| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2012 |
| Session | June |
| Marks | 16 |
| 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 standard M4 variable force question requiring Newton's second law for connected particles with air resistance, followed by solving a first-order linear differential equation and analyzing limiting behavior. While it involves multiple steps and careful algebra, the techniques are well-practiced in M4 and the question provides clear guidance through parts (a)-(c), making it moderately above average difficulty but not requiring novel insight. |
| Spec | 3.03k Connected particles: pulleys and equilibrium6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| \(2mg - T - kv^2 = 2ma\) | M1 A1 | Equation of motion for particle of mass \(2m\) aef |
| \(T - mg - kv^2 = ma\) | M1 A1 | Equation of motion for particle of mass \(m\) aef |
| Adding, \(mg - 2kv^2 = 3ma\) | ||
| \(\frac{2g}{3} - \frac{4kv^2}{3m} = 2v\frac{dv}{dx}\) | DM1 A1 | Eliminate \(T\), substitute for \(a\) and rearrange. Dependent on both previous M marks. Reach given answer correctly |
| Answer | Marks | Guidance |
|---|---|---|
| \(IF = e^{\int\frac{4k}{3m}dx} = e^{\frac{4kx}{3m}}\) | B1 | |
| \(v^2e^{\frac{4kx}{3m}} = \frac{2g}{3}\int e^{\frac{4kx}{3m}}dx = \frac{2g}{3} \cdot \frac{2k}{2k}e^{\frac{4kx}{3m}}(+C)\) | M1 A1 | Use integrating factor to obtain \(\frac{d}{dx}\left(v^2 e^{\frac{4kx}{3m}}\right) = \frac{2g}{3}e^{\frac{4kx}{3m}}\) and integrate |
| \(v^2 = \frac{mg}{2k} + Ce^{-\frac{4kx}{3m}}\) | ||
| \(x = 0, v = 0 \Rightarrow C = -\frac{mg}{2k}\) | M1 | Use initial values to evaluate \(C\) or as limits in a definite integral and find an expression for \(v^2\). aef. |
| \(v^2 = \frac{mg}{2k}\left(1 - e^{-\frac{4kx}{3m}}\right)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Separate variables: \(\int\frac{3m}{2mg-4kv^2}dv^2 = \int 1dx\) | B1 | CF: \(v^2 = Ae^{-\frac{4k}{3m}}\) |
| \(x = -\frac{3m}{4k}\ln\left | 2mg - 4kv^2\right | (+C)\) |
| \(x = -\frac{3m}{4k}\ln\left | \frac{2mg}{2mg-4kv^2}\right | \) |
| \(v^2 = \frac{mg}{2k}\left(1 - e^{-\frac{4kx}{3m}}\right)\) | A1 | \(v^2 = \frac{mg}{2k}\left(1 - e^{-\frac{4kx}{3m}}\right)\) |
| Answer | Marks | Guidance |
|---|---|---|
| When \(x = 0, T = \frac{4mg}{3}\) | M1 A1 | Substitute \(v = 0\) in the initial equations and solve for \(T\) |
| As \(x \to \infty, T \to \frac{9mg}{6} = \frac{3mg}{2}\) | M1 A1 | For large \(x\), \(v^2 \to \frac{mg}{2k}\). Substitute in the initial equations and solve for \(T\) cwo – answer is given. |
| Hence, \(\frac{4mg}{3} \leq T < \frac{3mg}{2}\) | A1 |
**Part (a)**
| $2mg - T - kv^2 = 2ma$ | M1 A1 | Equation of motion for particle of mass $2m$ aef |
| $T - mg - kv^2 = ma$ | M1 A1 | Equation of motion for particle of mass $m$ aef |
| Adding, $mg - 2kv^2 = 3ma$ | | |
| $\frac{2g}{3} - \frac{4kv^2}{3m} = 2v\frac{dv}{dx}$ | DM1 A1 | Eliminate $T$, substitute for $a$ and rearrange. Dependent on both previous M marks. Reach given answer correctly |
**(6) marks**
**Part (b)**
| $IF = e^{\int\frac{4k}{3m}dx} = e^{\frac{4kx}{3m}}$ | B1 | |
| $v^2e^{\frac{4kx}{3m}} = \frac{2g}{3}\int e^{\frac{4kx}{3m}}dx = \frac{2g}{3} \cdot \frac{2k}{2k}e^{\frac{4kx}{3m}}(+C)$ | M1 A1 | Use integrating factor to obtain $\frac{d}{dx}\left(v^2 e^{\frac{4kx}{3m}}\right) = \frac{2g}{3}e^{\frac{4kx}{3m}}$ and integrate |
| $v^2 = \frac{mg}{2k} + Ce^{-\frac{4kx}{3m}}$ | | |
| $x = 0, v = 0 \Rightarrow C = -\frac{mg}{2k}$ | M1 | Use initial values to evaluate $C$ or as limits in a definite integral and find an expression for $v^2$. aef. |
| $v^2 = \frac{mg}{2k}\left(1 - e^{-\frac{4kx}{3m}}\right)$ | A1 | |
**(5) marks**(13)
**OR**
| Separate variables: $\int\frac{3m}{2mg-4kv^2}dv^2 = \int 1dx$ | B1 | CF: $v^2 = Ae^{-\frac{4k}{3m}}$ |
| $x = -\frac{3m}{4k}\ln\left|2mg - 4kv^2\right| (+C)$ | M1A1 | PI: $v^2 = b = 0 + \frac{4k}{3m}b = \frac{2g}{3}$; GS: $v^2 = Ae^{-\frac{4k}{3m}} + \frac{mg}{2k}$ |
| $x = -\frac{3m}{4k}\ln\left|\frac{2mg}{2mg-4kv^2}\right|$ | M1 | $x = 0, v = 0 \Rightarrow A = -\frac{mg}{2k}$ |
| $v^2 = \frac{mg}{2k}\left(1 - e^{-\frac{4kx}{3m}}\right)$ | A1 | $v^2 = \frac{mg}{2k}\left(1 - e^{-\frac{4kx}{3m}}\right)$ |
**Part (c)**
| When $x = 0, T = \frac{4mg}{3}$ | M1 A1 | Substitute $v = 0$ in the initial equations and solve for $T$ |
| As $x \to \infty, T \to \frac{9mg}{6} = \frac{3mg}{2}$ | M1 A1 | For large $x$, $v^2 \to \frac{mg}{2k}$. Substitute in the initial equations and solve for $T$ cwo – answer is given. |
| Hence, $\frac{4mg}{3} \leq T < \frac{3mg}{2}$ | A1 | |
**(5)16 marks total for Question 3**
---\begin{enumerate}
\item Two particles, of masses $m$ and $2 m$, are connected to the ends of a long light inextensible string. The string passes over a small smooth fixed pulley and hangs vertically on either side. The particles are released from rest with the string taut. Each particle is subject to air resistance of magnitude $k v ^ { 2 }$, where $v$ is the speed of each particle after it has moved a distance $x$ from rest and $k$ is a positive constant.\\
(a) Show that $\frac { \mathrm { d } } { \mathrm { d } x } \left( v ^ { 2 } \right) + \frac { 4 k } { 3 m } v ^ { 2 } = \frac { 2 g } { 3 }$\\
(b) Find $v ^ { 2 }$ in terms of $x$.\\
(c) Deduce that the tension in the string, $T$, satisfies
\end{enumerate}
$$\frac { 4 m g } { 3 } \leqslant T < \frac { 3 m g } { 2 }$$
\hfill \mbox{\textit{Edexcel M4 2012 Q3 [16]}}