| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Prism or block on inclined plane |
| Difficulty | Challenging +1.2 This is a multi-part moments problem requiring finding the center of mass of a composite shape (rectangle + semicircle), applying toppling conditions, and resolving forces with friction. While it involves several steps and composite shapes, the techniques are standard M2 content: taking moments about a point, using standard center of mass formulas, and applying equilibrium conditions. The calculations are methodical rather than requiring novel insight. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| CoM semi-circle from \(DF = 4r/3\pi\) | B1 | |
| \((0.4 \times 1.8) \times 0.2 = (\pi r^2/2) \times (4r/3\pi)\) | M1 | Moments about \(A\) |
| \(r = 0.6\) | A1 AG | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(P\cos 60(0.4 + 0.6\cos 60)\) | B1 | Moment of vertical component |
| \(P\cos 30(1.8 - 0.6 + 0.6\sin 60)\) | B1 | Moment of horizontal component |
| \(15 \times 0.4 =\) | M1 | |
| \(P\cos 60(0.4 + 0.6\cos 60) + P\cos 30(1.8 - 0.6 + 0.6\sin 60)\) | ||
| \(P = 3.26\) N | A1 AG | \(3.2622\ldots\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\mu = 3.262\sin 60/(15 - 3.262\cos 60)\) | M1 | |
| \(\mu = 0.211\) | A1 | |
| [2] |
## Question 6:
### Part (i):
| Working/Answer | Mark | Guidance |
|---|---|---|
| CoM semi-circle from $DF = 4r/3\pi$ | B1 | |
| $(0.4 \times 1.8) \times 0.2 = (\pi r^2/2) \times (4r/3\pi)$ | M1 | Moments about $A$ |
| $r = 0.6$ | A1 AG | |
| | [3] | |
### Part (ii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $P\cos 60(0.4 + 0.6\cos 60)$ | B1 | Moment of vertical component |
| $P\cos 30(1.8 - 0.6 + 0.6\sin 60)$ | B1 | Moment of horizontal component |
| $15 \times 0.4 =$ | M1 | |
| $P\cos 60(0.4 + 0.6\cos 60) + P\cos 30(1.8 - 0.6 + 0.6\sin 60)$ | | |
| $P = 3.26$ N | A1 AG | $3.2622\ldots$ |
| | [4] | |
### Part (iii):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\mu = 3.262\sin 60/(15 - 3.262\cos 60)$ | M1 | |
| $\mu = 0.211$ | A1 | |
| | [2] | |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{d9970ad1-a7f4-429a-bad1-43e8d114b968-3_656_757_781_694}
The diagram shows the cross-section $A B C D E F$ through the centre of mass of a uniform prism which rests with $A B$ on rough horizontal ground. $A B C D$ is a rectangle with $A B = C D = 0.4 \mathrm {~m}$ and $B C = A D = 1.8 \mathrm {~m}$. The other part of the cross-section is a semicircle with diameter $D F$ and radius $r \mathrm {~m}$.\\
(i) Given that the prism is on the point of toppling, show that $r = 0.6$.
A force of magnitude $P \mathrm {~N}$ is applied to the prism, acting at $60 ^ { \circ }$ to the upwards vertical along a tangent to the semicircle at a point between $D$ and $E$. The prism has weight 15 N and is in equilibrium on the point of toppling about $B$.\\
(ii) Show that $P = 3.26$, correct to 3 significant figures.\\
(iii) Find the smallest possible value of the coefficient of friction between the prism and the ground.
\hfill \mbox{\textit{CAIE M2 2016 Q6 [9]}}