| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Elastic string – conical pendulum (string inclined to vertical) |
| Difficulty | Standard +0.8 This is a multi-part circular motion problem requiring elastic string mechanics, resolution of forces in 3D geometry, and energy considerations. Part (i) involves Hooke's law with geometric reasoning to find tension and mass. Part (ii) requires showing angular speed is independent of angle through force resolution. Part (iii) adds energy analysis comparing elastic PE and KE. The geometric setup with the string passing through a ring and the particle moving in a horizontal circle below requires careful spatial reasoning. While systematic, it demands integration of multiple M2 concepts and extended algebraic manipulation across 12 marks, placing it moderately above average difficulty. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks |
|---|---|
| (ii) | 0.4 |
| Answer | Marks |
|---|---|
| Both calculations give ω = 5 | M1 |
| Answer | Marks |
|---|---|
| B1 | λext |
| Answer | Marks |
|---|---|
| (iii) | 0.4 2 |
| Answer | Marks |
|---|---|
| θ = 45 | B1 |
| Answer | Marks |
|---|---|
| [4] | ft candidate’s value of ω |
Question 7:
--- 7 (i)
(ii) ---
7 (i)
(ii) | 0.4
15
cosθ
T =
2
3
T = AG
cosθ
Tcosθ = mg
m = 0.3
r = 0.4tanθ
0.3v2
=Tsinθ OR 0.3ω 2 r = Tsinθ
r
3
0.3ω 2 (0.4tanθ) = × sinθ
cosθ
ω = 5
SC
Candidates who choose at least two specific
values of θ:
Calculation of r twice
Both calculations give ω = 5 | M1
A1
M1
A1
[4]
B1
M1
A1
A1
[4]
B1
B1 | λext
Uses T =
2
Resolves vertically for P
nd
Newton’s 2 law with correct
expression for radial accn, ft cv(m(i))
(iii) | 0.4 2
15
cosθ
EPE =
2×2
( )2
0.3 5×0.4tanθ
KE =
2
0.4 2
15
cosθ 0.3(2tanθ)2
= ×2
2×2 2
2 2 2
cos θ tan θ = 0.5 OR sin θ = 0.5
θ = 45 | B1
B1
M1
A1
[4] | ft candidate’s value of ω
Award if × 2 is with wrong term
www\includegraphics{figure_7}
One end of a light elastic string with modulus of elasticity $15$ N is attached to a fixed point $A$ which is $2$ m vertically above a fixed small smooth ring $R$. The string has natural length $2$ m and it passes through $R$. The other end of the string is attached to a particle $P$ of mass $m$ kg which moves with constant angular speed $\omega$ rad s$^{-1}$ in a horizontal circle which has its centre $0.4$ m vertically below the ring. $PR$ makes an acute angle $\theta$ with the vertical (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Show that the tension in the string is $\frac{3}{\cos\theta}$ N and hence find the value of $m$. [4]
\item Show that the value of $\omega$ does not depend on $\theta$. [4]
\end{enumerate}
It is given that for one value of $\theta$ the elastic potential energy stored in the string is twice the kinetic energy of $P$.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Find this value of $\theta$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2014 Q7 [12]}}