| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Toppling and sliding of solids |
| Difficulty | Challenging +1.2 This is a multi-part mechanics question requiring center of mass calculations for composite bodies and toppling analysis. Part (i) is routine, part (ii) involves algebraic manipulation of standard formulas (though the 'show that' adds slight complexity), and part (iii) requires applying the toppling condition (vertical through COM passes through pivot point) with some trigonometry. While it requires multiple techniques and careful algebra, these are standard M2 procedures without requiring novel insight or particularly challenging problem-solving. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.05a Angular velocity: definitions |
| Answer | Marks | Guidance |
|---|---|---|
| \([T_1\sin(APN) = T_2\sin(BPN)]\) | M1 | For resolving forces horizontally |
| \((12 \div 13)T_1 = (15 \div 25)T_2\) or \(T_1\sin67.4° = T_2\sin36.9°\) | A1 | AEF |
| \([T_1\cos(APN) + T_2\cos(BPN) = 21]\) | M1 | For resolving forces vertically |
| \((5 \div 13)T_1 + (20 \div 25)T_2 = 21\) or \(T_1\cos67.4° + T_2\cos36.9° = 21\) | A1 | AEF |
| M1 | For solving for \(T_1\) and \(T_2\) | |
| Tension in \(S_1\) is 13 N, tension in \(S_2\) is 20 N | A1 | 6 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \([T_1/\sin(180 - BPN) = 21/\sin(APN + BPN)]\) | M1 | For using Lami's Theorem to form an equation in \(T_1\) |
| \(T_1/\sin(180 - \cos^{-1}(20/25)) = 21/\sin(\cos^{-1}(20/25) + \cos^{-1}(20/52))\) or \(T_1/\sin(180 - 36.9) = 21/\sin(36.9 + 67.4)\) | A1 | AEF |
| \([T_2/\sin(180 - APN) = 21/\sin(APN + BPN)]\) | M1 | For using Lami's Theorem to form an equation in \(T_2\) |
| \(T_2/\sin(180 - \cos^{-1}(20/25)) = 21/\sin(\cos^{-1}(20/25) + \cos^{-1}(20/52))\) or \(T_2/\sin(180 - 67.4) = 21/\sin(36.9 + 67.4)\) | A1 | AEF |
| M1 | For solving for \(T_1\) and \(T_2\) | |
| Tension in \(S_1\) is 13 N, tension in \(S_2\) is 20 N | A1 | 6 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| \([T_1/\sin(BPN) = 21/\sin(180 - (APN + BPN))]\) | M1 | For using the Sine Rule on a triangle of forces to form an equation in \(T_1\) |
| \(T_1/(15/25) = 21/\sin(\cos^{-1}(20/25) + \cos^{-1}(20/52))\) or \(T_1/\sin36.9° = 21/\sin(180 - (36.9 + 67.4))\) | A1 | AEF |
| \([T_2/\sin(APN) = 21/\sin(180 - (APN + BPN))]\) | M1 | For using the Sine Rule to form an equation in \(T_2\) |
| \(T_2/(12/13) = 21/\sin(\cos^{-1}(20/25) + \cos^{-1}(20/52))\) or \(T_2/\sin67.4° = 21/\sin(180 - (36.9 + 67.4))\) | A1 | AEF |
| M1 | For solving for \(T_1\) and \(T_2\) | |
| Tension in \(S_1\) is 13 N, tension in \(S_2\) is 20 N | A1 | 6 marks total |
| $[T_1\sin(APN) = T_2\sin(BPN)]$ | M1 | For resolving forces horizontally |
| $(12 \div 13)T_1 = (15 \div 25)T_2$ or $T_1\sin67.4° = T_2\sin36.9°$ | A1 | AEF |
| $[T_1\cos(APN) + T_2\cos(BPN) = 21]$ | M1 | For resolving forces vertically |
| $(5 \div 13)T_1 + (20 \div 25)T_2 = 21$ or $T_1\cos67.4° + T_2\cos36.9° = 21$ | A1 | AEF |
| | M1 | For solving for $T_1$ and $T_2$ |
| Tension in $S_1$ is 13 N, tension in $S_2$ is 20 N | A1 | 6 marks total |
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## Alternative solution using Lami's Theorem
| $[T_1/\sin(180 - BPN) = 21/\sin(APN + BPN)]$ | M1 | For using Lami's Theorem to form an equation in $T_1$ |
| $T_1/\sin(180 - \cos^{-1}(20/25)) = 21/\sin(\cos^{-1}(20/25) + \cos^{-1}(20/52))$ or $T_1/\sin(180 - 36.9) = 21/\sin(36.9 + 67.4)$ | A1 | AEF |
| $[T_2/\sin(180 - APN) = 21/\sin(APN + BPN)]$ | M1 | For using Lami's Theorem to form an equation in $T_2$ |
| $T_2/\sin(180 - \cos^{-1}(20/25)) = 21/\sin(\cos^{-1}(20/25) + \cos^{-1}(20/52))$ or $T_2/\sin(180 - 67.4) = 21/\sin(36.9 + 67.4)$ | A1 | AEF |
| | M1 | For solving for $T_1$ and $T_2$ |
| Tension in $S_1$ is 13 N, tension in $S_2$ is 20 N | A1 | 6 marks total |
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## Alternative solution using Sine Rule
| $[T_1/\sin(BPN) = 21/\sin(180 - (APN + BPN))]$ | M1 | For using the Sine Rule on a triangle of forces to form an equation in $T_1$ |
| $T_1/(15/25) = 21/\sin(\cos^{-1}(20/25) + \cos^{-1}(20/52))$ or $T_1/\sin36.9° = 21/\sin(180 - (36.9 + 67.4))$ | A1 | AEF |
| $[T_2/\sin(APN) = 21/\sin(180 - (APN + BPN))]$ | M1 | For using the Sine Rule to form an equation in $T_2$ |
| $T_2/(12/13) = 21/\sin(\cos^{-1}(20/25) + \cos^{-1}(20/52))$ or $T_2/\sin67.4° = 21/\sin(180 - (36.9 + 67.4))$ | A1 | AEF |
| | M1 | For solving for $T_1$ and $T_2$ |
| Tension in $S_1$ is 13 N, tension in $S_2$ is 20 N | A1 | 6 marks total |
---(i) Find $r$.
The upper cylinder is now fixed to the lower cylinder to create a uniform object.\\
(ii) Show that the centre of mass of the object is
$$\frac { 25 h ^ { 2 } + 180 h + 81 } { 50 h + 180 } \mathrm {~m}$$
from $A$.
The object is placed with the plane face containing $A$ in contact with a rough plane inclined at $\alpha ^ { \circ }$ to the horizontal, where $\tan \alpha = 0.5$. The object is on the point of toppling without sliding.\\
(iii) Calculate $h$.
\hfill \mbox{\textit{CAIE M2 2012 Q4 [8]}}