| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Elastic string – horizontal circle on surface |
| Difficulty | Standard +0.3 This is a standard circular motion problem with elastic strings requiring application of Hooke's law, centripetal force equation, and energy relationships. Part (i) is straightforward substitution; part (ii) requires forming and solving simultaneous equations using given energy condition. The mechanics are routine for M2 level with no novel insight needed, making it slightly easier than average. |
| 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 principle6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T = 0.2 \times 2.5^2/(0.4 + e)\) | B1 | Uses Newton's Second Law towards the centre of the circle. |
| \(T = 8e/0.4\) | B1 | Uses \(T = \lambda x / L\). |
| \(1.25/(0.4 + e) = 20e \rightarrow 20e^2 + 8e - 1.25 = 0\) | M1 | Eliminates \(T\) to find \(e\). |
| \(e = 0.12(0)\text{ m}\) | A1 | |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.2v^2/2 = 2[8x^2/(2 \times 0.4)]\) | B1 | Uses \(KE = 2EE\). |
| \(0.2v^2/(0.4 + x) = 8x/0.4\) | B1 | Uses \(T = \lambda x/L\) and \(T = mv^2/r\). |
| M1 | Attempts to solve the 2 equations to find \(v\) or \(x\). | |
| \(x = 0.4\) and \(v = 5.66\) or \(4\sqrt{2}\) | A1A1 | |
| Total | 5 |
## Question 6(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T = 0.2 \times 2.5^2/(0.4 + e)$ | B1 | Uses Newton's Second Law towards the centre of the circle. |
| $T = 8e/0.4$ | B1 | Uses $T = \lambda x / L$. |
| $1.25/(0.4 + e) = 20e \rightarrow 20e^2 + 8e - 1.25 = 0$ | M1 | Eliminates $T$ to find $e$. |
| $e = 0.12(0)\text{ m}$ | A1 | |
| **Total** | **4** | |
## Question 6(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.2v^2/2 = 2[8x^2/(2 \times 0.4)]$ | B1 | Uses $KE = 2EE$. |
| $0.2v^2/(0.4 + x) = 8x/0.4$ | B1 | Uses $T = \lambda x/L$ and $T = mv^2/r$. |
| | M1 | Attempts to solve the 2 equations to find $v$ or $x$. |
| $x = 0.4$ and $v = 5.66$ or $4\sqrt{2}$ | A1A1 | |
| **Total** | **5** | |
---6 One end of a light elastic string of natural length 0.4 m and modulus of elasticity 8 N is attached to a fixed point $O$ on a smooth horizontal plane. The other end of the string is attached to a particle $P$ of mass 0.2 kg which moves on the plane in a circular path with centre $O$. The speed of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the extension of the string is $x \mathrm {~m}$.\\
(i) Given that $v = 2.5$, find $x$.\\
It is given instead that the kinetic energy of $P$ is twice the elastic potential energy stored in the string.\\
(ii) Form two simultaneous equations and hence find $x$ and $v$.\\
\hfill \mbox{\textit{CAIE M2 2017 Q6 [9]}}