AQA M2 2008 January — Question 5 9 marks

Exam BoardAQA
ModuleM2 (Mechanics 2)
Year2008
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeTwo strings, two fixed points
DifficultyStandard +0.3 This is a standard M2 circular motion problem with two strings. Part (a) is direct application of a=v²/r, part (b) requires resolving forces vertically and horizontally with given answer to verify, and part (c) follows immediately. The geometry is straightforward (right-angled triangle), and all steps are routine applications of Newton's second law in circular motion. Slightly easier than average due to the 'show that' scaffold in part (b).
Spec6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks

5 Two light inextensible strings, of lengths 0.4 m and 0.2 m , each have one end attached to a particle, \(P\), of mass 4 kg . The other ends of the strings are attached to the points \(A\) and \(B\) respectively. The point \(A\) is vertically above the point \(B\). The particle moves in a horizontal circle, centre \(B\) and radius 0.2 m , at a speed of \(2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The particle and strings are shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{1bc18163-b20e-4dc6-bd35-496efec8dc73-4_396_558_587_735} $$\text { ← } 0.2 \mathrm {~m} \longrightarrow$$
  1. Calculate the magnitude of the acceleration of the particle.
  2. Show that the tension in string \(P A\) is 45.3 N , correct to three significant figures.
  3. Find the tension in string \(P B\).

(a) Acceleration is \(\frac{v^2}{r}\)
AnswerMarks Guidance
\(= \frac{2^2}{0.2} = 20\) m s\(^{-2}\)M1 A1 2 marks
(b) \(\theta = 30°\)
Resolve vertically:
\(T_1 \cos \theta = mg\)
\(T_1 \cos \theta = 4g\)
AnswerMarks Guidance
\(T_1 = 45.3\) NB1 M1 A1 A1 4 marks
(c) Resolve horizontally:
\(T_1 \sin \theta + T_2 = \frac{mv^2}{r}\)
\(45.3\sin \theta + T_2 = 4 \times 20\)
AnswerMarks Guidance
\(T_2 = 57.4\) NM1A1 A1 3 marks
Total: 9 marks
**(a)** Acceleration is $\frac{v^2}{r}$

$= \frac{2^2}{0.2} = 20$ m s$^{-2}$ | M1 A1 | 2 marks

**(b)** $\theta = 30°$

Resolve vertically:
$T_1 \cos \theta = mg$
$T_1 \cos \theta = 4g$
$T_1 = 45.3$ N | B1 M1 A1 A1 | 4 marks | AG

**(c)** Resolve horizontally:
$T_1 \sin \theta + T_2 = \frac{mv^2}{r}$

$45.3\sin \theta + T_2 = 4 \times 20$

$T_2 = 57.4$ N | M1A1 A1 | 3 marks | M1 for 3 terms, 2 correct; Condone 57.3 N

**Total: 9 marks**

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5 Two light inextensible strings, of lengths 0.4 m and 0.2 m , each have one end attached to a particle, $P$, of mass 4 kg . The other ends of the strings are attached to the points $A$ and $B$ respectively. The point $A$ is vertically above the point $B$. The particle moves in a horizontal circle, centre $B$ and radius 0.2 m , at a speed of $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The particle and strings are shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{1bc18163-b20e-4dc6-bd35-496efec8dc73-4_396_558_587_735}

$$\text { ← } 0.2 \mathrm {~m} \longrightarrow$$
\begin{enumerate}[label=(\alph*)]
\item Calculate the magnitude of the acceleration of the particle.
\item Show that the tension in string $P A$ is 45.3 N , correct to three significant figures.
\item Find the tension in string $P B$.
\end{enumerate}

\hfill \mbox{\textit{AQA M2 2008 Q5 [9]}}