| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2018 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable Force |
| Type | Given velocity function find force |
| Difficulty | Challenging +1.2 This is a standard M4 variable force question requiring differentiation of the given v²(x) relation to find resistance, finding maximum height by setting v=0, and integrating to find time. While it involves multiple steps and exponential/trigonometric functions, the techniques are routine for M4 students: applying F=ma with v(dv/dx), algebraic manipulation, and standard integration with substitution. The 'show that' format provides targets to work towards, reducing problem-solving demand. |
| Spec | 6.06a Variable force: dv/dt or v*dv/dx methods |
6. A particle of mass $m$ is projected vertically upwards in a resisting medium. As the particle moves upwards, the speed $v$ of the particle is given by
$$v ^ { 2 } = k g \left( 5 \mathrm { e } ^ { - \frac { x } { 2 k } } - 4 \right)$$
where $x$ is the distance of the particle above the point of projection and $k$ is a positive constant.
\begin{enumerate}[label=(\alph*)]
\item Show that the magnitude of the resistance to the motion of the particle is $\frac { m v ^ { 2 } } { 4 k }$.\\
(4)
\item Find, in terms of $k$, the greatest height reached by the particle above the point of projection.
\item Show that the time taken by the particle to reach its greatest height above the point of projection is $\sqrt { \frac { 4 k } { g } } \arctan \left( \frac { 1 } { 2 } \right)$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2018 Q6 [14]}}