| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Resistance as kv² - finding constant k |
| Difficulty | Standard +0.8 This FM2 question requires applying the work-energy principle with variable resistance, deriving a differential equation relating v and x, then integrating a rational function with cubic terms. The 'show that' part involves connecting P=Fv with F=ma in the chain rule form, which is non-trivial. Part (b) requires integrating 60v²/(200-3v³) using substitution—moderately challenging algebra but standard Further Maths technique. More demanding than typical A-level mechanics due to the calculus-dynamics integration and algebraic manipulation required. |
| Spec | 1.08h Integration by substitution6.02l Power and velocity: P = Fv6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Use of \(P = Fv\) | B1 | Seen or implied; need at least \(200 = Fv\); could be on its own, in an equation or on a diagram |
| Equation of motion: \(F - 3v^2 = 60a\) | M1 | Form equation of motion. Need all terms and dimensionally correct. Condone any correct form for acceleration and sign errors. Allow with \(m\) not substituted |
| \(\frac{200}{v} - 3v^2 = 60v\frac{dv}{dx}\) | A1 | Correct equation - any equivalent form with correct acceleration |
| \(\frac{dv}{dx} = \frac{200 - 3v^3}{60v^2}\) * | A1* | Obtain given answer from correct working. Must be as written in question but could swap LHS and RHS |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int \frac{60v^2}{200 - 3v^3}\, dv = \int 1\, dx \quad \left(-\frac{60}{9}\ln(200 - 3v^3) = x(+C)\right)\) | M1 | Separate variables and integrate to obtain \((x =) k\ln(\ldots)\). Constant of integration not required. Condone if \(x\) is not explicitly stated but M0 if incorrect function |
| \(D = \left[-\frac{60}{9}\ln(200 - 3v^3)\right]_2^4 = -\frac{60}{9}\ln\!\left(\frac{200 - 3\times 64}{200 - 3\times 8}\right)\) | M1 | Use limits correctly in an expression containing \(k\ln(200 - 3v^3)\) to find \(D\). Substitute and subtract in the correct order |
| \(= \frac{60}{9}\ln\frac{176}{8} = \frac{60}{9}\ln 22\) | A1 | Obtain exact answer from correct working. Any equivalent single term. No working seen is Max M1M0A0 |
## Question 2:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Use of $P = Fv$ | B1 | Seen or implied; need at least $200 = Fv$; could be on its own, in an equation or on a diagram |
| Equation of motion: $F - 3v^2 = 60a$ | M1 | Form equation of motion. Need all terms and dimensionally correct. Condone any correct form for acceleration and sign errors. Allow with $m$ not substituted |
| $\frac{200}{v} - 3v^2 = 60v\frac{dv}{dx}$ | A1 | Correct equation - any equivalent form with correct acceleration |
| $\frac{dv}{dx} = \frac{200 - 3v^3}{60v^2}$ * | A1* | Obtain given answer from correct working. Must be as written in question but could swap LHS and RHS |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int \frac{60v^2}{200 - 3v^3}\, dv = \int 1\, dx \quad \left(-\frac{60}{9}\ln(200 - 3v^3) = x(+C)\right)$ | M1 | Separate variables and integrate to obtain $(x =) k\ln(\ldots)$. Constant of integration not required. Condone if $x$ is not explicitly stated but M0 if incorrect function |
| $D = \left[-\frac{60}{9}\ln(200 - 3v^3)\right]_2^4 = -\frac{60}{9}\ln\!\left(\frac{200 - 3\times 64}{200 - 3\times 8}\right)$ | M1 | Use limits correctly in an expression containing $k\ln(200 - 3v^3)$ to find $D$. Substitute and subtract in the correct order |
| $= \frac{60}{9}\ln\frac{176}{8} = \frac{60}{9}\ln 22$ | A1 | Obtain exact answer from correct working. Any equivalent single term. No working seen is Max M1M0A0 |
---\begin{enumerate}
\item A cyclist and her cycle have a combined mass of 60 kg . The cyclist is moving along a straight horizontal road and is working at a constant rate of 200 W .
\end{enumerate}
When she has travelled a distance $x$ metres, her speed is $v \mathrm {~ms} ^ { - 1 }$ and the magnitude of the resistance to motion is $3 v ^ { 2 } \mathrm {~N}$.\\
(a) Show that $\frac { \mathrm { d } v } { \mathrm {~d} x } = \frac { 200 - 3 v ^ { 3 } } { 60 v ^ { 2 } }$
The distance travelled by the cyclist as her speed increases from $2 \mathrm {~ms} ^ { - 1 }$ to $4 \mathrm {~ms} ^ { - 1 }$ is $D$ metres.\\
(b) Find the exact value of $D$
\hfill \mbox{\textit{Edexcel FM2 2022 Q2 [7]}}