| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle on table with string above |
| Difficulty | Standard +0.3 This is a standard conical pendulum problem requiring resolution of forces and circular motion equations. Part (i) involves straightforward application of T cos 60° = mg and centripetal force. Part (ii) requires recognizing that maximum angular speed occurs when the normal reaction becomes zero, then calculating kinetic energy—slightly above average due to the conceptual step about the limiting case, but still a routine M2 question. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Radial acc\(^n = 1.2^2/(0.2\cos30)\) | B1 | Radial acc\(^n = 8.31...\text{ms}^{-2}\) |
| \(T\cos30 = 0.3 \times 1.2^2/(0.2\cos30)\) | M1 | Component of tension \(= m \times\) radial acc\(^n\) |
| \(T = 2.88\) N | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(T\cos60 = 0.3g\) | M1 | Uses \(T\) max in limiting case when \(R = 0\) |
| \(T = 6\) | A1 | May be implied |
| \(6\cos30 = 0.3\omega^2(0.2\cos30)\) | M1 | Component of max \(T = m \times\) maximum radial acc\(^n\) |
| \(\omega = 10\) AG | A1 [4] | From \(g = 10\) only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(T\cos30 = 0.3 \times 10^2(0.2\cos30)\) | M1 | Finds \(T\) max from \(m \times \max(RA)\) |
| \(T = 6\) | A1 | |
| \(R + 6\cos60 = 0.3g\) | M1 | Resolves vertically with \(T\) max |
| \(R = 0\) and a higher value of \(\omega\) makes \(R\) negative which is impossible | A1 | Additional justification needed of inequality |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(KE = 0.3(10 \times 0.2\cos30)^2/2\) | M1 | Attempt at KE with \(v = 10 \times\) radius |
| \(KE = 0.45\) J | A1 [2] |
## Question 6:
### Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Radial acc$^n = 1.2^2/(0.2\cos30)$ | B1 | Radial acc$^n = 8.31...\text{ms}^{-2}$ |
| $T\cos30 = 0.3 \times 1.2^2/(0.2\cos30)$ | M1 | Component of tension $= m \times$ radial acc$^n$ |
| $T = 2.88$ N | A1 [3] | |
### Part (ii)(a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $T\cos60 = 0.3g$ | M1 | Uses $T$ max in limiting case when $R = 0$ |
| $T = 6$ | A1 | May be implied |
| $6\cos30 = 0.3\omega^2(0.2\cos30)$ | M1 | Component of max $T = m \times$ maximum radial acc$^n$ |
| $\omega = 10$ AG | A1 [4] | From $g = 10$ only |
**OR**
| Answer | Mark | Guidance |
|--------|------|----------|
| $T\cos30 = 0.3 \times 10^2(0.2\cos30)$ | M1 | Finds $T$ max from $m \times \max(RA)$ |
| $T = 6$ | A1 | |
| $R + 6\cos60 = 0.3g$ | M1 | Resolves vertically with $T$ max |
| $R = 0$ and a higher value of $\omega$ makes $R$ negative which is impossible | A1 | Additional justification needed of inequality |
### Part (ii)(b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $KE = 0.3(10 \times 0.2\cos30)^2/2$ | M1 | Attempt at KE with $v = 10 \times$ radius |
| $KE = 0.45$ J | A1 [2] | |
**Total: [7]**
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{d6cb7a28-e8d7-4239-b9d3-120a284d7353-3_259_890_584_630}
One end of a light inextensible string of length 0.2 m is attached to a fixed point $A$ which is above a smooth horizontal table. A particle $P$ of mass 0.3 kg is attached to the other end of the string. $P$ moves on the table in a horizontal circle, with the string taut and making an angle of $60 ^ { \circ }$ with the downward vertical (see diagram).\\
(i) Calculate the tension in the string if the speed of $P$ is $1.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(ii) For the motion as described, show that the angular speed of $P$ cannot exceed $10 \mathrm { rad } \mathrm { s } ^ { - 1 }$, and hence find the greatest possible value for the kinetic energy of $P$.
\hfill \mbox{\textit{CAIE M2 2013 Q6 [9]}}