Edexcel M3 2022 January — Question 5 13 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2022
SessionJanuary
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 2
TypeVertical circle: string becomes slack
DifficultyChallenging +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.
Spec3.02h Motion under gravity: vector form6.05e Radial/tangential acceleration

5. \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Figure 4}
\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.
  1. 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 \(\theta = 30 ^ { \circ }\)
  2. show that \(T = \frac { 2 v } { g }\)
  3. Hence, show that the string goes taut again when it is next horizontal.

Question 5:
Part (a):
AnswerMarks Guidance
Working/AnswerMark 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):
AnswerMarks Guidance
Working/AnswerMark 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):
AnswerMarks Guidance
Working/AnswerMark 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]}}