| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2010 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: tension at specific point |
| Difficulty | Challenging +1.2 This is a standard M3 vertical circular motion problem requiring energy conservation and Newton's second law in the radial direction. Part (a) is a straightforward 'show that' using conservation of energy with a given initial position and speed. Part (b) requires finding tension at extreme positions (top and bottom), which is routine application of T = mv²/r ± mg. The geometry is slightly non-standard (starting 2a below horizontal) but the methods are textbook applications with no novel insight required. |
| Spec | 3.03k Connected particles: pulleys and equilibrium6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\frac{1}{2}m \times 2ag - \frac{1}{2}mv^2 = mg(2a - 3a\sin\theta)\) | M1 A1=A1 | Energy equation |
| Leading to \(v^2 = 2ga(3\sin\theta - 1)\) | M1 A1 | cso |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Minimum value of \(T\) is when \(v = 0 \Rightarrow \sin\theta = \frac{1}{3}\) | B1 | |
| \(T = mg\sin\theta = \frac{mg}{3}\) | M1 A1 | |
| Maximum value of \(T\) is when \(\theta = \frac{\pi}{2}\), \((v^2 = 4ag)\) | ||
| \(\uparrow \quad T = \frac{mv^2}{3a} + mg\) | M1 A1 | |
| \(= \frac{7mg}{3}\) | A1 | (6) |
| \(\left(\frac{mg}{3} \leq T \leq \frac{7mg}{3}\right)\) | [11] |
# Question 5:
## Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\frac{1}{2}m \times 2ag - \frac{1}{2}mv^2 = mg(2a - 3a\sin\theta)$ | M1 A1=A1 | Energy equation |
| Leading to $v^2 = 2ga(3\sin\theta - 1)$ | M1 A1 | cso |
| | **(5)** | |
## Part (b)
| Working | Marks | Guidance |
|---------|-------|----------|
| Minimum value of $T$ is when $v = 0 \Rightarrow \sin\theta = \frac{1}{3}$ | B1 | |
| $T = mg\sin\theta = \frac{mg}{3}$ | M1 A1 | |
| Maximum value of $T$ is when $\theta = \frac{\pi}{2}$, $(v^2 = 4ag)$ | | |
| $\uparrow \quad T = \frac{mv^2}{3a} + mg$ | M1 A1 | |
| $= \frac{7mg}{3}$ | A1 | **(6)** |
| $\left(\frac{mg}{3} \leq T \leq \frac{7mg}{3}\right)$ | | **[11]** |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d831556d-fdf3-4639-9a89-6d3b372d3446-10_590_858_242_575}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
One end $A$ of a light inextensible string of length $3 a$ is attached to a fixed point. A particle of mass $m$ is attached to the other end $B$ of the string. The particle is held in equilibrium at a distance $2 a$ below the horizontal through $A$, with the string taut. The particle is then projected with speed $\sqrt { } ( 2 a g )$, in the direction perpendicular to $A B$, in the vertical plane containing $A$ and $B$, as shown in Figure 4. In the subsequent motion the string remains taut. When $A B$ is at an angle $\theta$ below the horizontal, the speed of the particle is $v$ and the tension in the string is $T$.
\begin{enumerate}[label=(\alph*)]
\item Show that $v ^ { 2 } = 2 \operatorname { ag } ( 3 \sin \theta - 1 )$.
\item Find the range of values of $T$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2010 Q5 [11]}}