2018 Paper 1 Q10
Year: 2018
Paper: 1
Question Number: 10
Course: LFM Pure and Mechanics
Section: Newton's laws and connected particles
Problem
- Show that \[ T = \frac{n(mD+MR)}{nm+2M}\,. \]
- Show that \(T\) is greater than the tension in any other coupling provided that \(k> \frac12n\,\).
- Show also that, if \(k> \frac12n\,\), then at least one of the couplings is in compression (that is, there is a negative tension in the coupling).
Solution
- \(\,\)
\begin{align*} \text{N2}(\leftarrow, \text{first engine}): && D-T &= Ma \\ \text{N2}(\leftarrow, \text{rest of train}): && T-nR+D &= (M+nm)a \\ \Rightarrow && \frac{D-T}{M} &= \frac{T+D-nR}{M+nm} \\ \Rightarrow && T \left ( \frac{1}{M+nm}+\frac{1}{M} \right) &= \frac{D}{M} + \frac{nR-D}{M+nm} \\ \Rightarrow && T \left ( 2M+nm\right) &= DM +Dnm + nRM - DM \\ &&&= n(mD+MR) \\ \Rightarrow && T &= \frac{n(mD+MR)}{2M+nm} \end{align*}
- The greatest coupling must occur behind an engine, because each carriage behind an engine acts as a drag.
Therefore we need only consider the couple between the second engine and the rest of the carriages:
\begin{align*} \text{N2}(\leftarrow, \text{up to second engine}): && 2D - T_2 - kR &= (2M+km)a \\ \text{N2}(\leftarrow, \text{everything else}): && T_2 - (n-k)R &= (n-k)ma \\ \Rightarrow && \frac{2D-T_2-kR}{2M+km} &= \frac{T_2-(n-k)R}{(n-k)m} \\ \Rightarrow && T_2 \left (\frac{1}{(n-k)m} + \frac{1}{2M+km} \right) &= \frac{2D-kR}{2M+km} + \frac{R}{m} \\ \Rightarrow && T_2 \left (2M+ nm \right) &= (2D-kR)m(n-k) + R(2M+km)(n-k) \\ \Rightarrow && T_2 &= \frac{(n-k)\left (2Dm+2RM \right)}{2M+nm} \\ &&&= \frac{2(n-k)(mD + MR)}{2M+nm} \end{align*} Therefore \(T > T_2\) provided \(2(n-k) < n \Leftrightarrow k > \frac12n\)
- If there is a coupling which is in negative tension, it must be between the two engines. In particular, if there is one, there must be one directly in front of the first engine.
\begin{align*} \text{N2}(\leftarrow, \text{before second engine}): && D - T_3 - kR &= (M+km)a \\ \text{N2}(\leftarrow, \text{everything else}): && T_3 +D- (n-k)R &= (M+(n-k)m)a \\ \Rightarrow && \frac{D-T_3-kR}{M+km} &= \frac{T_3+D-(n-k)R}{M+(n-k)m} \\ \Rightarrow && T_3 \left ( \frac{1}{M+(n-k)m} + \frac{1}{M+km} \right) &= \frac{D-(n-k)R}{M+(n-k)m}+\frac{kR-D}{M+km} \\ \Rightarrow && T_3 (2M+nm) &= (D-(n-k)R)(M+km)+(kR-D)(M+(n-k)m) \\ &&&= D(M+km-M-(n-k)m) + R(kM+k(n-k)m-(n-k)M-k(n-k)m) \\ &&&= D(n-2k)m+RM(2k-n)m \\ &&&= (n-2k)m(D-RM) \end{align*} Therefore \(T_3\) is negative if \(k > \frac12n\) so there are some connections in compression.
Examiner's report
— 2018 STEP 1, Question 10In order to get the fullest picture, this document should be read in conjunction with the question paper, the marking scheme and (for comments on the underlying purpose and motivation for finding the right solution-approaches to questions) the Hints and Solutions document. The purpose of the STEPs is to learn what students are able to achieve mathematically when applying the knowledge, skills and techniques that they have learned within their standard A-level (or equivalent) courses … but seldom within the usual range of familiar settings. STEP questions require candidates to work at an extended piece of mathematics, often with the minimum of specific guidance, and to make the necessary connections. This requires a very different mind-set to that which is sufficient for success at A-level, and the requisite skills tend only to develop with prolonged and determined practice at such longer questions. One of the most crucial features of the STEPs is that the routine technical and manipulative skills are almost taken for granted; it is necessary for candidates to produce them with both speed and accuracy so that the maximum amount of time can be spent in thinking their way through the problem and the various hurdles and obstacles that have been set before them. Most STEP questions begin by asking the solver to do something relatively routine or familiar before letting them loose on the real problem. Almost always, such an opening has not been put there to allow one to pick up a few easy marks, but rather to point the solver in the right direction for what follows. Very often, the opening result or technique will need to be used, adapted or extended in the later parts of the question, with the demands increasing the further on that one goes. So a candidate should never think that they are simply required to 'go through the motions'; rather they will, sooner or later, be required to show either genuine skill or real insight in order to make a reasonably complete effort. The more successful candidates are the ones who manage to figure out how to move on from the given starting-point. Finally, reading through a finished solution is often misleading – even unhelpful – unless you have attempted the problem for yourself. This is because the thinking has been done for you. So, when you read through the report and look at the solutions (either in the mark scheme or the Hints and Solutions booklet), try to figure out how you could have arrived at the solution, learn from your mistakes and pick up as many tips as you can whilst working through past paper questions. This year far too many candidates wasted time by attempting more than six questions, with many of these candidates picking up 0-4 marks on several 'false starts' which petered out the moment some understanding was required. There were almost 2000 candidates for this SI paper. Almost one-sixth of candidates failed to reach a total of 30 and around two-thirds fell below half-marks overall. This highlights the fact that many candidates don't find this test an easy one. At the other end of the spectrum, almost one-in-ten managed a total of 84 out of 120 – these candidates usually marked out by their ability to complete whole questions – with almost 4% of the entry achieving the highly praiseworthy feat of getting into three-figures with their overall score. The paper is constructed so that question 1 is very approachable indeed, the intention being to get everyone started with some measure of success; unsurprisingly, Q1 was the most popular question of all, with almost all candidates attempting it, and it also turned out to be the most successful question on the paper with a mean score of more than 15 out of 20. Around 7% of candidates didn't make any kind of attempt at it at all. In order of popularity, Q1 was followed by Qs. 2, 7, 4 and 3. Indeed, it was the pure maths questions in Section A that attracted the majority of attention from candidates, with the most popular applied question (Q9, mechanics) still getting fewer 'hits' than the least popular pure question (Q5). Questions 10, 11 and 13 proved to attract very little attention from candidates and many of the attempts were minimal.
Rating Information
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1471.6
Banger Comparisons: 2
Show LaTeX source
Problem source
A train is made up of two engines, each of mass $M$, and $n$ carriages, each of mass $m$. One of the engines is at
the front of the train, and the other is coupled between the $k$th and $(k+1)$th carriages. When the train is accelerating along a straight, horizontal track, the resistance to the motion of each carriage is $R$ and the driving force on each engine is $D$, where $2D >nR\,$.
The tension in the coupling between the engine at the front
and the first carriage is $T$.
\begin{questionparts}
\item Show that
\[
T = \frac{n(mD+MR)}{nm+2M}\,.
\]
\item Show that $T$ is greater than the tension in any other
coupling provided that $k> \frac12n\,$.
\item Show also that, if $k> \frac12n\,$, then at least one of the couplings is in compression (that is, there is a negative tension in the coupling).
\end{questionparts}Solution source
\begin{questionparts}
\item $\,$
\begin{center}
\begin{tikzpicture}
\draw (-4,0) -- (4,0);
\draw (-2,0) rectangle (-1, 1);
\draw (1,0) rectangle (3, 1);
\draw[dashed] (-1, 0.5) -- (1, 0.5);
\draw[-latex, ultra thick, blue] (-2, 0.5) -- ++(-1, 0) node[pos=0.5, above] {$D$};
% \draw[-latex, ultra thick, blue] (-1.5, 0) -- ++(0.5, 0) node[pos=0.5, above] {$D$};
\node at (-1.5, 0.5) {$M$};
\node at (2, 0.5) {$M+nm$};
\draw[-latex, ultra thick, blue] (-1, 0.5) -- ++(0.75, 0) node[pos=0.5, above] {$T$};
\draw[-latex, ultra thick, blue] (1, 0.5) -- ++(-0.75, 0) node[pos=0.5, above] {$T$};
\draw[-latex, ultra thick, blue] (2, 0) -- ++(+0.75, 0) node[pos=0.5, below] {$nR - D$};
\end{tikzpicture}
\end{center}
\begin{align*}
\text{N2}(\leftarrow, \text{first engine}): && D-T &= Ma \\
\text{N2}(\leftarrow, \text{rest of train}): && T-nR+D &= (M+nm)a \\
\Rightarrow && \frac{D-T}{M} &= \frac{T+D-nR}{M+nm} \\
\Rightarrow && T \left ( \frac{1}{M+nm}+\frac{1}{M} \right) &= \frac{D}{M} + \frac{nR-D}{M+nm} \\
\Rightarrow && T \left ( 2M+nm\right) &= DM +Dnm + nRM - DM \\
&&&= n(mD+MR) \\
\Rightarrow && T &= \frac{n(mD+MR)}{2M+nm}
\end{align*}
\item
The greatest coupling must occur behind an engine, because each carriage behind an engine acts as a drag.
Therefore we need only consider the couple between the second engine and the rest of the carriages:
\begin{center}
\begin{tikzpicture}
\draw (-4,0) -- (4,0);
\draw (-3,0) rectangle (-1, 1);
\draw (1,0) rectangle (3, 1);
\draw[dashed] (-1, 0.5) -- (1, 0.5);
\draw[-latex, ultra thick, blue] (-3, 0.5) -- ++(-1, 0) node[pos=0.5, above] {$2D$};
% \draw[-latex, ultra thick, blue] (-1.5, 0) -- ++(0.5, 0) node[pos=0.5, above] {$D$};
\node at (-2, 0.5) {$2M+km$};
\node at (2, 0.5) {$(n-k)m$};
\draw[-latex, ultra thick, blue] (-1, 0.5) -- ++(0.75, 0) node[pos=0.5, above] {$T_2$};
\draw[-latex, ultra thick, blue] (1, 0.5) -- ++(-0.75, 0) node[pos=0.5, above] {$T_2$};
\draw[-latex, ultra thick, blue] (2, 0) -- ++(+0.75, 0) node[pos=0.5, below] {$(n-k)R$};
\draw[-latex, ultra thick, blue] (-2, 0) -- ++(+0.75, 0) node[pos=0.5, below] {$kR$};
\end{tikzpicture}
\end{center}
\begin{align*}
\text{N2}(\leftarrow, \text{up to second engine}): && 2D - T_2 - kR &= (2M+km)a \\
\text{N2}(\leftarrow, \text{everything else}): && T_2 - (n-k)R &= (n-k)ma \\
\Rightarrow && \frac{2D-T_2-kR}{2M+km} &= \frac{T_2-(n-k)R}{(n-k)m} \\
\Rightarrow && T_2 \left (\frac{1}{(n-k)m} + \frac{1}{2M+km} \right) &= \frac{2D-kR}{2M+km} + \frac{R}{m} \\
\Rightarrow && T_2 \left (2M+ nm \right) &= (2D-kR)m(n-k) + R(2M+km)(n-k) \\
\Rightarrow && T_2 &= \frac{(n-k)\left (2Dm+2RM \right)}{2M+nm} \\
&&&= \frac{2(n-k)(mD + MR)}{2M+nm}
\end{align*}
Therefore $T > T_2$ provided $2(n-k) < n \Leftrightarrow k > \frac12n$
\item
If there is a coupling which is in negative tension, it must be between the two engines. In particular, if there is one, there must be one directly in front of the first engine.
\begin{center}
\begin{tikzpicture}
\draw (-4,0) -- (4,0);
\draw (-3,0) rectangle (-1, 1);
\draw (1,0) rectangle (4, 1);
\draw[dashed] (-1, 0.5) -- (1, 0.5);
\draw[-latex, ultra thick, blue] (-3, 0.5) -- ++(-1, 0) node[pos=0.5, above] {$D$};
% \draw[-latex, ultra thick, blue] (-1.5, 0) -- ++(0.5, 0) node[pos=0.5, above] {$D$};
\node at (-2, 0.5) {$M+km$};
\node at (2.5, 0.5) {$M+(n-k)m$};
\draw[-latex, ultra thick, blue] (-1, 0.5) -- ++(0.75, 0) node[pos=0.5, above] {$T_3$};
\draw[-latex, ultra thick, blue] (1, 0.5) -- ++(-0.75, 0) node[pos=0.5, above] {$T_3$};
\draw[-latex, ultra thick, blue] (2.5, 0) -- ++(+0.75, 0) node[pos=0.5, below] {$(n-k)R-D$};
\draw[-latex, ultra thick, blue] (-2, 0) -- ++(+0.75, 0) node[pos=0.5, below] {$kR$};
\end{tikzpicture}
\end{center}
\begin{align*}
\text{N2}(\leftarrow, \text{before second engine}): && D - T_3 - kR &= (M+km)a \\
\text{N2}(\leftarrow, \text{everything else}): && T_3 +D- (n-k)R &= (M+(n-k)m)a \\
\Rightarrow && \frac{D-T_3-kR}{M+km} &= \frac{T_3+D-(n-k)R}{M+(n-k)m} \\
\Rightarrow && T_3 \left ( \frac{1}{M+(n-k)m} + \frac{1}{M+km} \right) &= \frac{D-(n-k)R}{M+(n-k)m}+\frac{kR-D}{M+km} \\
\Rightarrow && T_3 (2M+nm) &= (D-(n-k)R)(M+km)+(kR-D)(M+(n-k)m) \\
&&&= D(M+km-M-(n-k)m) + R(kM+k(n-k)m-(n-k)M-k(n-k)m) \\
&&&= D(n-2k)m+RM(2k-n)m \\
&&&= (n-2k)m(D-RM)
\end{align*}
Therefore $T_3$ is negative if $k > \frac12n$ so there are some connections in compression.
\end{questionparts}
This question was neither popular nor well-handled by those starting it; part of the problem was that there appears to be too many things to consider … two engines and an unspecified number of carriages. A mean score of only just over 4 out of 20 is actually rather misleading here, since most of those attempts consisted of little more than an unused (and usually incomplete) diagram. Of those who made a serious attempt, most started this question well, considering Newton's 2nd Law for the front engine and the whole train. However, in later parts very few candidates considered a general carriage between the two engines which made any argument quite difficult. Candidates generally had difficulty in linking the condition that k > 12 n in any meaningful way because they were not sufficiently systematic in their consideration of carriages between the two engines and carriages after the two engines. Some candidates effectively did part (iii) before part (ii) and this was perfectly acceptable.