Edexcel M4 2006 June — Question 5 11 marks

Exam BoardEdexcel
ModuleM4 (Mechanics 4)
Year2006
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicPower and driving force
TypeVariable resistance: find constant speed
DifficultyStandard +0.8 Part (a) is a standard derivation using F=ma and power=force×velocity (3 marks of routine setup). Part (b) requires separating variables in a non-trivial differential equation, partial fractions decomposition, integration, and algebraic manipulation to reach a final answer—this is a substantial M4 problem requiring multiple techniques and careful algebra, placing it moderately above average difficulty.
Spec4.10b Model with differential equations: kinematics and other contexts6.02l Power and velocity: P = Fv

A train of mass \(m\) is moving along a straight horizontal railway line. A time \(t\), the train is moving with speed \(v\) and the resistance to motion has magnitude \(kv\), where \(k\) is a constant. The engine of the train is working at a constant rate \(P\).
  1. Show that, when \(v > 0\), \(mv\frac{dv}{dt} + kv^2 = P\). [3]
When \(t = 0\), the speed of the train is \(\frac{1}{3}\sqrt{\frac{P}{k}}\).
  1. Find, in terms of \(m\) and \(k\), the time taken for the train to double its initial speed. [8]

A train of mass $m$ is moving along a straight horizontal railway line. A time $t$, the train is moving with speed $v$ and the resistance to motion has magnitude $kv$, where $k$ is a constant. The engine of the train is working at a constant rate $P$.

\begin{enumerate}[label=(\alph*)]
\item Show that, when $v > 0$, $mv\frac{dv}{dt} + kv^2 = P$.
[3]
\end{enumerate}

When $t = 0$, the speed of the train is $\frac{1}{3}\sqrt{\frac{P}{k}}$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find, in terms of $m$ and $k$, the time taken for the train to double its initial speed.
[8]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M4 2006 Q5 [11]}}