2012 Paper 3 Q11

Year: 2012
Paper: 3
Question Number: 11

Course: UFM Mechanics
Section: Work, energy and Power 2

Difficulty: 1700.0 Banger: 1500.0

Problem

One end of a thin heavy uniform inextensible perfectly flexible rope of length \(2L\) and mass \(2M\) is attached to a fixed point \(P\). A particle of mass \(m\) is attached to the other end. Initially, the particle is held at \(P\) and the rope hangs vertically in a loop below \(P\). The particle is then released so that it and a section of the rope (of decreasing length) fall vertically as shown in the diagram.
\psset{xunit=1.0cm,yunit=0.9cm,algebraic=true,dimen=middle,dotstyle=o,dotsize=3pt 0,linewidth=0.3pt,arrowsize=3pt 2,arrowinset=0.25} \begin{pspicture*}(0.13,-0.26)(3.26,5.51) \psline(1,5)(3,5) \psline[linewidth=0.1pt,linestyle=dashed,dash=2pt 2pt]{<->}(1.52,0)(1.52,5) \psline[linewidth=0.1pt,linestyle=dashed,dash=2pt 2pt]{<->}(2.53,3.2)(2.53,5) \psline(2.1,3.18)(2.06,0.25) \psline(2,5)(2.02,0.26) \psline(2.02,0.26)(2.03,0) \psline(2.03,0)(2.06,0.25) \rput[tl](1.94,5.45){\(P\)} \rput[tl](2.6,4.25){\(x\)} \rput[tl](0.2,2.85){\(L+\frac{1}{2}x\)} \begin{scriptsize} \psdots[dotsize=4pt 0,dotstyle=*](2.1,3.18) \end{scriptsize} \end{pspicture*}
You may assume that each point on the moving section of the rope falls at the same speed as the particle. Given that energy is conserved, show that, when the particle has fallen a distance \(x\) (where \(x< 2L\)), its speed \(v\) is given by \[ v^2 = \frac { 2g x \big( mL +ML - \frac14 Mx)}{mL +ML - \frac12 Mx}\,. \] Hence show that the acceleration of the particle is \[ g + \frac{ Mgx\big(mL+ML- \frac14 Mx\big)}{2\big(mL +ML -\frac12 Mx\big)^2}\, \,.\] Deduce that the acceleration of the particle after it is released is greater than \(g\).

No solution available for this problem.

Examiner's report
— 2012 STEP 3, Question 11
Mean: ~8.8 / 20 (inferred) ~15% attempted (inferred) Inferred ~8.8/20: 'slightly less success than Q10' (9.8) → 9.8 − 1.0 = 8.8. Inferred ~15% from 'slightly less popular than Q10' (17%) → 17 − 2 ≈ 15

This was slightly less popular than question 10, and slightly less success was achieved. Most candidates correctly evaluated the kinetic and potential energies of the particle, and the kinetic energy of the rope. However they had more difficulty finding the potential energy of the rope, and put themselves at an unnecessary disadvantage by not explaining their logic. There were different ways of splitting up the rope, which one they used they frequently failed to make clear, and likewise those calculating potential energy relative to a reference point failed to make the choice of that point clear. The second part of the question was done very well using the result given for the first part. The last part was fairly easy, but quite a few candidates did not justify the logic fully.

The number of candidates attempting more than six questions was, as last year, about 25%, though most of these extra attempts achieved little credit.

Source: Cambridge STEP 2012 Examiner's Report · 2012-full.pdf
Rating Information

Difficulty Rating: 1700.0

Difficulty Comparisons: 0

Banger Rating: 1500.0

Banger Comparisons: 0

Show LaTeX source
Problem source
One end of a thin heavy uniform inextensible perfectly flexible rope of length $2L$
and mass $2M$
is attached to a fixed point $P$. A particle of mass $m$ is 
attached to the other end. Initially, the particle is held at
$P$ and the rope hangs vertically in a loop below $P$. The particle is then released
so that it and a section of the rope (of decreasing length)
fall vertically as shown in the diagram. 
\begin{center}
\psset{xunit=1.0cm,yunit=0.9cm,algebraic=true,dimen=middle,dotstyle=o,dotsize=3pt 0,linewidth=0.3pt,arrowsize=3pt 2,arrowinset=0.25}
\begin{pspicture*}(0.13,-0.26)(3.26,5.51)
\psline(1,5)(3,5)
\psline[linewidth=0.1pt,linestyle=dashed,dash=2pt 2pt]{<->}(1.52,0)(1.52,5)
\psline[linewidth=0.1pt,linestyle=dashed,dash=2pt 2pt]{<->}(2.53,3.2)(2.53,5)
\psline(2.1,3.18)(2.06,0.25)
\psline(2,5)(2.02,0.26)
\psline(2.02,0.26)(2.03,0)
\psline(2.03,0)(2.06,0.25)
\rput[tl](1.94,5.45){$P$}
\rput[tl](2.6,4.25){$x$}
\rput[tl](0.2,2.85){$L+\frac{1}{2}x$}
\begin{scriptsize}
\psdots[dotsize=4pt 0,dotstyle=*](2.1,3.18)
\end{scriptsize}
\end{pspicture*}
\end{center}
You may assume that
each point on the moving section of the rope falls at the same speed as the 
particle. Given that energy is conserved, show 
that, when  the particle has fallen a distance $x$ (where $x<   2L$),
its speed $v$ is given by
\[
v^2 = \frac { 2g x \big( mL +ML - \frac14 Mx)}{mL +ML - \frac12 Mx}\,.
\]  
Hence show that the acceleration of the particle is 
\[
 g +
 \frac{ Mgx\big(mL+ML- \frac14 Mx\big)}{2\big(mL +ML -\frac12 Mx\big)^2}\,
\,.\] 
Deduce that the acceleration of the particle after it is 
released is greater than $g$.