Edexcel M3 2002 June — Question 3 10 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2002
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable Force
TypeMotion with exponential force
DifficultyStandard +0.3 This is a standard M3 variable force question requiring application of F=ma with v dv/dx, followed by integration of an exponential function and interpretation of a limiting value. The techniques are routine for this module, though the exponential makes it slightly above average difficulty compared to simpler polynomial force questions.
Spec3.02f Non-uniform acceleration: using differentiation and integration3.03d Newton's second law: 2D vectors6.06a Variable force: dv/dt or v*dv/dx methods

A particle \(P\) of mass 2.5 kg moves along the positive \(x\)-axis. It moves away from a fixed origin \(O\), under the action of a force directed away from \(O\). When \(OP = x\) metres the magnitude of the force is \(2e^{-0.1x}\) newtons and the speed of \(P\) is \(v\) m s\(^{-1}\). When \(x = 0\), \(v = 2\). Find
  1. \(v^2\) in terms of \(x\), [6]
  2. the value of \(x\) when \(v = 4\). [3]
  3. Give a reason why the speed of \(P\) does not exceed \(\sqrt{20}\) m s\(^{-1}\). [1]

A particle $P$ of mass 2.5 kg moves along the positive $x$-axis. It moves away from a fixed origin $O$, under the action of a force directed away from $O$. When $OP = x$ metres the magnitude of the force is $2e^{-0.1x}$ newtons and the speed of $P$ is $v$ m s$^{-1}$. When $x = 0$, $v = 2$. Find

\begin{enumerate}[label=(\alph*)]
\item $v^2$ in terms of $x$,
[6]
\item the value of $x$ when $v = 4$.
[3]
\item Give a reason why the speed of $P$ does not exceed $\sqrt{20}$ m s$^{-1}$.
[1]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3 2002 Q3 [10]}}