| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2022 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: string becomes slack |
| Difficulty | Challenging +1.2 This is a standard M3 vertical circle problem requiring energy conservation, circular motion equations (tension=0 when slack), and projectile motion. Part (a) uses energy methods, part (b) applies projectile equations with given angle, and part (c) combines results. While multi-step, it follows a well-established template with clear signposting and no novel insight required—slightly above average due to the algebraic manipulation and integration of multiple mechanics topics. |
| Spec | 3.02h Motion under gravity: vector form6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\frac{1}{2}mu^2 - \frac{1}{2}mv^2 = mga(1+\cos\theta)\) | M1A2,1,0 | M1 for energy equation with correct terms; A2 correct, A1 for at most one error |
| \(mg\cos\theta = \frac{mv^2}{a}\) | M1A1 | M1 for equation of motion towards \(O\) with correct terms, condone sign errors and sin/cos confusion (\(R\) may appear); A1 correct (\(R=0\) must be used) |
| Eliminate \(\theta\) | M1 | |
| \(3v^2 = u^2 - 2ag\)* | A1* | Given answer correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Vertical motion: \(-\dfrac{a\sqrt{3}}{2} = (v\sin 30°)T - \dfrac{1}{2}gT^2\) | M1A1 | M1 correct terms, condone sign errors and sin/cos confusion; A1 correct equation |
| Solve for \(T\): \(T = \dfrac{v \pm \sqrt{v^2 + 4ag\sqrt{3}}}{2g}\) | M1 | For solving for \(T\) |
| Use \(v^2 = ag\dfrac{\sqrt{3}}{2}\) and \(T > 0\) to show \(T = \dfrac{2v}{g}\) | A1* | Given answer correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Horizontal motion: \(x = v\cos 30° \times \dfrac{2v}{g}\) | M1 | For equation for horizontal motion, condone sign errors and sin/cos confusion |
| \(= \dfrac{3a}{2}\) and hence taut \((= a + a\sin 30°)\)* | A1* | Given answer correctly justified |
# Question 5:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}mu^2 - \frac{1}{2}mv^2 = mga(1+\cos\theta)$ | M1A2,1,0 | M1 for energy equation with correct terms; A2 correct, A1 for at most one error |
| $mg\cos\theta = \frac{mv^2}{a}$ | M1A1 | M1 for equation of motion towards $O$ with correct terms, condone sign errors and sin/cos confusion ($R$ may appear); A1 correct ($R=0$ must be used) |
| Eliminate $\theta$ | M1 | |
| $3v^2 = u^2 - 2ag$* | A1* | Given answer correctly obtained |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Vertical motion: $-\dfrac{a\sqrt{3}}{2} = (v\sin 30°)T - \dfrac{1}{2}gT^2$ | M1A1 | M1 correct terms, condone sign errors and sin/cos confusion; A1 correct equation |
| Solve for $T$: $T = \dfrac{v \pm \sqrt{v^2 + 4ag\sqrt{3}}}{2g}$ | M1 | For solving for $T$ |
| Use $v^2 = ag\dfrac{\sqrt{3}}{2}$ and $T > 0$ to show $T = \dfrac{2v}{g}$ | A1* | Given answer correctly obtained |
## Part (c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Horizontal motion: $x = v\cos 30° \times \dfrac{2v}{g}$ | M1 | For equation for horizontal motion, condone sign errors and sin/cos confusion |
| $= \dfrac{3a}{2}$ and hence taut $(= a + a\sin 30°)$* | A1* | Given answer correctly justified |
5.
\begin{figure}[h]
\begin{center}
\begin{tikzpicture}[>=latex, line join=round, line cap=round]
% Define parameters for easy adjustment
\def\HeightA{3.6} % Height of point A
\def\Px{3.4} % X-coordinate of point P
\def\Pradius{0.14} % Radius of the black dot P
% Define main coordinates
\coordinate (O) at (0, 0);
\coordinate (A) at (0, \HeightA);
\coordinate (P) at (\Px, \Pradius); % Center of P is above the line by its radius
% Draw horizontal line (ground)
\draw (-4.5, 0) -- (4.2, 0);
% Draw vertical dashed line
\draw[densely dashed, semithick] (O) -- (A);
% Draw the solid slanted line
\draw[semithick] (A) -- (P) node[midway, above right=1pt] {$a$};
% Draw the angle arc and label
% Calculate the angle of the line AP from the vertical
\pgfmathsetmacro{\myangle}{atan(\Px / (\HeightA - \Pradius))}
\begin{scope}[shift={(A)}]
% Draw arc starting from the vertical dashed line (-90 degrees)
\draw (0, -1.6) arc (-90:{-90 + \myangle}:1.6);
% Place the angle label roughly bisecting the arc
\node at ({-90 + \myangle/2 + 2}:2.0) {$\theta^\circ$};
\end{scope}
% Labels for A and O
\node[above=2pt] at (A) {$A$};
\node[below=2pt] at (O) {$O$};
% Draw the particle P (resting exactly on the horizontal line) and its label
\fill (P) circle (\Pradius);
\node[right=3pt] at (P) {$P$};
\end{tikzpicture}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\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 $P$ is held at rest vertically below $O$, with the string taut, as shown in Figure 4.
The particle is then projected horizontally with speed $u$, where $u > \sqrt { 2 a g }$\\
Air resistance is modelled as being negligible.\\
At the instant when the string makes an angle $\theta$ with the upward vertical through $O$, the speed of $P$ is $v$ and the string goes slack.
\begin{enumerate}[label=(\alph*)]
\item Show that $3 v ^ { 2 } = u ^ { 2 } - 2 a g$
From the instant when the string goes slack to the instant when $O P$ is next horizontal, $P$ moves as a projectile.
The time from the instant when the string goes slack to the instant when $O P$ is next horizontal is $T$
Given that $\theta = 30 ^ { \circ }$
\item show that $T = \frac { 2 v } { g }$
\item Hence, show that the string goes taut again when it is next horizontal.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2022 Q5 [13]}}