| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Acceleration as function of displacement (v dv/dx method) |
| Difficulty | Standard +0.8 This is a non-constant acceleration problem requiring the chain rule technique (a = v dv/dx) and integration, which is beyond standard M1 content. While the calculus itself is straightforward, recognizing that maximum velocity occurs when a=0 and handling acceleration as a function of position (rather than time) requires solid understanding of mechanics principles and is less routine than typical A-level questions. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(a = 0\) when \(x = 2.5\) | B1 | |
| \(v\,dv/dx = 15 - 6x\) | M1 | |
| \(\int v\,dv = \int(15-6x)\,dx\) | ||
| \(v^2/2 = \left[15x - 3x^2\right] (+c)\) | A1 | |
| M1 | For use of limits \(0\) and \(2.5\) or evaluating \(c(=0)\) | |
| \(v = 6.12\) | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Solves \(15x - 3x^2 = 0\) | M1 | \(x = 5\). Accept assumption \(c = 0\) |
| \(a = (15 - 6 \times 5) = -15\text{ ms}^{-2}\) | A1 [2] |
## Question 4:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $a = 0$ when $x = 2.5$ | B1 | |
| $v\,dv/dx = 15 - 6x$ | M1 | |
| $\int v\,dv = \int(15-6x)\,dx$ | | |
| $v^2/2 = \left[15x - 3x^2\right] (+c)$ | A1 | |
| | M1 | For use of limits $0$ and $2.5$ or evaluating $c(=0)$ |
| $v = 6.12$ | A1 [5] | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Solves $15x - 3x^2 = 0$ | M1 | $x = 5$. Accept assumption $c = 0$ |
| $a = (15 - 6 \times 5) = -15\text{ ms}^{-2}$ | A1 [2] | |
---4 A particle $P$ starts from rest at a point $O$ and travels in a straight line. The acceleration of $P$ is $( 15 - 6 x ) \mathrm { m } \mathrm { s } ^ { - 2 }$, where $x \mathrm {~m}$ is the displacement of $P$ from $O$.\\
(i) Find the value of $x$ for which $P$ reaches its maximum velocity, and calculate this maximum velocity.\\
(ii) Calculate the acceleration of $P$ when it is at instantaneous rest and $x > 0$.
\hfill \mbox{\textit{CAIE M2 2011 Q4 [7]}}