| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Friction inequality derivation |
| Difficulty | Standard +0.8 This is a two-part friction problem requiring resolution of forces, application of friction laws, and algebraic manipulation to derive inequalities for the coefficient of friction. Part (i) involves showing a specific inequality (likely from equilibrium or limiting friction), while part (ii) requires analyzing a different scenario with an upward force component. The multi-step nature, need to set up force equations correctly, and derivation of inequalities rather than simple calculation places this above average difficulty, though it follows standard M1 friction methodology. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(F = 12\cos\alpha\) | B1 | |
| M1 | For resolving forces vertically | |
| \(R_1 = 2g + 12\sin\alpha\) | A1 | |
| \([12 \times 0.8 \leq \mu(2g + 12 \times 0.6)]\) | M1 | For using \(F_1 \leq \mu R\) |
| \(\mu \geq 9.6/27.2 = 6/17\) | A1 | 5 AG |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) \(12\cos\alpha > \mu R_2\) | B1 | |
| \(R_2 = 2g - 12 \times 0.6\) | B1 | |
| \(\mu < 9.6/12.8 = 3/4\) | B1 | 3 |
| **(i)** $F = 12\cos\alpha$ | B1 | |
| | M1 | For resolving forces vertically |
| $R_1 = 2g + 12\sin\alpha$ | A1 | |
| $[12 \times 0.8 \leq \mu(2g + 12 \times 0.6)]$ | M1 | For using $F_1 \leq \mu R$ |
| $\mu \geq 9.6/27.2 = 6/17$ | A1 | 5 AG |
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## Question 5 (continued):
| **(ii)** $12\cos\alpha > \mu R_2$ | B1 | |
| $R_2 = 2g - 12 \times 0.6$ | B1 | |
| $\mu < 9.6/12.8 = 3/4$ | B1 | 3 |
---(i) Show that $\mu \geqslant \frac { 6 } { 17 }$.
When the applied force acts upwards as in Fig. 2 the block slides along the floor.\\
(ii) Find another inequality for $\mu$.
\hfill \mbox{\textit{CAIE M1 2011 Q5 [8]}}