| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2005 |
| 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 requires using the chain rule technique a = v(dv/dx) for acceleration as a function of position, then solving a separable differential equation and applying initial conditions. While the integration is straightforward, recognizing the method and finding the minimum velocity requires solid understanding of non-constant acceleration—more demanding than routine mechanics but standard for Further Maths M2. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using \(\alpha = v\dfrac{\text{d}v}{\text{d}x}\) and attempting to separate the variables | |
| \(\int v\,\text{d}v = \int(x - 2.4)\,\text{d}x\) | A1 | |
| \(\frac{1}{2}v^2 = \frac{1}{2}x^2 - 2.4x\ (+C)\) | A1 | Only allow second M1 if \(v = f(x)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{2}(2.5)^2 = 0 - 0 + C\) | M1 | For using \(v = 2.5\) when \(x = 0\) to find \(C\) (or equivalent using limits) |
| \(v = \sqrt{x^2 - 4.8x + 6.25}\) | A1 | Allow \(v^2 = x^2 - 4.8x + 6.25\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dv}{dx} = 0 \rightarrow x = 2.4 \rightarrow v_{\min} = v(2.4)\) | M1 | For any complete method for finding \(v_{\min}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Minimum value of \(v\) is \(0.7\) | A1 | 2 |
## Question 5:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using $\alpha = v\dfrac{\text{d}v}{\text{d}x}$ and attempting to separate the variables |
| $\int v\,\text{d}v = \int(x - 2.4)\,\text{d}x$ | A1 | |
| $\frac{1}{2}v^2 = \frac{1}{2}x^2 - 2.4x\ (+C)$ | A1 | Only allow second M1 if $v = f(x)$ |
## Question 5 (continued):
$\frac{1}{2}(2.5)^2 = 0 - 0 + C$ | M1 | For using $v = 2.5$ when $x = 0$ to find $C$ (or equivalent using limits)
$v = \sqrt{x^2 - 4.8x + 6.25}$ | A1 | Allow $v^2 = x^2 - 4.8x + 6.25$
**(ii)**
$\frac{dv}{dx} = 0 \rightarrow x = 2.4 \rightarrow v_{\min} = v(2.4)$ | M1 | For any complete method for finding $v_{\min}$
$\left(= \sqrt{2.4^2 - 4.8(2.4) + 6.25}\right)$ or $v = \sqrt{(x-2.4)^2 + 0.7^2} \rightarrow v^{\min} = v(2.4)$
Minimum value of $v$ is $0.7$ | A1 | 2
---5 The acceleration of a particle moving in a straight line is $( x - 2.4 ) \mathrm { m } \mathrm { s } ^ { - 2 }$ when its displacement from a fixed point $O$ of the line is $x \mathrm {~m}$. The velocity of the particle is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, and it is given that $v = 2.5$ when $x = 0$. Find\\
(i) an expression for $v$ in terms of $x$,\\
(ii) the minimum value of $v$.
\hfill \mbox{\textit{CAIE M2 2005 Q5 [7]}}