| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2019 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Variable force (velocity v) - use v dv/dx |
| Difficulty | Challenging +1.2 This is a Further Mechanics 2 question requiring the v dv/dx technique with F=k/v, leading to a separable differential equation. Part (a) involves standard integration and using two boundary conditions to find constants—straightforward for FM students. Part (b) requires integrating dt = dx/v with a cubic relationship, which is algebraically involved but follows a clear method. The 'show that' format reduces difficulty. Harder than typical A-level mechanics due to the FM2 content, but represents a standard variable force question for this module. |
| Spec | 1.08h Integration by substitution6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Complete strategy to find \(v^3\) in terms of \(x\) | M1 | Complete strategy e.g. use of \(F=ma\) with appropriate form for \(a\), and use boundary conditions to confirm given result |
| \(0.4v\frac{dv}{dx} = \frac{k}{v}\) | M1 | Separate variables and integrate. Condone missing \(C\) |
| \(0.4v^2\frac{dv}{dx} = k\), \(\quad \frac{0.4}{3}v^3 = kx + C\) | A1 | Correct integration. Accept equivalent forms. Condone missing \(C\) |
| \(x=3, v=2\): \(\frac{3.2}{3} = 3k+C\) and \(x=6, v=2.5\): \(\frac{25}{12} = 6k+C\) | M1 | Use boundary conditions to form 2 equations in 2 unknowns and solve for \(k\) or \(C\) |
| \(3k = \frac{25}{12} - \frac{3.2}{3}\), \(\quad k = \frac{61}{180}\), \(\quad C = \frac{1}{20}\) | A1 | Obtain correct values for the constants |
| \(v^3 = \frac{3}{0.4}\left(\frac{61x}{180}+\frac{1}{20}\right) = \frac{61x+9}{24}\) * | A1* | Obtain given answer from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{5k}{2v} = \frac{dv}{dt} = \left(\frac{61}{72v}\right)\) | M1 | Select correct form for derivative and form a correct differential equation in \(v\) and \(t\) – follow their \(k\) |
| \(\int 2v\,dv = \int 5k\,dt \implies v^2 = 5kt + C'\) \(\quad (36v^2 = 61t + C')\) | M1 | Separate and integrate. Condone with no \(+C'\) – follow their \(k\) |
| \(\left[v^2\right]_2^{2.5} = \left[5kt\right]_0^T\), \(\quad \left(61T = 36(2.5^2 - 2^2)\right)\) | M1 | Evaluate definite integral of the form \(pv^2 = qt + C'\) or use limits to find value of constant of integration – follow their \(k\) |
| \(T = \frac{180}{61}\left(\frac{9}{20}\right) = \frac{81}{61}\) * | A1* | Obtain given answer from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dx}{dt} = \sqrt[3]{\frac{61x+9}{24}}\) | M1 | Select correct form for derivative and form a correct differential equation in \(x\) and \(t\) |
| \(\int(61x+9)^{\frac{1}{3}}dx = \int\frac{1}{\sqrt[3]{24}}dt\), \(\quad \frac{3}{2\times61}(61x+9)^{\frac{2}{3}} = \frac{t}{\sqrt[3]{24}} + C''\) | M1 | Separate and integrate. Condone with no \(+C''\) |
| \(T = 2\times\sqrt[3]{3}\times\frac{3}{2\times61}\left(375^{\frac{2}{3}}-192^{\frac{2}{3}}\right) = \frac{3\times\sqrt[3]{3}}{61}\left(\left(5\sqrt[3]{3}\right)^2 - \left(4\sqrt[3]{3}\right)^2\right)\) | M1 | Evaluate definite integral of the form \(pt = q(61x+9)^{\frac{2}{3}} + C''\) or use limits to find value of constant of integration |
| \(T = \frac{9}{61}(25-16) = \frac{81}{61}\) * | A1* | Obtain given answer from correct working |
## Question 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Complete strategy to find $v^3$ in terms of $x$ | M1 | Complete strategy e.g. use of $F=ma$ with appropriate form for $a$, and use boundary conditions to confirm given result |
| $0.4v\frac{dv}{dx} = \frac{k}{v}$ | M1 | Separate variables and integrate. Condone missing $C$ |
| $0.4v^2\frac{dv}{dx} = k$, $\quad \frac{0.4}{3}v^3 = kx + C$ | A1 | Correct integration. Accept equivalent forms. Condone missing $C$ |
| $x=3, v=2$: $\frac{3.2}{3} = 3k+C$ and $x=6, v=2.5$: $\frac{25}{12} = 6k+C$ | M1 | Use boundary conditions to form 2 equations in 2 unknowns and solve for $k$ or $C$ |
| $3k = \frac{25}{12} - \frac{3.2}{3}$, $\quad k = \frac{61}{180}$, $\quad C = \frac{1}{20}$ | A1 | Obtain correct values for the constants |
| $v^3 = \frac{3}{0.4}\left(\frac{61x}{180}+\frac{1}{20}\right) = \frac{61x+9}{24}$ * | A1* | Obtain **given answer** from correct working |
---
## Question 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{5k}{2v} = \frac{dv}{dt} = \left(\frac{61}{72v}\right)$ | M1 | Select correct form for derivative and form a correct differential equation in $v$ and $t$ – follow their $k$ |
| $\int 2v\,dv = \int 5k\,dt \implies v^2 = 5kt + C'$ $\quad (36v^2 = 61t + C')$ | M1 | Separate and integrate. Condone with no $+C'$ – follow their $k$ |
| $\left[v^2\right]_2^{2.5} = \left[5kt\right]_0^T$, $\quad \left(61T = 36(2.5^2 - 2^2)\right)$ | M1 | Evaluate definite integral of the form $pv^2 = qt + C'$ or use limits to find value of constant of integration – follow their $k$ |
| $T = \frac{180}{61}\left(\frac{9}{20}\right) = \frac{81}{61}$ * | A1* | Obtain **given answer** from correct working |
---
## Question 2(b) alt:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dx}{dt} = \sqrt[3]{\frac{61x+9}{24}}$ | M1 | Select correct form for derivative and form a correct differential equation in $x$ and $t$ |
| $\int(61x+9)^{\frac{1}{3}}dx = \int\frac{1}{\sqrt[3]{24}}dt$, $\quad \frac{3}{2\times61}(61x+9)^{\frac{2}{3}} = \frac{t}{\sqrt[3]{24}} + C''$ | M1 | Separate and integrate. Condone with no $+C''$ |
| $T = 2\times\sqrt[3]{3}\times\frac{3}{2\times61}\left(375^{\frac{2}{3}}-192^{\frac{2}{3}}\right) = \frac{3\times\sqrt[3]{3}}{61}\left(\left(5\sqrt[3]{3}\right)^2 - \left(4\sqrt[3]{3}\right)^2\right)$ | M1 | Evaluate definite integral of the form $pt = q(61x+9)^{\frac{2}{3}} + C''$ or use limits to find value of constant of integration |
| $T = \frac{9}{61}(25-16) = \frac{81}{61}$ * | A1* | Obtain **given answer** from correct working |
---\begin{enumerate}
\item A particle, $P$, of mass 0.4 kg is moving along the positive $x$-axis, in the positive $x$ direction under the action of a single force. At time $t$ seconds, $t > 0 , P$ is $x$ metres from the origin $O$ and the speed of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The force is acting in the direction of $x$ increasing and has magnitude $\frac { k } { v }$ newtons, where $k$ is a constant.
\end{enumerate}
At $x = 3 , v = 2$ and at $x = 6 , v = 2.5$\\
(a) Show that $v ^ { 3 } = \frac { 61 x + 9 } { 24 }$
The time taken for the speed of $P$ to increase from $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ to $2.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ is $T$ seconds.\\
(b) Use algebraic integration to show that $T = \frac { 81 } { 61 }$
\hfill \mbox{\textit{Edexcel FM2 2019 Q2 [10]}}