| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle with peg/obstacle |
| Difficulty | Standard +0.8 This is a multi-part vertical circle problem requiring energy conservation, circular motion dynamics, and impulse analysis. Part (a) requires showing the particle reaches the top (standard but requires careful energy work), part (b) applies energy conservation across the full circle, and part (c) involves post-impulse motion with a novel 'instantaneous rest' condition requiring both energy and tension considerations. The combination of techniques and the non-standard impulse element elevates this above routine M3 questions. |
| Spec | 3.02h Motion under gravity: vector form6.02i Conservation of energy: mechanical energy principle6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| Energy to \(B\): \(\frac{1}{2}m(3\sqrt{ag})^2 - \frac{1}{2}mV^2 = mag\) leading to \(V^2 = 7ag\) | M1 A1 | |
| NL2 along radius at \(B\): \(T_B + mg = m\frac{V^2}{a}\) | M1 A1 | |
| \(T_B + mg = 7mg\) | M1 | |
| \(T_B = 6mg\); \(T_B > 0 \Rightarrow\) particle reaches \(B\) | A1 | (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| Energy to \(C\): \(\frac{1}{2}mU^2 - \frac{1}{2}m(3\sqrt{ag})^2 = mag\) | M1 | |
| \(U^2 = 2ag + 9ag\) | ||
| \(U = \sqrt{11ga}\) | A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Notes |
| Energy from \(C\) to rest: \(\frac{1}{2}m\left(\frac{5}{12}\sqrt{11ag}\right)^2 = mga(1-\cos\theta)\) | M1 A1 | |
| \(\frac{25}{144}\times 11ag = 2ga(1-\cos\theta)\) | ||
| \(\cos\theta = \frac{1}{2}\left(2-\frac{25\times 11}{144}\right)\) | M1 | |
| \(\theta = 87.4...° \approx 87°\) (or 1.5 rad) or better | A1 | (4) |
# Question 6:
## Part (a):
| Working/Answer | Marks | Notes |
|---|---|---|
| Energy to $B$: $\frac{1}{2}m(3\sqrt{ag})^2 - \frac{1}{2}mV^2 = mag$ leading to $V^2 = 7ag$ | M1 A1 | |
| NL2 along radius at $B$: $T_B + mg = m\frac{V^2}{a}$ | M1 A1 | |
| $T_B + mg = 7mg$ | M1 | |
| $T_B = 6mg$; $T_B > 0 \Rightarrow$ particle reaches $B$ | A1 | (6) |
## Part (b):
| Working/Answer | Marks | Notes |
|---|---|---|
| Energy to $C$: $\frac{1}{2}mU^2 - \frac{1}{2}m(3\sqrt{ag})^2 = mag$ | M1 | |
| $U^2 = 2ag + 9ag$ | | |
| $U = \sqrt{11ga}$ | A1 | (2) |
## Part (c):
| Working/Answer | Marks | Notes |
|---|---|---|
| Energy from $C$ to rest: $\frac{1}{2}m\left(\frac{5}{12}\sqrt{11ag}\right)^2 = mga(1-\cos\theta)$ | M1 A1 | |
| $\frac{25}{144}\times 11ag = 2ga(1-\cos\theta)$ | | |
| $\cos\theta = \frac{1}{2}\left(2-\frac{25\times 11}{144}\right)$ | M1 | |
| $\theta = 87.4...° \approx 87°$ (or 1.5 rad) or better | A1 | (4) |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{826ad8ff-6e5c-4224-88ba-e78b79d1bc21-11_574_540_226_701}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
A particle $P$ 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 point $B$ is vertically above $O$ and the point $C$ is vertically below $O$, with $O B = O C = a$, as shown in Figure 5. The particle is projected vertically upwards with speed $3 \sqrt { } ( a g )$.
\begin{enumerate}[label=(\alph*)]
\item Show that $P$ will pass through $B$.
\item Find the speed of $P$ as it reaches $C$.
As $P$ passes through $C$ it receives an impulse. Immediately after this, the speed of $P$ is $\frac { 5 } { 12 } \sqrt { } ( 11 a g )$ and the direction of motion of $P$ is unchanged.
\item Find the angle between the string and the downward vertical when $P$ comes to instantaneous rest.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2011 Q6 [12]}}