\max} - 3mg = 3m\frac{V^2}{a}$ | M1 | Attempting equation of motion along radius at the bottom |
| $T_{\max} = 3mg + 6mg + 3m\frac{u^2}{a}$ | A1 | Correct expression for the max tension |
| $T_{\min} = 3m\frac{u^2}{a} - 9mg$ | — | (Expression for min tension at top) |
| $9mg + 3m\frac{u^2}{a} = 3\!\left(3m\frac{u^2}{a} - 9mg\right)$ | dM1 | Forming equation connecting their tension at top with their tension at bottom; if 3 is multiplying wrong tension this mark can still be gained. Dependent on both previous M marks |
| $u^2 = 6ag$ | A1 cso | Obtaining the given answer |
**(5 marks) — Total: 12 marks**4.
\begin{figure}[h]
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\includegraphics[alt={},max width=\textwidth]{bb73b211-7629-4ed7-9b71-91841c29bb85-12_403_497_251_712}
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\caption{Figure 2}
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\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 $u ^ { 2 } = 6 a g$.\\
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VIIIV SIHI NI JIIYM ION OC & VIIVV SIHI NI JIIIAM ION OO & VEYV SIHIL NI JIIIM ION OO \\
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\hfill \mbox{\textit{Edexcel M3 2018 Q4 [12]}}