| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Motion on a slope |
| Type | Motion up rough slope |
| Difficulty | Moderate -0.3 This is a straightforward mechanics problem requiring standard application of Newton's second law on an inclined plane with friction. Part (i) involves resolving forces (weight components and friction) and is scaffolded by giving the answer. Part (ii) is a routine kinematics calculation using constant acceleration equations. The numerical values are clean and the problem follows a standard textbook template with no novel insight required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([-(1 \div 3)(W\cos\alpha) - W\sin\alpha = (W/g)a]\) | M1 | For using Newton's 2nd law and \(F = \mu R\) |
| \((-0.32 - 0.28)g = a\) | A1 | |
| \(a = -6\) | A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \([0 = 5.4^2 + 2(-6)s]\) or \([mgs(0.28) = \frac{1}{2}m(5.4)^2 - mgs(0.96)/3]\) | M1 | For using \(0 = u^2 + 2as\) or PE gain = KE loss – WD against friction |
| Distance is 2.43 m | A1 | 2 marks |
## Question 1:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[-(1 \div 3)(W\cos\alpha) - W\sin\alpha = (W/g)a]$ | M1 | For using Newton's 2nd law and $F = \mu R$ |
| $(-0.32 - 0.28)g = a$ | A1 | |
| $a = -6$ | A1 | 3 marks | AG |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $[0 = 5.4^2 + 2(-6)s]$ or $[mgs(0.28) = \frac{1}{2}m(5.4)^2 - mgs(0.96)/3]$ | M1 | For using $0 = u^2 + 2as$ or PE gain = KE loss – WD against friction |
| Distance is 2.43 m | A1 | 2 marks |
---1 A particle moves up a line of greatest slope of a rough plane inclined at an angle $\alpha$ to the horizontal, where $\sin \alpha = 0.28$. The coefficient of friction between the particle and the plane is $\frac { 1 } { 3 }$.\\
(i) Show that the acceleration of the particle is $- 6 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(ii) Given that the particle's initial speed is $5.4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, find the distance that the particle travels up the plane.
\hfill \mbox{\textit{CAIE M1 2013 Q1 [5]}}