| Exam Board | Edexcel |
|---|---|
| Module | FM2 AS (Further Mechanics 2 AS) |
| Year | 2018 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Acceleration as function of velocity (separation of variables) |
| Difficulty | Standard +0.8 This is a Further Maths mechanics question requiring separation of variables to solve a differential equation, then integration involving exponentials and logarithms. While the techniques are standard for FM2, it requires multiple steps (solving DE, finding constant, then integrating v to find displacement with substitution) and careful algebraic manipulation, placing it moderately above average difficulty. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\int \frac{1}{4}dt = \int \frac{1}{50-10v}dv\) | M1 | Strategy to find \(v\) and attempt the integration |
| \(\frac{1}{4}t = -\frac{1}{10}\ln(50-10v)(+C)\) | A1 | Correct integration |
| \(\frac{1}{4}t = -\frac{1}{10}\ln(50-10v) + \frac{1}{10}\ln 50\) | M1 | Use boundary conditions as limits or evaluate constant |
| \(-\frac{5t}{2} = \ln\!\left(\frac{5-v}{5}\right)\) | M1 | Remove logarithm to express \(v\) in terms of \(t\) |
| \(v = 5\!\left(1 - e^{-2.5t}\right)\) | A1* | Obtain given answer from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Limiting value is 5 | B1 | Correct answer from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equation in \(x\) and \(t\): \(\dfrac{dx}{dt} = 5\!\left(1-e^{-2.5t}\right)\) | M1 | Set up equation of motion in terms of \(x\) and \(t\) |
| \(\Rightarrow \int 1\, dx = \int 5\!\left(1-e^{-2.5t}\right)dt\) | M1 | Separate variables and attempt integration of both sides |
| \(x = 5t + 2e^{-2.5t}(+C)\) | A1 | Any equivalent form; condone if \(+C\) not seen |
| Use \(v = 2.5\) and \(v = 5(1-e^{-2.5t})\) to find value of \(t\) | M1 | Use \(v=2.5\) to find limit for \(t\) |
| \(1 - \frac{2.5}{5} = e^{-2.5t} \Rightarrow t = \frac{2}{5}\ln 2\) | A1 | Any equivalent exact form (0.277) |
| \(\left[x\right]_0^d = \left[5t + 2e^{-2.5t}\right]_0^{\frac{2}{5}\ln 2}\) | M1 | Use boundary conditions as limits or evaluate constant |
| \(d = 2\ln 2 - 1\) | A1* | Sufficient correct working to justify given answer |
## Question 4:
**Part 4(a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\int \frac{1}{4}dt = \int \frac{1}{50-10v}dv$ | M1 | Strategy to find $v$ and attempt the integration |
| $\frac{1}{4}t = -\frac{1}{10}\ln(50-10v)(+C)$ | A1 | Correct integration |
| $\frac{1}{4}t = -\frac{1}{10}\ln(50-10v) + \frac{1}{10}\ln 50$ | M1 | Use boundary conditions as limits or evaluate constant |
| $-\frac{5t}{2} = \ln\!\left(\frac{5-v}{5}\right)$ | M1 | Remove logarithm to express $v$ in terms of $t$ |
| $v = 5\!\left(1 - e^{-2.5t}\right)$ | A1* | Obtain given answer from correct working |
**Part 4(b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Limiting value is 5 | B1 | Correct answer from correct working |
**Part 4(c):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Equation in $x$ and $t$: $\dfrac{dx}{dt} = 5\!\left(1-e^{-2.5t}\right)$ | M1 | Set up equation of motion in terms of $x$ and $t$ |
| $\Rightarrow \int 1\, dx = \int 5\!\left(1-e^{-2.5t}\right)dt$ | M1 | Separate variables and attempt integration of both sides |
| $x = 5t + 2e^{-2.5t}(+C)$ | A1 | Any equivalent form; condone if $+C$ not seen |
| Use $v = 2.5$ and $v = 5(1-e^{-2.5t})$ to find value of $t$ | M1 | Use $v=2.5$ to find limit for $t$ |
| $1 - \frac{2.5}{5} = e^{-2.5t} \Rightarrow t = \frac{2}{5}\ln 2$ | A1 | Any equivalent exact form (0.277) |
| $\left[x\right]_0^d = \left[5t + 2e^{-2.5t}\right]_0^{\frac{2}{5}\ln 2}$ | M1 | Use boundary conditions as limits or evaluate constant |
| $d = 2\ln 2 - 1$ | A1* | Sufficient correct working to justify given answer |\begin{enumerate}
\item A particle, $P$, moves on the $x$-axis. At time $t$ seconds, $t \geqslant 0$, the velocity of $P$ is $v \mathrm {~ms} ^ { - 1 }$ in the direction of $x$ increasing and the acceleration of $P$ is $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$ in the direction of $x$ increasing.
\end{enumerate}
When $t = 0$ the particle is at rest at the origin $O$.\\
Given that $a = \frac { 5 } { 2 } ( 5 - v )$\\
(a) show that $v = 5 \left( 1 - \mathrm { e } ^ { - 2.5 t } \right)$\\
(b) state the limiting value of $v$ as $t$ increases.
At the instant when $v = 2.5$, the particle is $d$ metres from $O$.\\
(c) Show that $d = 2 \ln 2 - 1$
\hfill \mbox{\textit{Edexcel FM2 AS 2018 Q4 [13]}}