| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Prism or block on inclined plane |
| Difficulty | Standard +0.3 This is a standard M2 moments question involving a uniform prism on an inclined plane with four parts likely covering: (i) finding normal reaction, (ii) finding friction force, (iii) determining limiting equilibrium angle, and (iv) possibly finding the position where the prism would topple. These are routine applications of resolving forces, taking moments about an edge, and using standard friction laws—slightly easier than average due to the structured multi-part format guiding students through the solution. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(T = 42m(0.5 - 0.4)/0.4\) | B1 | \(T = 10.5m\) |
| \(10m = Y + T \times (0.3/0.5)\) | M1 | |
| \(Y = 3.7m\) AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(m\omega^2 \times 0.4 = T \times (0.4/0.5)\) | M1 | ft cv of T |
| \(\omega = 4.58 \text{ rads}^{-1}\) | A1\(\checkmark\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(mv^2/0.4 = T \times (0.4/0.5) \pm 2m\) | M1 | Either case |
| \(v = 2.04 \text{ ms}^{-1}\) | A1 | |
| or \(v = 1.6 \text{ ms}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(Y = 10m\) | B1 | Fresh value of Y not Y(i); Reject or ignore \(-2\) |
| \(10m = mv^2/0.4\) | M1 | |
| \(v = 2 \text{ ms}^{-1}\) | A1 |
## Question 7:
### Part (i):
| Working | Marks | Guidance |
|---------|-------|----------|
| $T = 42m(0.5 - 0.4)/0.4$ | B1 | $T = 10.5m$ |
| $10m = Y + T \times (0.3/0.5)$ | M1 | |
| $Y = 3.7m$ AG | A1 | |
**Total: 3 marks**
### Part (ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| $m\omega^2 \times 0.4 = T \times (0.4/0.5)$ | M1 | ft cv of T |
| $\omega = 4.58 \text{ rads}^{-1}$ | A1$\checkmark$ | |
**Total: 2 marks**
### Part (iii):
| Working | Marks | Guidance |
|---------|-------|----------|
| $mv^2/0.4 = T \times (0.4/0.5) \pm 2m$ | M1 | Either case |
| $v = 2.04 \text{ ms}^{-1}$ | A1 | |
| or $v = 1.6 \text{ ms}^{-1}$ | A1 | |
**Total: 3 marks**
### Part (iv):
| Working | Marks | Guidance |
|---------|-------|----------|
| $Y = 10m$ | B1 | Fresh value of Y not Y(i); Reject or ignore $-2$ |
| $10m = mv^2/0.4$ | M1 | |
| $v = 2 \text{ ms}^{-1}$ | A1 | |
**Total: 3 marks**7\\
\includegraphics[max width=\textwidth, alt={}, center]{5998f4b1-21da-4c25-8b09-91a1cb1eee42-4_357_776_260_680}
A small bead $B$ of mass $m \mathrm {~kg}$ moves with constant speed in a horizontal circle on a fixed smooth wire. The wire is in the form of a circle with centre $O$ and radius 0.4 m . One end of a light elastic string of natural length 0.4 m and modulus of elasticity $42 m \mathrm {~N}$ is attached to $B$. The other end of the string is attached to a fixed point $A$ which is 0.3 m vertically above $O$ (see diagram).\\
(i) Show that the vertical component of the contact force exerted by the wire on the bead is 3.7 mN upwards.\\
(ii) Given that the contact force has zero horizontal component, find the angular speed of $B$.\\
(iii) Given instead that the horizontal component of the contact force has magnitude $2 m \mathrm {~N}$, find the two possible speeds of $B$.
The string is now removed. $B$ again moves on the wire in a horizontal circle with constant speed. It is given that the vertical and horizontal components of the contact force exerted by the wire on the bead have equal magnitudes.\\
(iv) Find the speed of $B$.
\end{document}
\hfill \mbox{\textit{CAIE M2 2014 Q7}}