| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | March |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Block on horizontal plane motion |
| Difficulty | Standard +0.3 This is a standard two-part friction problem requiring resolution of forces, use of F=μR at limiting equilibrium, and then F=ma. The angle is given in a convenient form (3-4-5 triangle), making resolution straightforward. While it requires multiple steps and careful bookkeeping of forces, it follows a completely standard template with no novel insight required, making it slightly easier than average. |
| Spec | 1.05g Exact trigonometric values: for standard angles3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(5\cos\alpha = F \Rightarrow [F = 4]\) | M1 | For resolving forces horizontally; allow use of \(\alpha = 36.9°\) throughout |
| \(R + 5\sin\alpha = 8 \Rightarrow [R = 5]\) | M1 | For resolving forces vertically |
| \(4 = 5\mu\) | M1 | For using \(F = \mu R\) |
| \(\mu = 0.8\) | A1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R + 10\sin\alpha = 8 \Rightarrow [R = 2]\) and \(F = 0.8 \times R \Rightarrow [F = 1.6]\) | B1 | For resolving forces vertically to find new value of \(R\) and using \(F = \mu R\) |
| \(10\cos\alpha - F = 0.8a\) | M1 | For resolving horizontally |
| \(a = 8\ \text{ms}^{-2}\) | A1 | 3 |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $5\cos\alpha = F \Rightarrow [F = 4]$ | M1 | For resolving forces horizontally; allow use of $\alpha = 36.9°$ throughout |
| $R + 5\sin\alpha = 8 \Rightarrow [R = 5]$ | M1 | For resolving forces vertically |
| $4 = 5\mu$ | M1 | For using $F = \mu R$ |
| $\mu = 0.8$ | A1 | 4 | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R + 10\sin\alpha = 8 \Rightarrow [R = 2]$ **and** $F = 0.8 \times R \Rightarrow [F = 1.6]$ | B1 | For resolving forces vertically to find new value of $R$ **and** using $F = \mu R$ |
| $10\cos\alpha - F = 0.8a$ | M1 | For resolving horizontally |
| $a = 8\ \text{ms}^{-2}$ | A1 | 3 | |
---4 A particle $P$ of mass 0.8 kg is placed on a rough horizontal table. The coefficient of friction between $P$ and the table is $\mu$. A force of magnitude 5 N , acting upwards at an angle $\alpha$ above the horizontal, where $\tan \alpha = \frac { 3 } { 4 }$, is applied to $P$. The particle is on the point of sliding on the table.\\
(i) Find the value of $\mu$.\\
(ii) The magnitude of the force acting on $P$ is increased to 10 N , with the direction of the force remaining the same. Find the acceleration of $P$.
\hfill \mbox{\textit{CAIE M1 2016 Q4 [7]}}