| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv² - horizontal motion or engine power |
| Difficulty | Challenging +1.2 This is a standard Further Mechanics 2 variable force question with air resistance proportional to v². While it requires setting up F=ma with resistance, separating variables, and integrating (including partial fractions for part a), these are well-practiced FM2 techniques. The 'show that' format provides target answers to work towards, reducing problem-solving demand. More routine than average FM2 content but harder than typical single-maths mechanics. |
| Spec | 1.08h Integration by substitution6.02a Work done: concept and definition6.02l Power and velocity: P = Fv6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equation of motion: \(1800 - 2v^2 = 500a\) | B1 | All three terms dimensionally correct |
| Select form \(a = 500\dfrac{dv}{dt}\) | M1 | Use correct form for acceleration to give equation in \(v\), \(t\) only |
| \(\int \dfrac{2}{500}dt = \int \dfrac{1}{900-v^2}dv = \dfrac{1}{60}\int \dfrac{1}{30+v} + \dfrac{1}{30-v}dv\) | M1 | Separate variables and integrate |
| \(\dfrac{t}{250} = \dfrac{1}{60}\ln(30+v) - \dfrac{1}{60}\ln(30-v)\ (+C)\) | A1 | Condone missing \(C\) |
| Use boundary conditions correctly | M1 | |
| \(T = \dfrac{25}{6}\ln\!\left(\dfrac{30+10}{30-10}\right) = \dfrac{25}{6}\ln 2\) | M1, A1* | Show sufficient working and statement that result has been achieved |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equation of motion: \(500v\dfrac{dv}{dx} = 1800 - 2v^2\) | M1 | Correct form of acceleration giving equation in \(v\), \(x\) only |
| \(\int \dfrac{500v}{1800 - 2v^2}dv = \int 1\,dx\) | M1 | Separate variables and integrate |
| \(-125\ln(1800 - 2v^2) = x\ (+C)\) | A1 | Condone missing \(C\) |
| Use boundary conditions: \(x = -125\ln 1600 + 125\ln 1800\) | M1 | Extract and use boundary conditions |
| \(x = 125\ln\dfrac{9}{8}\ \text{(m)}\) | A1* | Show sufficient working and statement that result has been achieved |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion: $1800 - 2v^2 = 500a$ | B1 | All three terms dimensionally correct |
| Select form $a = 500\dfrac{dv}{dt}$ | M1 | Use correct form for acceleration to give equation in $v$, $t$ only |
| $\int \dfrac{2}{500}dt = \int \dfrac{1}{900-v^2}dv = \dfrac{1}{60}\int \dfrac{1}{30+v} + \dfrac{1}{30-v}dv$ | M1 | Separate variables and integrate |
| $\dfrac{t}{250} = \dfrac{1}{60}\ln(30+v) - \dfrac{1}{60}\ln(30-v)\ (+C)$ | A1 | Condone missing $C$ |
| Use boundary conditions correctly | M1 | |
| $T = \dfrac{25}{6}\ln\!\left(\dfrac{30+10}{30-10}\right) = \dfrac{25}{6}\ln 2$ | M1, A1* | Show sufficient working and statement that result has been achieved |
---
## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion: $500v\dfrac{dv}{dx} = 1800 - 2v^2$ | M1 | Correct form of acceleration giving equation in $v$, $x$ only |
| $\int \dfrac{500v}{1800 - 2v^2}dv = \int 1\,dx$ | M1 | Separate variables and integrate |
| $-125\ln(1800 - 2v^2) = x\ (+C)$ | A1 | Condone missing $C$ |
| Use boundary conditions: $x = -125\ln 1600 + 125\ln 1800$ | M1 | Extract and use boundary conditions |
| $x = 125\ln\dfrac{9}{8}\ \text{(m)}$ | A1* | Show sufficient working and statement that result has been achieved |
---\begin{enumerate}
\item A car of mass 500 kg moves along a straight horizontal road.
\end{enumerate}
The engine of the car produces a constant driving force of 1800 N .\\
The car accelerates from rest from the fixed point $O$ at time $t = 0$ and at time $t$ seconds the car is $x$ metres from $O$, moving with speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
When the speed of the car is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the resistance to the motion of the car has magnitude $2 v ^ { 2 } \mathrm {~N}$.
At time $T$ seconds, the car is at the point $A$, moving with speed $10 \mathrm {~ms} ^ { - 1 }$.\\
(a) Show that $T = \frac { 25 } { 6 } \ln 2$\\
(b) Show that the distance from $O$ to $A$ is $125 \ln \frac { 9 } { 8 } \mathrm {~m}$.
\hfill \mbox{\textit{Edexcel FM2 Q4 [11]}}