| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given velocity function find force |
| Difficulty | Standard +0.3 This is a standard M3 variable force question requiring chain rule differentiation (v dv/dx = a), substitution, and algebraic manipulation. Part (a) is a 'show that' requiring one differentiation step, part (b) uses terminal velocity (a=0), and part (c) involves straightforward substitution and rearrangement. All techniques are routine for M3 students with no novel insight required, making it slightly easier than average. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(v^2 = kg - kge^{-\frac{2t}{k}} \therefore 2v\frac{dv}{dx} = 2ge^{-\frac{2t}{k}}\) | M1 A2 | |
| \(f = \) accel. \(= v\frac{dv}{dx} = ge^{-\frac{2t}{k}}\) | A1 | |
| (b) when \(x\) is large, \(e^{-\frac{2t}{k}} \to 0\) | M1 | |
| \(\therefore 49^2 = kg\) giving \(k = \frac{49^2}{9.8} = 245\) | M1 A1 | |
| (c) \(v^2 = kg - kge^{-\frac{2t}{k}} = kg - kf\) | M1 A1 | |
| \(\therefore f = g - \frac{1}{k}v^2 = 9.8 - \frac{1}{245}v^2\) | M1 A1 | (11) |
(a) $v^2 = kg - kge^{-\frac{2t}{k}} \therefore 2v\frac{dv}{dx} = 2ge^{-\frac{2t}{k}}$ | M1 A2
$f = $ accel. $= v\frac{dv}{dx} = ge^{-\frac{2t}{k}}$ | A1
(b) when $x$ is large, $e^{-\frac{2t}{k}} \to 0$ | M1
$\therefore 49^2 = kg$ giving $k = \frac{49^2}{9.8} = 245$ | M1 A1
(c) $v^2 = kg - kge^{-\frac{2t}{k}} = kg - kf$ | M1 A1
$\therefore f = g - \frac{1}{k}v^2 = 9.8 - \frac{1}{245}v^2$ | M1 A1 | (11)
---4. Whilst in free-fall a parachutist falls vertically such that his velocity, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, when he is $x$ metres below his initial position is given by
$$v ^ { 2 } = k g \left( 1 - \mathrm { e } ^ { - \frac { 2 x } { k } } \right) ,$$
where $k$ is a constant.\\
Given that he experiences an acceleration of $\mathrm { f } \mathrm { m } \mathrm { s } ^ { - 2 }$,
\begin{enumerate}[label=(\alph*)]
\item show that $f = g \mathrm { e } ^ { - \frac { 2 x } { k } }$.
After falling a large distance, his velocity is constant at $49 \mathrm {~ms} ^ { - 1 }$.
\item Find the value of $k$.
\item Hence, express $f$ in the form ( $\lambda - \mu v ^ { 2 }$ ) where $\lambda$ and $\mu$ are constants which you should find.\\
(4 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q4 [11]}}