2014 Paper 3 Q9

Year: 2014
Paper: 3
Question Number: 9

Course: UFM Mechanics
Section: Variable Force

Difficulty: 1700.0 Banger: 1500.0

Problem

A particle of mass \(m\) is projected with velocity \(\+ u\). It is acted upon by the force \(m\+g\) due to gravity and by a resistive force \(-mk \+v\), where \(\+v\) is its velocity and \(k\) is a positive constant. Given that, at time \(t\) after projection, its position \(\+r\) relative to the point of projection is given by \[ \+r = \frac{kt -1 +\.e^{-kt}} {k^2} \, \+g + \frac{ 1-\.e^{-kt}}{k} \, \+u \,, \] find an expression for \(\+v\) in terms of \(k\), \(t\), \(\+g\) and \(\+u\). Verify that the equation of motion and the initial conditions are satisfied. Let \(\+u = u\cos\alpha \, \+i + u \sin\alpha \, \+j\) and $\+g = -g\, \+j\(, where \)0<\alpha<90^\circ\(, and let \)T$ be the time after projection at which \(\+r \,.\, \+j =0\). Show that \[ uk \sin\alpha = \left(\frac{kT}{1-\.e^{-kT}} -1\right)g\,. \] Let \(\beta\) be the acute angle between \(\+v\) and \(\+i\) at time \(T\). Show that \[ \tan\beta = \frac{(\.e^{kT}-1)g}{uk\cos\alpha}-\tan\alpha \,. \] Show further that \(\tan\beta >\tan\alpha\) (you may assume that \(\sinh kT >kT\)) and deduce that~\(\beta >\alpha\).

No solution available for this problem.

Examiner's report
— 2014 STEP 3, Question 9
Mean: ~7 / 20 (inferred) 20% attempted Inferred 7.0/20 from 'scoring at the same level as Q8 (7/20)'

A fifth attempted this, scoring at the same level as question 8. The first differentiation and the verification of the initial conditions were managed but very few bothered to check that the equation of motion was satisfied. Most obtained the first displayed result but few realised that θ was the angle of depression rather than elevation and this generated plenty of sign errors. A few did achieve the very final result.

A 10% increase in the number of candidates and the popularity of all questions ensured that all questions had a good number of attempts, though the first two questions were very much the most popular. Every question received at least one absolutely correct solution. In most cases when candidates submitted more than six solutions, the extra ones were rarely substantial attempts. Five sixths gave in at least six attempts.

Source: Cambridge STEP 2014 Examiner's Report · 2014-full.pdf
Rating Information

Difficulty Rating: 1700.0

Difficulty Comparisons: 0

Banger Rating: 1500.0

Banger Comparisons: 0

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Problem source
A particle of mass $m$ is projected with velocity $\+ u$. It is
  acted upon by the force $m\+g$ due to gravity and by a resistive
  force $-mk \+v$, where $\+v$ is its velocity and $k$ is a positive
  constant.
  Given that, at time $t$ after projection, its position $\+r$
  relative to the point of projection is given by
  \[
  \+r = \frac{kt -1 +\.e^{-kt}} {k^2} \, \+g + \frac{ 1-\.e^{-kt}}{k}
  \, \+u \,,
  \]
  find an expression for $\+v$ in terms of $k$, $t$, $\+g$ and
  $\+u$. Verify that the equation of motion and the initial conditions
  are satisfied.
  Let $\+u = u\cos\alpha \, \+i + u \sin\alpha \, \+j$ and $\+g = -g\,
  \+j$, where $0<\alpha<90^\circ$, and let $T$ be the time after projection at
  which $\+r \,.\, \+j =0$. Show that
  \[
  uk \sin\alpha = \left(\frac{kT}{1-\.e^{-kT}} -1\right)g\,.
  \]
  Let $\beta$ be the acute angle between $\+v$ and $\+i$ at time
  $T$. Show that
  \[
  \tan\beta = \frac{(\.e^{kT}-1)g}{uk\cos\alpha}-\tan\alpha \,.
  \]
  Show further that $\tan\beta >\tan\alpha$ (you may assume that
 $\sinh kT >kT$) and deduce that~$\beta >\alpha$.