2016 Paper 1 Q10
Year: 2016
Paper: 1
Question Number: 10
Course: UFM Mechanics
Section: Momentum and Collisions 1
Problem
- Show that \(e = \dfrac {\lambda-1}{3\lambda+1}\) and deduce that \(e < \frac 13\,\).
- Given also that \(C\) and \(D\) move towards each other with the same speed, find the value of \(\lambda\) and of \(e\).
Solution
- \begin{align*} \text{NEL}: && w_C &= e(v_B - v_C) \\ \text{COM}: && mv_B+ mv_C &= m w_C \\ \Rightarrow && w_C &= v_B + v_C\\ \Rightarrow && e(v_B - v_C) &= (v_B + v_C) \\ \Rightarrow && (1-e)v_B &= -(1+e)v_C \\ \Rightarrow && (1-e) \frac{\lambda(1+ e)}{1+\lambda} &= (1+e) \frac{1+e}{2} \\ \Rightarrow && 2\lambda - 2\lambda e &= 1+\lambda + e + \lambda e \\ \Rightarrow && (3\lambda +1)e &= \lambda - 1 \\ \Rightarrow && e &= \frac{\lambda -1}{3\lambda + 1} \\ &&&< \frac{\lambda - 1 + \frac{4}{3}}{3\lambda + 1} \\ &&& = \frac13 \end{align*}
- Since they move towards each other at the same speed \(w_C = - v_D\) \begin{align*} && w_C &= - v_D \\ \Rightarrow && v_B + v_C &= -(v_C+eu) \\ \Rightarrow && -eu &= v_B +2v_C \\ &&&= \frac{\lambda(1+ e)}{1+\lambda} u -(1+e)u \\ \Rightarrow && 1 &= \frac{\lambda(1+e)}{1+\lambda} \\ \Rightarrow && 1+\lambda &= \lambda \left ( 1 + \frac{\lambda -1}{3\lambda+1} \right) \\ &&&= \lambda \frac{4\lambda}{3\lambda +1} \\ \Rightarrow && 1+4\lambda + 3\lambda^2 &= 4\lambda^2 \\ \Rightarrow && 0 &= \lambda^2 - 4\lambda - 1 \\ \Rightarrow && \lambda &= \frac{4 \pm \sqrt{20}}{2} \\ &&&= 2\pm \sqrt{5} \\ \Rightarrow && \lambda &= 2 + \sqrt{5} \\ && e &= \frac{1+\sqrt{5}}{7+3\sqrt{5}} \\ &&&=\sqrt{5}-2 \end{align*}
Examiner's report
— 2016 STEP 1, Question 10This year, more than 2000 candidates signed up to sit this paper, though just under 2000 actually sat it. This figure is about the same as the entry figure for 2015, though the number of candidates opting to sit STEP I has risen significantly over recent years; for instance, it was around 1500 in 2013. There is no doubt that the purpose of the STEPs is to learn which students can genuinely use their mathematical knowledge, skills and techniques in an arena that demands of them a level of performance that exceeds anything they will have encountered within the standard A-level (or equivalent) assessments. The ability to work at an extended piece of mathematical work, often with the minimum of specific guidance, allied to the need for both determination and the ability to "make connections" at speed and under considerable time pressure, are characteristics that only follow from careful preparation, and there is a great benefit to be had from an early encounter with, and subsequent prolonged exposure to, these kinds of questions. It is not always easy to say what level of preparation has been undertaken by candidates, but the minimum expected requirement is the ability to undertake routine A-level-standard tasks and procedures with speed and accuracy. At the top end of the scale, almost 100 candidates produced a three-figure score to the paper, which is a phenomenal achievement; and around 250 others scored a mark of 70+, which is also exceptionally impressive. At the other end of the scale, over 400 candidates failed to reach a total of 40 marks out of the 120 available. For STEP I, the most approachable questions are always set as Qs.1 & 2 on the paper, with Q1 in particular intended to afford every candidate the opportunity to get something done successfully. So it is perfectly reasonable for a candidate, upon opening the paper, to make an immediate start at the first and/or second question(s) before looking around to decide which of the remaining 10 or 11 questions they feel they can tackle. It is very important that candidates spend a few minutes – possibly at the beginning – reading through the questions to decide which six they intend to work, since they will ultimately only be credited with their best six question marks. Many students spend time attempting seven, eight, or more questions and find themselves giving up too easily on a question the moment the going gets tough, and this is a great pity, since they are not allowing themselves thinking time, either on the paper as a whole or on individual questions. The other side to the notion of strategy is that most candidates clearly believe that they need to attempt (at least) six questions when, in fact, four questions (almost) completely done would guarantee a Grade 1 (Distinction), especially if their score on these first four questions were then to be supplemented by a couple of early attempts at the starting parts of a couple more questions (for the first five or six marks); attempts which need not take longer than, say, ten minutes of their time. It is thus perfectly reasonable to suggest to candidates, in their preparations, that they can spend more than 30 minutes on a question, but only IF they think they are going to finish it off satisfactorily, although it might be best if they were advised to spend absolutely no more than 40-45 minutes on any single question; if they haven't finished by then, it really is time to move on. Curve-sketching skills are usually a common weakness, but were only tested on this paper in Q3. The other common area of weakness – algebra – was tested relatively frequently, and proved to be as testing as usual. Calculus skills were generally "okay" although the integration of first-order differential equations by the separation of variables, as appearing repeatedly in Q4, was found challenging by many of the candidates who attempted this question. The most noticeable deficiency, however, was in the widespread inability to construct an argument, particularly in Qs. 5, 7 & 8. Vectors are often poorly handled, and this year proved no exception.
Rating Information
Difficulty Rating: 1484.0
Difficulty Comparisons: 1
Banger Rating: 1500.0
Banger Comparisons: 0
Show LaTeX source
Problem source
Four particles $A$, $B$, $C$ and $D$ are initially at rest on a smooth horizontal table. They lie equally spaced a small distance apart, in the order $ABCD$, in a straight line. Their masses are $\lambda m$, $m$, $m$ and $m$, respectively, where $\lambda>1$.
Particles $A$ and $D$ are simultaneously projected, both at speed $u$, so that they collide with $B$ and $C$ (respectively). In the following collision between $B$ and $C$, particle $B$ is brought to rest. The coefficient of restitution in each collision is $e$.
\begin{questionparts}
\item Show that $e = \dfrac {\lambda-1}{3\lambda+1}$ and deduce that $e < \frac 13\,$.
\item Given also that $C$ and $D$ move towards each other with the same speed, find the value of $\lambda$ and of $e$.
\end{questionparts}Solution source
\begin{center}
\begin{tikzpicture}[scale=0.7]
\def\h{2.75};
\def\arrowheight{1.5};
\def\arrowwidth{0.4};
\def\massa{$\lambda m$};
\def\massb{$m$};
\def\velocityua{$u$};
\def\velocityub{$0$};
\def\velocityva{$v_A$};
\def\velocityvb{$v_B$};
\draw (0,0) circle (1);
\draw (2.5,0) circle (1);
\draw (8,0) circle (1);
\draw (10.5,0) circle (1);
\node at (1.25, \h) {before collision};
\node at (9.25, \h) {after collision};
\draw[dashed] (5.25,-\arrowheight) -- (5.25, \arrowheight);
\node at (0,0) {\massa};
\draw[->, thick, red] (-\arrowwidth, \arrowheight) -- (\arrowwidth, \arrowheight);
\node[above, red] at (0, \arrowheight) {\velocityua};
\node at (2.5,0) {\massb};
\draw[->, thick, red] (2.5-\arrowwidth, \arrowheight) -- (2.5+\arrowwidth, \arrowheight); \node[above, red] at (2.5, \arrowheight) {\velocityub};
\node at (8,0) {\massa};
\draw[->, thick, red] (8-\arrowwidth, \arrowheight) -- (8+\arrowwidth, \arrowheight);
\node[red,above] at (8, \arrowheight) {\velocityva};
\node at (10.5,0 ) {\massb};
\draw[->, thick, red] (10.5-\arrowwidth, \arrowheight) -- (10.5+\arrowwidth, \arrowheight);
\node[red,above] at (10.5, \arrowheight) {\velocityvb};
\end{tikzpicture}
\end{center}
Collision between A & B. Since the speed of approach is $u$ and the coefficient of restitution is $e$ we must have $v_B = v_A + eu$.
\begin{align*}
\text{COM}: && \lambda m u &= \lambda m (v_B - eu) + m v_B \\
\Rightarrow && v_B(\lambda + 1) &=\lambda (1+ e) u \\
\Rightarrow && v_B &= \frac{\lambda(1+ e)}{1+\lambda} u
\end{align*}
\begin{center}
\begin{tikzpicture}[scale=0.7]
\def\h{2.75};
\def\arrowheight{1.5};
\def\arrowwidth{0.4};
\def\massa{$m$};
\def\massb{$m$};
\def\velocityua{$0$};
\def\velocityub{$u$};
\def\velocityva{$v_C$};
\def\velocityvb{$v_D$};
\draw (0,0) circle (1);
\draw (2.5,0) circle (1);
\draw (8,0) circle (1);
\draw (10.5,0) circle (1);
\node at (1.25, \h) {before collision};
\node at (9.25, \h) {after collision};
\draw[dashed] (5.25,-\arrowheight) -- (5.25, \arrowheight);
\node at (0,0) {\massa};
\draw[<-, thick, red] (-\arrowwidth, \arrowheight) -- (\arrowwidth, \arrowheight);
\node[above, red] at (0, \arrowheight) {\velocityua};
\node at (2.5,0) {\massb};
\draw[<-, thick, red] (2.5-\arrowwidth, \arrowheight) -- (2.5+\arrowwidth, \arrowheight); \node[above, red] at (2.5, \arrowheight) {\velocityub};
\node at (8,0) {\massa};
\draw[<-, thick, red] (8-\arrowwidth, \arrowheight) -- (8+\arrowwidth, \arrowheight);
\node[red,above] at (8, \arrowheight) {\velocityva};
\node at (10.5,0 ) {\massb};
\draw[<-, thick, red] (10.5-\arrowwidth, \arrowheight) -- (10.5+\arrowwidth, \arrowheight);
\node[red,above] at (10.5, \arrowheight) {\velocityvb};
\end{tikzpicture}
\end{center}
Collision between A & B. Since the speed of approach is $u$ and the coefficient of restitution is $e$ we must have $v_D = v_C + eu$.
\begin{align*}
\text{COM}: && m(-u) &= mv_C + m(v_C + eu) \\
\Rightarrow && 2v_C &= -(1+e)u \\
\Rightarrow && v_C &= -\frac{1+e}{2} u
\end{align*}
\begin{questionparts}
\item
\begin{center}
\begin{tikzpicture}[scale=0.7]
\def\h{2.75};
\def\arrowheight{1.5};
\def\arrowwidth{0.4};
\def\massa{$m$};
\def\massb{$m$};
\def\velocityua{$v_B$};
\def\velocityub{$v_C$};
\def\velocityva{$0$};
\def\velocityvb{$w_C$};
\draw (0,0) circle (1);
\draw (2.5,0) circle (1);
\draw (8,0) circle (1);
\draw (10.5,0) circle (1);
\node at (1.25, \h) {before collision};
\node at (9.25, \h) {after collision};
\draw[dashed] (5.25,-\arrowheight) -- (5.25, \arrowheight);
\node at (0,0) {\massa};
\draw[->, thick, red] (-\arrowwidth, \arrowheight) -- (\arrowwidth, \arrowheight);
\node[above, red] at (0, \arrowheight) {\velocityua};
\node at (2.5,0) {\massb};
\draw[->, thick, red] (2.5-\arrowwidth, \arrowheight) -- (2.5+\arrowwidth, \arrowheight); \node[above, red] at (2.5, \arrowheight) {\velocityub};
\node at (8,0) {\massa};
\draw[->, thick, red] (8-\arrowwidth, \arrowheight) -- (8+\arrowwidth, \arrowheight);
\node[red,above] at (8, \arrowheight) {\velocityva};
\node at (10.5,0 ) {\massb};
\draw[->, thick, red] (10.5-\arrowwidth, \arrowheight) -- (10.5+\arrowwidth, \arrowheight);
\node[red,above] at (10.5, \arrowheight) {\velocityvb};
\end{tikzpicture}
\end{center}
\begin{align*}
\text{NEL}: && w_C &= e(v_B - v_C)
\\
\text{COM}: && mv_B+ mv_C &= m w_C \\
\Rightarrow && w_C &= v_B + v_C\\
\Rightarrow && e(v_B - v_C) &= (v_B + v_C) \\
\Rightarrow && (1-e)v_B &= -(1+e)v_C \\
\Rightarrow && (1-e) \frac{\lambda(1+ e)}{1+\lambda} &= (1+e) \frac{1+e}{2} \\
\Rightarrow && 2\lambda - 2\lambda e &= 1+\lambda + e + \lambda e \\
\Rightarrow && (3\lambda +1)e &= \lambda - 1 \\
\Rightarrow && e &= \frac{\lambda -1}{3\lambda + 1} \\
&&&< \frac{\lambda - 1 + \frac{4}{3}}{3\lambda + 1} \\
&&& = \frac13
\end{align*}
\item Since they move towards each other at the same speed $w_C = - v_D$
\begin{align*}
&& w_C &= - v_D \\
\Rightarrow && v_B + v_C &= -(v_C+eu) \\
\Rightarrow && -eu &= v_B +2v_C \\
&&&= \frac{\lambda(1+ e)}{1+\lambda} u -(1+e)u \\
\Rightarrow && 1 &= \frac{\lambda(1+e)}{1+\lambda} \\
\Rightarrow && 1+\lambda &= \lambda \left ( 1 + \frac{\lambda -1}{3\lambda+1} \right) \\
&&&= \lambda \frac{4\lambda}{3\lambda +1} \\
\Rightarrow && 1+4\lambda + 3\lambda^2 &= 4\lambda^2 \\
\Rightarrow && 0 &= \lambda^2 - 4\lambda - 1 \\
\Rightarrow && \lambda &= \frac{4 \pm \sqrt{20}}{2} \\
&&&= 2\pm \sqrt{5} \\
\Rightarrow && \lambda &= 2 + \sqrt{5} \\
&& e &= \frac{1+\sqrt{5}}{7+3\sqrt{5}} \\
&&&=\sqrt{5}-2
\end{align*}
\end{questionparts}
This question was a relatively undemanding collisions question, a fact that was apparent to a lot of the candidates, almost half of whom made an attempt at the question. It turned out to be only one of three questions on the paper for which the mean achieved mark was over 10 out of 20 (after Qs. 1 & 3), and this is no doubt partly due to the fact that there are generally only three physical "laws" or "principles" to be applied on this type of mechanics problem. On this occasion, no energy considerations were involved, which meant that careful application of Conservation of Linear Momentum and Newton's (Experimental) Law of Restitution pretty much saw the solver through the whole thing. The only minor hurdle is that candidates often get confused about which directions are being taken as positive, and this is usually down to the lack of a suitably-marked diagram. Other than that, it was only a few algebraic slips here and there that prevented full and successful attempts from being made.