| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Block on horizontal plane motion |
| Difficulty | Moderate -0.3 This is a straightforward mechanics problem requiring resolution of forces, application of F=ma, and friction calculations using standard formulas. The 'show that' format and given numerical answer reduce difficulty. All steps are routine for M1 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(3.5 = 10a \rightarrow a = 0.35 \text{ ms}^{-2}\) | B1 | Allow \(a = 3.5/10\) |
| \([10\cos15 - F = 2 \times 0.35]\) | M1 | For applying Newton's 2nd law to the particle |
| \(F = 8.96 \text{ N}\) | A1 [3] | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(s = \frac{1}{2}(0 + 3.5) \times 10 = 17.5 \text{ m}\) | B1 | Distance moved in 10 secs |
| \([10\cos15 \times 17.5 = F \times 17.5 + \frac{1}{2}(2)(3.5)^2]\) | M1 | Work done by 10 N force \(=\) WD against \(F\) + KE gain |
| \(F = 8.96 \text{ N}\) | A1 [3] | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([R = 2g - 10\sin15]\) | M1 | Resolving forces vertically |
| \([\mu = 8.96/(2g - 10\sin15)]\) | M1 | Using \(F = \mu R\) |
| \(\mu = 0.515\) | A1 [3] |
## Question 1:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $3.5 = 10a \rightarrow a = 0.35 \text{ ms}^{-2}$ | B1 | Allow $a = 3.5/10$ |
| $[10\cos15 - F = 2 \times 0.35]$ | M1 | For applying Newton's 2nd law to the particle |
| $F = 8.96 \text{ N}$ | A1 [3] | AG |
**Alternative to 1(i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $s = \frac{1}{2}(0 + 3.5) \times 10 = 17.5 \text{ m}$ | B1 | Distance moved in 10 secs |
| $[10\cos15 \times 17.5 = F \times 17.5 + \frac{1}{2}(2)(3.5)^2]$ | M1 | Work done by 10 N force $=$ WD against $F$ + KE gain |
| $F = 8.96 \text{ N}$ | A1 [3] | AG |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[R = 2g - 10\sin15]$ | M1 | Resolving forces vertically |
| $[\mu = 8.96/(2g - 10\sin15)]$ | M1 | Using $F = \mu R$ |
| $\mu = 0.515$ | A1 [3] | |
---1 A particle of mass 2 kg is initially at rest on a rough horizontal plane. A force of magnitude 10 N is applied to the particle at $15 ^ { \circ }$ above the horizontal. It is given that 10 s after the force is applied, the particle has a speed of $3.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Show that the magnitude of the frictional force is 8.96 N , correct to 3 significant figures.\\
(ii) Find the coefficient of friction between the particle and the plane.
\hfill \mbox{\textit{CAIE M1 2016 Q1 [6]}}