Edexcel M3 2010 June — Question 1 7 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2010
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeConical pendulum – horizontal circle in free space (no surface)
DifficultyModerate -0.8 This is a standard conical pendulum problem requiring basic resolution of forces and application of circular motion formula (T cos θ = mg, T sin θ = mv²/r). The geometry is given (5-12-13 triangle), making it straightforward with no problem-solving insight needed—just routine application of two equations.
Spec6.05c Horizontal circles: conical pendulum, banked tracks

1. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{049ea68c-d15f-41f8-860e-0816d36a2748-02_458_516_281_712} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} A garden game is played with a small ball \(B\) of mass \(m\) attached to one end of a light inextensible string of length 13l. The other end of the string is fixed to a point \(A\) on a vertical pole as shown in Figure 1. The ball is hit and moves with constant speed in a horizontal circle of radius \(5 l\) and centre \(C\), where \(C\) is vertically below \(A\). Modelling the ball as a particle, find
  1. the tension in the string,
  2. the speed of the ball.

Question 1:
Part (a)
AnswerMarks Guidance
WorkingMarks Notes
\(\cos\alpha = \dfrac{12}{13}\)B1
\(R(\uparrow)\quad T\cos\alpha = mg\)M1 Resolving vertically
\(T \times \dfrac{12}{13} = mg\)
\(T = \dfrac{13}{12}mg\)A1 (3) oe
Part (b)
AnswerMarks Guidance
WorkingMarks Notes
Equation of motion: \(T\sin\alpha = m\dfrac{v^2}{5l}\)M1 A1
\(\dfrac{13mg}{12} \times \dfrac{5}{13} = m\dfrac{v^2}{5l}\)M1 dep
\(v^2 = \dfrac{25gl}{12}\)
\(v = \dfrac{5}{2}\sqrt{\dfrac{gl}{3}}\)A1 (4) Accept \(5\sqrt{\dfrac{gl}{12}}\) or \(\sqrt{\dfrac{25gl}{12}}\) or any equivalent
Total: [7]
# Question 1:

## Part (a)

| Working | Marks | Notes |
|---------|-------|-------|
| $\cos\alpha = \dfrac{12}{13}$ | B1 | |
| $R(\uparrow)\quad T\cos\alpha = mg$ | M1 | Resolving vertically |
| $T \times \dfrac{12}{13} = mg$ | | |
| $T = \dfrac{13}{12}mg$ | A1 | (3) oe |

## Part (b)

| Working | Marks | Notes |
|---------|-------|-------|
| Equation of motion: $T\sin\alpha = m\dfrac{v^2}{5l}$ | M1 A1 | |
| $\dfrac{13mg}{12} \times \dfrac{5}{13} = m\dfrac{v^2}{5l}$ | M1 dep | |
| $v^2 = \dfrac{25gl}{12}$ | | |
| $v = \dfrac{5}{2}\sqrt{\dfrac{gl}{3}}$ | A1 | (4) Accept $5\sqrt{\dfrac{gl}{12}}$ or $\sqrt{\dfrac{25gl}{12}}$ or any equivalent |

**Total: [7]**

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1.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{049ea68c-d15f-41f8-860e-0816d36a2748-02_458_516_281_712}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

A garden game is played with a small ball $B$ of mass $m$ attached to one end of a light inextensible string of length 13l. The other end of the string is fixed to a point $A$ on a vertical pole as shown in Figure 1. The ball is hit and moves with constant speed in a horizontal circle of radius $5 l$ and centre $C$, where $C$ is vertically below $A$. Modelling the ball as a particle, find
\begin{enumerate}[label=(\alph*)]
\item the tension in the string,
\item the speed of the ball.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3 2010 Q1 [7]}}