| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2019 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Conical pendulum (horizontal circle) |
| Difficulty | Standard +0.3 This is a standard conical pendulum problem requiring resolution of forces and circular motion equations. Part (i) involves resolving vertically to find tension (straightforward geometry and equilibrium), while part (ii) applies F=mrω² horizontally. The geometry is given clearly, making this slightly easier than average but still requiring proper method across two connected parts. |
| Spec | 3.03b Newton's first law: equilibrium6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T\cos\theta\left(=T\times\frac{0.15}{0.8}\right)=0.3g\) | M1 | Resolve vertically. \(\theta\) is the angle between the string and the vertical |
| \(T=16\text{ N}\) (AG) | A1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(r^2=0.8^2-0.15^2\) | B1 | \(r=0.78581...\) |
| \(16\sin\theta\left(=16\times\frac{0.78581...}{0.8}\right)=\frac{0.3v^2}{0.78581...}\) | M1 | Use Newton's Second Law horizontally |
| \(v=6.416\) | A1 | |
| Total: 3 |
## Question 1:
**Part (i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T\cos\theta\left(=T\times\frac{0.15}{0.8}\right)=0.3g$ | M1 | Resolve vertically. $\theta$ is the angle between the string and the vertical |
| $T=16\text{ N}$ (AG) | A1 | |
| **Total: 2** | | |
**Part (ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $r^2=0.8^2-0.15^2$ | B1 | $r=0.78581...$ |
| $16\sin\theta\left(=16\times\frac{0.78581...}{0.8}\right)=\frac{0.3v^2}{0.78581...}$ | M1 | Use Newton's Second Law horizontally |
| $v=6.416$ | A1 | |
| **Total: 3** | | |
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\includegraphics[max width=\textwidth, alt={}, center]{111bcbf6-daaf-4d8d-9299-d591ac7369f1-03_231_970_258_591}
A particle $P$ of mass 0.3 kg is attached to a fixed point $A$ by a light inextensible string of length 0.8 m . The fixed point $O$ is 0.15 m vertically below $A$. The particle $P$ moves with constant speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a horizontal circle with centre $O$ (see diagram).\\
(i) Show that the tension in the string is 16 N .\\
(ii) Find the value of $v$.\\
\hfill \mbox{\textit{CAIE M2 2019 Q1 [5]}}