| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2019 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Two strings, two fixed points |
| Difficulty | Standard +0.8 This is a non-trivial circular motion problem requiring geometric analysis to find the radius and vertical positions, then applying Newton's second law in both vertical and horizontal directions with two unknowns (tensions). Part (i) involves solving simultaneous equations with the constraint that tensions are equal, while part (ii) requires finding two different tensions. The geometry with two angles and the need to resolve forces in two directions elevates this above standard single-string conical pendulum problems. |
| Spec | 3.03b Newton's first law: equilibrium3.03e Resolve forces: two dimensions6.05a Angular velocity: definitions6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(r = 0.5\sin30 = 0.25\) m | B1 | |
| \(T\cos30 - T\cos70 = 0.4g\) | M1 | Resolve vertically |
| \(T = 7.6335\) | A1 | |
| \(7.6335\sin30 + 7.6335\sin70 = 0.4v^2 / 0.25\) | M1 | Use Newton's Second Law with \(a = \frac{v^2}{r}\) |
| \(v = 2.62\) m s\(^{-1}\) | A1 | |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(A\cos30 - B\cos70 = 0.4g\) and \(A\sin30 + B\sin70 = 0.4 \times 12^2 \times 0.5\sin30\) | M1 | Resolves vertically and uses Newton's Second Law with \(a = r\omega^2\) |
| Both correct | A1 | |
| Attempt to solve for \(A\) or \(B\) | M1 | |
| \(A = 8.82\) N | A1 | |
| \(B = 10.6\) N | A1 | |
| Total: 5 |
## Question 5(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $r = 0.5\sin30 = 0.25$ m | B1 | |
| $T\cos30 - T\cos70 = 0.4g$ | M1 | Resolve vertically |
| $T = 7.6335$ | A1 | |
| $7.6335\sin30 + 7.6335\sin70 = 0.4v^2 / 0.25$ | M1 | Use Newton's Second Law with $a = \frac{v^2}{r}$ |
| $v = 2.62$ m s$^{-1}$ | A1 | |
| **Total: 5** | | |
---
## Question 5(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $A\cos30 - B\cos70 = 0.4g$ and $A\sin30 + B\sin70 = 0.4 \times 12^2 \times 0.5\sin30$ | M1 | Resolves vertically and uses Newton's Second Law with $a = r\omega^2$ |
| Both correct | A1 | |
| Attempt to solve for $A$ or $B$ | M1 | |
| $A = 8.82$ N | A1 | |
| $B = 10.6$ N | A1 | |
| **Total: 5** | | |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{9daebcbe-826e-4eda-afa7-c935c6ea2bfc-06_671_504_255_824}\\
$A$ and $B$ are two fixed points on a vertical axis with $A$ above $B$. A particle $P$ of mass 0.4 kg is attached to $A$ by a light inextensible string of length 0.5 m . The particle $P$ is attached to $B$ by another light inextensible string. $P$ moves with constant speed in a horizontal circle with centre $O$ between $A$ and $B$. Angle $B A P = 30 ^ { \circ }$ and angle $A B P = 70 ^ { \circ }$ (see diagram).\\
(i) Given that the tensions in the two strings are equal, find the speed of $P$.\\
(ii) Given instead that the angular speed of $P$ is $12 \mathrm { rad } \mathrm { s } ^ { - 1 }$, find the tensions in the strings.\\
\hfill \mbox{\textit{CAIE M2 2019 Q5 [10]}}