Edexcel M3 2006 January — Question 7 15 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2006
SessionJanuary
Marks15
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeParticle on table with string above
DifficultyStandard +0.3 This is a standard M3 circular motion problem with clear geometry and straightforward application of Newton's second law in both horizontal and vertical directions. Part (a) is a 'show that' requiring resolving forces, part (b) follows directly, part (c) requires recognizing R≥0, and part (d) involves recalculating with new conditions. All steps are routine for M3 students with no novel insights required.
Spec3.03n Equilibrium in 2D: particle under forces6.05c Horizontal circles: conical pendulum, banked tracks

7. \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Figure 3} \includegraphics[alt={},max width=\textwidth]{67a9cf74-833f-4b4a-9fde-3c62dcc08e8c-5_625_1141_319_424}
\end{figure} A particle \(P\) of mass \(m\) is attached to one end of a light inextensible string of length \(2 a\). The other end of the string is fixed to a point \(A\) which is vertically above the point \(O\) on a smooth horizontal table. The particle \(P\) remains in contact with the surface of the table and moves in a circle with centre \(O\) and with angular speed \(\sqrt { \left( \frac { k g } { 3 a } \right) }\), where \(k\) is a constant. Throughout the motion the string remains taut and \(\angle A P O = 30 ^ { \circ }\), as shown in Figure 3.
  1. Show that the tension in the string is \(\frac { 2 k m g } { 3 }\).
  2. Find, in terms of \(m , g\) and \(k\), the normal reaction between \(P\) and the table.
  3. Deduce the range of possible values of \(k\). The angular speed of \(P\) is changed to \(\sqrt { \left( \frac { 2 g } { a } \right) }\). The particle \(P\) now moves in a horizontal circle above the table. The centre of this circle is \(X\).
  4. Show that \(X\) is the mid-point of \(O A\).

Question 7:
Part (a)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
N2L \(\leftarrow\): \(T\cos 30° = m(2a\cos 30°)\left(\dfrac{kg}{3a}\right)\)M1 A1
\(T = \dfrac{2kmg}{3}\) ✱A1 cso
(3 marks)
Part (b)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(\uparrow\; R = mg - T\sin 30°\)M1 A1
\(= mg\left(1 - \dfrac{k}{3}\right)\)A1
(3 marks)
Part (c)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\((R \geq 0) \Rightarrow k \leq 3\)M1 A1 ignore \(k > 0\), accept \(k < 3\)
(2 marks)
Part (d)
AnswerMarks Guidance
Answer/WorkingMarks Guidance
N2L \(\leftarrow\): \(T\cos\theta = m(2a\cos\theta)\left(\dfrac{2g}{a}\right)\), \((T = 4mg)\)M1 A1
\(\uparrow\; T\sin\theta = mg\)M1
Eliminating \(T\)M1
\(AX = 2a\sin\theta = \dfrac{1}{2}a\)A1
\(AO = 2a\sin 30° = a \Rightarrow AX = \dfrac{1}{2}AO\) ✱B1, A1 cso
(7 marks) — Total 15 marks
## Question 7:

**Part (a)**

| Answer/Working | Marks | Guidance |
|---|---|---|
| N2L $\leftarrow$: $T\cos 30° = m(2a\cos 30°)\left(\dfrac{kg}{3a}\right)$ | M1 A1 | |
| $T = \dfrac{2kmg}{3}$ ✱ | A1 | cso |

**(3 marks)**

**Part (b)**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $\uparrow\; R = mg - T\sin 30°$ | M1 A1 | |
| $= mg\left(1 - \dfrac{k}{3}\right)$ | A1 | |

**(3 marks)**

**Part (c)**

| Answer/Working | Marks | Guidance |
|---|---|---|
| $(R \geq 0) \Rightarrow k \leq 3$ | M1 A1 | ignore $k > 0$, accept $k < 3$ |

**(2 marks)**

**Part (d)**

| Answer/Working | Marks | Guidance |
|---|---|---|
| N2L $\leftarrow$: $T\cos\theta = m(2a\cos\theta)\left(\dfrac{2g}{a}\right)$, $(T = 4mg)$ | M1 A1 | |
| $\uparrow\; T\sin\theta = mg$ | M1 | |
| Eliminating $T$ | M1 | |
| $AX = 2a\sin\theta = \dfrac{1}{2}a$ | A1 | |
| $AO = 2a\sin 30° = a \Rightarrow AX = \dfrac{1}{2}AO$ ✱ | B1, A1 | cso |

**(7 marks) — Total 15 marks**
7.

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 3}
  \includegraphics[alt={},max width=\textwidth]{67a9cf74-833f-4b4a-9fde-3c62dcc08e8c-5_625_1141_319_424}
\end{center}
\end{figure}

A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $2 a$. The other end of the string is fixed to a point $A$ which is vertically above the point $O$ on a smooth horizontal table. The particle $P$ remains in contact with the surface of the table and moves in a circle with centre $O$ and with angular speed $\sqrt { \left( \frac { k g } { 3 a } \right) }$, where $k$ is a constant. Throughout the motion the string remains taut and $\angle A P O = 30 ^ { \circ }$, as shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\item Show that the tension in the string is $\frac { 2 k m g } { 3 }$.
\item Find, in terms of $m , g$ and $k$, the normal reaction between $P$ and the table.
\item Deduce the range of possible values of $k$.

The angular speed of $P$ is changed to $\sqrt { \left( \frac { 2 g } { a } \right) }$. The particle $P$ now moves in a horizontal circle above the table. The centre of this circle is $X$.
\item Show that $X$ is the mid-point of $O A$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3 2006 Q7 [15]}}