| Exam Board | Edexcel |
|---|---|
| Module | M5 (Mechanics 5) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Variable mass problems (mass increasing) |
| Difficulty | Challenging +1.8 This is a challenging M5 variable mass problem requiring application of Newton's second law from first principles with momentum considerations, followed by separating variables and integrating a non-trivial expression involving inverse tanh. It demands strong conceptual understanding beyond standard mechanics and careful algebraic manipulation, placing it well above average difficulty. |
| Spec | 6.03b Conservation of momentum: 1D two particles |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Using impulse-momentum: \(\frac{d(mv)}{dt} = mg\) | M1 | Newton's second law from first principles |
| \(m\frac{dv}{dt} + v\frac{dm}{dt} = mg\) | M1 | Product rule correctly applied |
| \(\frac{dm}{dt} = mkv\) | B1 | Statement of mass increase rate |
| \(m\frac{dv}{dt} + v(mkv) = mg\) | M1 | Substitution |
| \(\frac{dv}{dt} = g - kv^2\) | A1 | Correct completion (given) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\int \frac{dv}{g-kv^2} = \int dt\) | M1 | Separating variables |
| \(\frac{1}{2\sqrt{gk}}\ln\left(\frac{\sqrt{g}+\sqrt{k}v}{\sqrt{g}-\sqrt{k}v}\right) = t\) (or equivalent \(\tanh\) form) | M1 A1 | Correct integration |
| Substitute \(v = \frac{1}{2}\sqrt{\frac{g}{k}}\): \(t = \frac{1}{2\sqrt{gk}}\ln\left(\frac{3}{1}\right) = \frac{\ln 3}{2\sqrt{gk}}\) | A1 | Correct final answer |
## Question 7:
**(a) Show $\frac{dv}{dt} = g - kv^2$**
| Working/Answer | Mark | Guidance |
|---|---|---|
| Using impulse-momentum: $\frac{d(mv)}{dt} = mg$ | M1 | Newton's second law from first principles |
| $m\frac{dv}{dt} + v\frac{dm}{dt} = mg$ | M1 | Product rule correctly applied |
| $\frac{dm}{dt} = mkv$ | B1 | Statement of mass increase rate |
| $m\frac{dv}{dt} + v(mkv) = mg$ | M1 | Substitution |
| $\frac{dv}{dt} = g - kv^2$ | A1 | Correct completion (given) |
**(b) Time to reach speed $\frac{1}{2}\sqrt{\frac{g}{k}}$**
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\int \frac{dv}{g-kv^2} = \int dt$ | M1 | Separating variables |
| $\frac{1}{2\sqrt{gk}}\ln\left(\frac{\sqrt{g}+\sqrt{k}v}{\sqrt{g}-\sqrt{k}v}\right) = t$ (or equivalent $\tanh$ form) | M1 A1 | Correct integration |
| Substitute $v = \frac{1}{2}\sqrt{\frac{g}{k}}$: $t = \frac{1}{2\sqrt{gk}}\ln\left(\frac{3}{1}\right) = \frac{\ln 3}{2\sqrt{gk}}$ | A1 | Correct final answer |
I can see this is an exam paper (Question 8) about moments of inertia and rotation of a lamina. Let me extract the mark scheme content based on what can be inferred from the question structure. However, I should note that **these pages shown are answer/working pages and the final question page** — they do not actually contain a mark scheme.
What is visible is the **question content** for Q8. I can outline the solution approach for each part:
---7. A raindrop absorbs water as it falls vertically under gravity through a cloud. In a model of the motion the cloud is assumed to consist of stationary water particles. At time $t$, the mass of the raindrop is $m$ and the speed of the raindrop is $v$. At time $t = 0$, the raindrop is at rest. The rate of increase of the mass of the raindrop with respect to time is modelled as being $m k v$, where $k$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Ignoring air resistance, show from first principles, that
$$\frac { \mathrm { d } v } { \mathrm {~d} t } = g - k v ^ { 2 }$$
\item Find the time taken for the raindrop to reach a speed of $\frac { 1 } { 2 } \sqrt { } \left( \frac { g } { k } \right)$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M5 2014 Q7 [9]}}