| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Particle on inclined plane motion |
| Difficulty | Standard +0.3 This is a standard two-part mechanics problem involving motion on an inclined plane with friction. Part (i) likely involves basic kinematics or force resolution on a smooth section, while part (ii) requires applying equations of motion with friction over a known distance and time. The problem is slightly easier than average as it follows a predictable structure with given numerical values and standard M1 techniques, requiring no novel insight beyond textbook methods. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Acceleration for \(t < 0.8\) is \(4/0.8\) | B1 | |
| \([5 = 10\sin\theta]\) | M1 | For using \(a = g\sin\theta\) |
| \(\theta = 30°\) | A1 | [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([mgh = \frac{1}{2}m4^2\) and \(s = \{(0+4) \div 2\} \times 0.8]\) | M1 | For using \(PE\ \text{loss} = KE\ \text{gain}\) and \(s \div t = (u+v) \div 2\ (A\ \text{to}\ B)\) |
| \(\sin\theta = 0.8/1.6\) | A1 | |
| \(\theta = 30°\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Acceleration for \(0.8 < t < 4.8\) is \(-4/(4.8 - 0.8)\) | B1 | |
| \([mg\sin 30° - F = m(-1)]\) | M1 | For using Newton's second law |
| M1 | For using \(\mu = F/R\) | |
| \(\mu = \dfrac{mg\sin 30° + m}{mg\cos 30°}\) | A1ft | ft following a wrong answer for \(\theta\) in part (i) |
| Coefficient is 0.693 | A1 | [5] Accept 0.69 |
# Question 5:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Acceleration for $t < 0.8$ is $4/0.8$ | B1 | |
| $[5 = 10\sin\theta]$ | M1 | For using $a = g\sin\theta$ |
| $\theta = 30°$ | A1 | **[3]** |
### Alternative for part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[mgh = \frac{1}{2}m4^2$ and $s = \{(0+4) \div 2\} \times 0.8]$ | M1 | For using $PE\ \text{loss} = KE\ \text{gain}$ and $s \div t = (u+v) \div 2\ (A\ \text{to}\ B)$ |
| $\sin\theta = 0.8/1.6$ | A1 | |
| $\theta = 30°$ | A1 | |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Acceleration for $0.8 < t < 4.8$ is $-4/(4.8 - 0.8)$ | B1 | |
| $[mg\sin 30° - F = m(-1)]$ | M1 | For using Newton's second law |
| | M1 | For using $\mu = F/R$ |
| $\mu = \dfrac{mg\sin 30° + m}{mg\cos 30°}$ | A1ft | ft following a wrong answer for $\theta$ in part (i) |
| Coefficient is 0.693 | A1 | **[5]** Accept 0.69 |(i) Find the value of $\theta$.
At time 4.8 s after leaving $A$, the particle comes to rest at $C$.\\
(ii) Find the coefficient of friction between $P$ and the rough part of the plane.
\hfill \mbox{\textit{CAIE M1 2012 Q5 [8]}}