| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Air resistance kv² - falling from rest or projected downward |
| Difficulty | Challenging +1.2 This is a standard Further Mechanics 2 question on variable force with air resistance proportional to v². While it requires integration techniques (separation of variables, partial fractions for part a, and chain rule manipulation for part c), these are well-practiced methods in FM2. The 'show that' format provides the target, reducing problem-solving demand. The interpretation in part (b) is straightforward (terminal velocity). This is moderately above average difficulty due to the algebraic manipulation and being Further Maths content, but remains a textbook-style exercise without novel insight required. |
| Spec | 1.08h Integration by substitution6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(mg - kv^2 = m\frac{\mathrm{d}v}{\mathrm{d}t}\) | M1 | Equation of motion with correct form for the acceleration |
| Separate variables and integrate | M1 | Separate the variables and integrate ('standard integral') |
| \(t = \frac{m}{k} \cdot \frac{1}{2\sqrt{\frac{mg}{k}}} \ln\left(\frac{\sqrt{\frac{mg}{k}}+v}{\sqrt{\frac{mg}{k}}-v}\right) \; (+C)\) | A1 | Correct equation in any form (ignoring constant or limits) |
| \(t = \frac{V}{2g}\ln\left(\frac{V+v}{V-v}\right)\) where \(V^2 = \frac{mg}{k}\) | A1* | Correctly obtain the printed answer including dealing with constant or limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(V^2 = \frac{mg}{k} \Rightarrow kV^2 = mg\) i.e. resistance = weight OR using answer to (a): as \(t\to\infty\), \(v\to V\) from below | B1 | Correctly rearrange and interpret OR correctly argue and interpret |
| Hence \(V\) is the terminal velocity of the stone | B1 | Correct statement or equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(mg - kv^2 = mv\frac{\mathrm{d}v}{\mathrm{d}s}\) | M1 | Equation of motion with correct form for the acceleration |
| Separate variables and integrate | M1 | Separate the variables and integrate ('standard integral') |
| \(s = -\frac{m}{2k}\ln\left(\frac{mg}{k} - v^2\right) \; (+D)\) | A1 | Correct equation in any form (ignoring constant or limits) |
| \(s = \frac{V^2}{2g}\ln\left(\frac{V^2}{V^2-v^2}\right)\) | A1* | Correctly obtain the printed answer including dealing with constant or limits |
## Question 2:
### Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $mg - kv^2 = m\frac{\mathrm{d}v}{\mathrm{d}t}$ | M1 | Equation of motion with correct form for the acceleration |
| Separate variables and integrate | M1 | Separate the variables and integrate ('standard integral') |
| $t = \frac{m}{k} \cdot \frac{1}{2\sqrt{\frac{mg}{k}}} \ln\left(\frac{\sqrt{\frac{mg}{k}}+v}{\sqrt{\frac{mg}{k}}-v}\right) \; (+C)$ | A1 | Correct equation in any form (ignoring constant or limits) |
| $t = \frac{V}{2g}\ln\left(\frac{V+v}{V-v}\right)$ where $V^2 = \frac{mg}{k}$ | A1* | Correctly obtain the printed answer including dealing with constant or limits |
### Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V^2 = \frac{mg}{k} \Rightarrow kV^2 = mg$ i.e. resistance = weight **OR** using answer to (a): as $t\to\infty$, $v\to V$ from below | B1 | Correctly rearrange and interpret OR correctly argue and interpret |
| Hence $V$ is the terminal velocity of the stone | B1 | Correct statement or equivalent |
### Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $mg - kv^2 = mv\frac{\mathrm{d}v}{\mathrm{d}s}$ | M1 | Equation of motion with correct form for the acceleration |
| Separate variables and integrate | M1 | Separate the variables and integrate ('standard integral') |
| $s = -\frac{m}{2k}\ln\left(\frac{mg}{k} - v^2\right) \; (+D)$ | A1 | Correct equation in any form (ignoring constant or limits) |
| $s = \frac{V^2}{2g}\ln\left(\frac{V^2}{V^2-v^2}\right)$ | A1* | Correctly obtain the printed answer including dealing with constant or limits |\begin{enumerate}
\item At time $t = 0$, a small stone $P$ of mass $m$ is released from rest and falls vertically through the air. At time $t$, the speed of $P$ is $v$ and the resistance to the motion of $P$ from the air is modelled as a force of magnitude $k v ^ { 2 }$, where $k$ is a constant.\\
(a) Show that $t = \frac { V } { 2 g } \ln \left( \frac { V + v } { V - v } \right)$ where $V ^ { 2 } = \frac { m g } { k }$\\
(b) Give an interpretation of the value of $V$, justifying your answer.
\end{enumerate}
At time $t , P$ has fallen a distance $s$.\\
(c) Show that $s = \frac { V ^ { 2 } } { 2 g } \ln \left( \frac { V ^ { 2 } } { V ^ { 2 } - v ^ { 2 } } \right)$
\hfill \mbox{\textit{Edexcel FM2 2021 Q2 [10]}}