| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Block on rough horizontal surface – equilibrium (finding friction, normal reaction, or coefficient of friction) |
| Difficulty | Moderate -0.3 This is a straightforward two-part equilibrium problem requiring resolution of forces in two perpendicular directions. Part (i) involves basic trigonometry and force balance on a smooth surface, while part (ii) adds friction at limiting equilibrium with the friction force already given. Both parts are standard M1 exercises with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([X\cos30° = 40\cos60°]\) | M1 | For resolving forces horizontally |
| \(X = 23.1\ (= 40/\sqrt{3})\) | A1 | 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \([X\cos30° - 10 = 40\cos60°]\) | M1 | For resolving forces horizontally |
| \(X = 60 \div \sqrt{3}\) or \(34.6\) | A1 | |
| \([R + X\sin30° + 40\sin60° = 15g]\) | M1 | For resolving forces vertically \((R = 98.038)\) |
| \([\mu = 10 \div (150 - 30/\sqrt{3} - 20\sqrt{3})]\) | M1 | For using \(F = \mu R\) |
| Coefficient is \(0.102\) | A1 | 5 marks total |
## Question 4:
### Part (i):
| Working | Mark | Guidance |
|---------|------|----------|
| $[X\cos30° = 40\cos60°]$ | M1 | For resolving forces horizontally |
| $X = 23.1\ (= 40/\sqrt{3})$ | A1 | 2 marks total |
### Part (ii):
| Working | Mark | Guidance |
|---------|------|----------|
| $[X\cos30° - 10 = 40\cos60°]$ | M1 | For resolving forces horizontally |
| $X = 60 \div \sqrt{3}$ or $34.6$ | A1 | |
| $[R + X\sin30° + 40\sin60° = 15g]$ | M1 | For resolving forces vertically $(R = 98.038)$ |
| $[\mu = 10 \div (150 - 30/\sqrt{3} - 20\sqrt{3})]$ | M1 | For using $F = \mu R$ |
| Coefficient is $0.102$ | A1 | 5 marks total |
---4\\
\includegraphics[max width=\textwidth, alt={}, center]{c7133fc4-9a14-43fd-b5ed-788da72291cd-3_383_791_262_678}
Forces of magnitude $X \mathrm {~N}$ and 40 N act on a block $B$ of mass 15 kg , which is in equilibrium in contact with a horizontal surface between points $A$ and $C$ on the surface. The forces act in the same vertical plane and in the directions shown in the diagram.\\
(i) Given that the surface is smooth, find the value of $X$.\\
(ii) It is given instead that the surface is rough and that the block is in limiting equilibrium. The frictional force acting on the block has magnitude 10 N in the direction towards $A$. Find the coefficient of friction between the block and the surface.
\hfill \mbox{\textit{CAIE M1 2014 Q4 [7]}}