| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2009 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given acceleration function find velocity |
| Difficulty | Standard +0.3 This is a standard M2 variable acceleration problem requiring the chain rule identity v dv/dx = a, followed by straightforward integration of 1/(x+2) and applying initial conditions. Part (ii) is simple substitution. The techniques are routine for M2 students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y)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 |
| \(\left[v\dfrac{dv}{dx} = \dfrac{1}{x+2} \Rightarrow \int v\,dv = \int\dfrac{dx}{x+2}\right]\) | M1 | For using \(a = v\,dv/dx\), separating variables and attempting to integrate |
| \(v^2/2 = \ln(x+2) \quad (+A)\) | A1 | |
| \([2 = \ln 2 + A]\) | M1 | For using \(v(0) = 2\) |
| \(v^2 = 2\ln(0.5x + 1) + 4\) | A1 [4] | AG |
| (ii) \(\dfrac{1}{x+2} = \dfrac{1}{4} \Rightarrow x = 2\) | B1 | |
| Value of \(v\) is \(2.32\ \text{ms}^{-1}\) | B1 [2] |
## Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\left[v\dfrac{dv}{dx} = \dfrac{1}{x+2} \Rightarrow \int v\,dv = \int\dfrac{dx}{x+2}\right]$ | M1 | For using $a = v\,dv/dx$, separating variables and attempting to integrate |
| $v^2/2 = \ln(x+2) \quad (+A)$ | A1 | |
| $[2 = \ln 2 + A]$ | M1 | For using $v(0) = 2$ |
| $v^2 = 2\ln(0.5x + 1) + 4$ | A1 [4] | AG |
| **(ii)** $\dfrac{1}{x+2} = \dfrac{1}{4} \Rightarrow x = 2$ | B1 | |
| Value of $v$ is $2.32\ \text{ms}^{-1}$ | B1 [2] | |
---3 A particle $P$ starts from a fixed point $O$ and moves in a straight line. When the displacement of $P$ from $O$ is $x \mathrm {~m}$, its velocity is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and its acceleration is $\frac { 1 } { x + 2 } \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(i) Given that $v = 2$ when $x = 0$, use integration to show that $v ^ { 2 } = 2 \ln \left( \frac { 1 } { 2 } x + 1 \right) + 4$.\\
(ii) Find the value of $v$ when the acceleration of $P$ is $\frac { 1 } { 4 } \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\hfill \mbox{\textit{CAIE M2 2009 Q3 [6]}}