| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | String through hole – lower particle also moves in horizontal circle (conical pendulum below) |
| Difficulty | Challenging +1.2 This is a multi-part circular motion problem requiring force analysis in two connected systems (particle on surface and hanging particle), with elastic string mechanics. Part (i) involves finding limiting conditions when the elastic string reaches natural length, requiring equilibrium of Q and circular motion of P. Part (ii) requires solving simultaneous equations involving tension, elastic force (Hooke's law), and circular motion at given angular speed. While it involves multiple concepts and careful geometric reasoning, the techniques are standard M2 material with straightforward application of T=mrω² and elastic string formulae. More challenging than basic circular motion but not requiring novel insight. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| 6(i) | r [= 0.6 – (0.4-0.3)]= 0.5 | B1 |
| T = 0.3g | B1 | Resolve vertically for Q |
| 0.2v2 /0.5=0.3g | M1 | Use Newton’s Second Law |
| Answer | Marks |
|---|---|
| v = 2.74 ms−1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 6(ii) | r = 0.5 + e | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 0.3 | B1 | Use T = λx/l |
| 0.2×82(5+e)=50e+0.3g | M1 | Use Newton’s Second Law |
| Answer | Marks |
|---|---|
| 50−12.8 | A1 |
| HP = 0.591 m | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 6:
--- 6(i) ---
6(i) | r [= 0.6 – (0.4-0.3)]= 0.5 | B1
T = 0.3g | B1 | Resolve vertically for Q
0.2v2 /0.5=0.3g | M1 | Use Newton’s Second Law
horizontally for P
v = 2.74 ms−1 | A1
4
--- 6(ii) ---
6(ii) | r = 0.5 + e | B1 | e = extension of the string
15e
T = =50e
0.3 | B1 | Use T = λx/l
0.2×82(5+e)=50e+0.3g | M1 | Use Newton’s Second Law
horizontally with a=rω2
( )
6.4−3
e= ( = 0.0914)
( )
50−12.8 | A1
HP = 0.591 m | A1
5
Question | Answer | Marks | Guidance\includegraphics{figure_6}
A particle $P$ of mass $0.2 \text{ kg}$ is attached to one end of a light inextensible string of length $0.6 \text{ m}$. The other end of the string is attached to a particle $Q$ of mass $0.3 \text{ kg}$. The string passes through a small hole $H$ in a smooth horizontal surface. A light elastic string of natural length $0.3 \text{ m}$ and modulus of elasticity $15 \text{ N}$ joins $Q$ to a fixed point $A$ which is $0.4 \text{ m}$ vertically below $H$. The particle $P$ moves on the surface in a horizontal circle with centre $H$ (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Calculate the greatest possible speed of $P$ for which the elastic string is not extended. [4]
\item Find the distance $HP$ given that the angular speed of $P$ is $8 \text{ rad s}^{-1}$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2018 Q6 [9]}}