| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Two strings/rods system |
| Difficulty | Challenging +1.2 This is a standard circular motion problem with two strings requiring resolution of forces, geometry to find angles, and application of F=mrω². Part (i) involves straightforward force resolution and circular motion equations. Part (ii) requires finding the critical angular speed when the lower string becomes slack, which is a common exam technique. The geometry is given clearly (3-4-5 triangle), making this more routine than problems requiring complex geometric insight. Slightly above average due to the two-part nature and need to handle vertical/horizontal components carefully, but well within standard M2 scope. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(r = 0.3\text{ m}\) | B1 | Can be implied |
| \(0.4T/0.5 - 2(0.4/0.5) = 6\) | M1 | Resolving vertically for the particle |
| \(T = 9.5\text{ N}\) | A1 | |
| \(9.5(0.3/0.5) + 2(0.3/0.5) = 6v^2/(0.3g)\) | M1 | Newton's Second Law radially for P |
| \(v = 1.86\text{ ms}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \([0.4T/0.5 = 6],\ T = 7.5\) | B1 | Uses tension in \(BP = 0\) and resolves vertically |
| \(7.5(0.3/0.5) = (6/g)\ \omega^2(0.3)\) | M1 | Newton's Second Law radially for P |
| \(\omega = 5\text{ rad s}^{-1}\) AG | A1 |
## Question 4:
### Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $r = 0.3\text{ m}$ | B1 | Can be implied |
| $0.4T/0.5 - 2(0.4/0.5) = 6$ | M1 | Resolving vertically for the particle |
| $T = 9.5\text{ N}$ | A1 | |
| $9.5(0.3/0.5) + 2(0.3/0.5) = 6v^2/(0.3g)$ | M1 | Newton's Second Law radially for P |
| $v = 1.86\text{ ms}^{-1}$ | A1 | |
**Total: 5 marks**
### Part (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $[0.4T/0.5 = 6],\ T = 7.5$ | B1 | Uses tension in $BP = 0$ and resolves vertically |
| $7.5(0.3/0.5) = (6/g)\ \omega^2(0.3)$ | M1 | Newton's Second Law radially for P |
| $\omega = 5\text{ rad s}^{-1}$ AG | A1 | |
**Total: 3 marks**
---4 One end of a light inextensible string of length 0.5 m is attached to a fixed point $A$. The other end of the string is attached to a particle $P$ of weight 6 N . Another light inextensible string of length 0.5 m connects $P$ to a fixed point $B$ which is 0.8 m vertically below $A$. The particle $P$ moves with constant speed in a horizontal circle with centre at the mid-point of $A B$. Both strings are taut.\\
(i) Calculate the speed of $P$ when the tension in the string $B P$ is 2 N .\\
(ii) Show that the angular speed of $P$ must exceed $5 \mathrm { rad } \mathrm { s } ^ { - 1 }$.
\hfill \mbox{\textit{CAIE M2 2015 Q4 [8]}}