| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Variable force (position x) - find velocity |
| Difficulty | Challenging +1.2 This is a variable force mechanics problem requiring Newton's second law with v dv/dx, finding maximum speed (where acceleration = 0) and using work-energy principles. While it involves multiple steps and careful setup of forces on an incline, the mathematical techniques are standard M2 content with straightforward integration and equation solving. More challenging than routine constant acceleration problems but less demanding than questions requiring novel geometric insight or complex proof. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.4v\,dv/dx = 0.4g\sin30 - 0.6x\) | B1 | Newton's Second Law, – sign essential |
| \(\int v\,dv = \int(5-1.5x)\,dx\) | M1 | Accept uncancelled integration |
| \(v^2/2 = 5x - 1.5x^2/2\ (+c)\) | A1 | Accept omission of \(c\) |
| \(0.4g\sin30 - 0.6x = 0\) | M1 | Maximum speed when acceleration \(= 0\) |
| \(x = 3\tfrac{1}{3}\) | A1 | Accept \(10/3\) |
| \(v^2/2 = 5\times10/3 - 1.5\times(10/3)^2/2\) | M1 | |
| \(v = 4.08\text{ ms}^{-1}\) | A1 [7] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0 = 5x - 1.5x^2/2\) | M1 | Uses \(v = 0\) appropriately |
| \(x = 6\tfrac{2}{3} = 6.67\) | A1 [2] | Not \(20/3\) |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.4v\,dv/dx = 0.4g\sin30 - 0.6x$ | B1 | Newton's Second Law, – sign essential |
| $\int v\,dv = \int(5-1.5x)\,dx$ | M1 | Accept uncancelled integration |
| $v^2/2 = 5x - 1.5x^2/2\ (+c)$ | A1 | Accept omission of $c$ |
| $0.4g\sin30 - 0.6x = 0$ | M1 | Maximum speed when acceleration $= 0$ |
| $x = 3\tfrac{1}{3}$ | A1 | Accept $10/3$ |
| $v^2/2 = 5\times10/3 - 1.5\times(10/3)^2/2$ | M1 | |
| $v = 4.08\text{ ms}^{-1}$ | A1 [7] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 = 5x - 1.5x^2/2$ | M1 | Uses $v = 0$ appropriately |
| $x = 6\tfrac{2}{3} = 6.67$ | A1 [2] | Not $20/3$ |
---5 A particle $P$ of mass 0.4 kg is released from rest at the top of a smooth plane inclined at $30 ^ { \circ }$ to the horizontal. The motion of $P$ down the slope is opposed by a force of magnitude $0.6 x \mathrm {~N}$, where $x \mathrm {~m}$ is the distance $P$ has travelled down the slope. $P$ comes to rest before reaching the foot of the slope. Calculate\\
(i) the greatest speed of $P$ during its motion,\\
(ii) the distance travelled by $P$ during its motion.
\hfill \mbox{\textit{CAIE M2 2012 Q5 [9]}}