Year: 2009
Paper: 1
Question Number: 11
Course: UFM Mechanics
Section: Momentum and Collisions 1
There were significantly more candidates attempting this paper again this year (over 900 in total), and the scores were pleasing: fewer than 5% of candidates failed to get at least 20 marks, and the median mark was 48. The pure questions were the most popular as usual; about two-thirds of candidates attempted each of the pure questions, with the exceptions of question 2 (attempted by about 90%) and question 5 (attempted by about one third). The mechanics questions were only marginally more popular than the probability and statistics questions this year; about one quarter of the candidates attempted each of the mechanics questions, while the statistics questions were attempted by about one fifth of the candidates. A significant number of candidates ignored the advice on the front cover and attempted more than six questions. In general, those candidates who submitted answers to eight or more questions did fairly poorly; very few people who tackled nine or more questions gained more than 60 marks overall (as only the best six questions are taken for the final mark). This suggests that a skill lacking in many students attempting STEP is the ability to pick questions effectively. This is not required for A-levels, so must become an important part of STEP preparation. Another "rubric"-type error was failing to follow the instructions in the question. In particular, when a question says "Hence", the candidate must make (significant) use of the preceding result(s) in their answer if they wish to gain any credit. In some questions (such as question 2), many candidates gained no marks for the final part (which was worth 10 marks) as they simply quoted an answer without using any of their earlier work. There were a number of common errors which appeared across the whole paper. These included a noticeable weakness in algebraic manipulations, sometimes indicating a serious lack of understanding of the mathematics involved. As examples, one candidate tried to use the misremembered identity cos β = sin √(1 − β²), while numerous candidates made deductions of the form "if a² + b² = c², then a + b = c" at some point in their work. Fraction manipulations are also notorious in the school classroom; the effects of this weakness were felt here, too. Another common problem was a lack of direction; writing a whole page of algebraic manipulations with no sense of purpose was unlikely to either reach the requested answer or gain the candidate any marks. It is a good idea when faced with a STEP question to ask oneself, "What is the point of this (part of the) question?" or "Why has this (part of the) question been asked?" Thinking about this can be a helpful guide. One aspect of this is evidenced by pages of formulæ and equations with no explanation. It is very good practice to explain why you are doing the calculation you are, and to write sentences in English to achieve this. It also forces one to focus on the purpose of the calculations, and may help avoid some dead ends. Finally, there is a tendency among some candidates when short of time to write what they would do at this point, rather than using the limited time to actually try doing it. Such comments gain no credit; marks are only awarded for making progress in a question. STEP questions do require a greater facility with mathematics and algebraic manipulation than the A-level examinations, as well as a depth of understanding which goes beyond that expected in a typical sixth-form classroom.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Two particles move on a smooth horizontal table and collide. The masses of the particles are $m$ and $M$. Their velocities before the collision are $u{\bf i}$ and $v{\bf i}\,$, respectively, where $\bf i$ is a unit vector and $u>v$.
Their velocities after the collision are $p{\bf i}$ and $q{\bf i}\,$, respectively.
The coefficient of restitution between the two particles is $e$, where $e<1$.
\begin{questionparts}
\item Show that the loss of kinetic energy due to the collision is
\[
\tfrac12 m (u-p)(u-v)(1-e)\,,
\]
and deduce that $u\ge p$.
\item
Given that each particle loses the same (non-zero) amount of kinetic energy in the collision, show that
\[
u+v+p+q=0\,,
\]
and that, if $m\ne M$,
\[
e= \frac{(M+3m)u + (3M+m)v}{(M-m)(u-v)}\,.
\]
\end{questionparts}\begin{questionparts}
\item \begin{align*}
\text{COM}: && mu + Mv &= mp + Mq \\
\Rightarrow && m(u-p) &= M(q-v) \\
\text{NEL}: && q-p &= e(u-v) \\
&& q +ev &= p+eu
\\
&& \Delta \text{ k.e.} &= \frac12 m u^2 + \frac12 M v^2 -\frac12 m p^2 - \frac12 M q^2 \\
&&&= \frac12m (u^2 - p^2)+\frac12M(v^2-q^2) \\
&&&= \frac12m (u^2 - p^2)+\frac12M(v-q)(v+q) \\
&&&= \frac12m(u^2-p^2) - \frac12 m(u-p)(v+q) \\
&&&= \frac12 m(u-p) \left ( u+p-v-q\right) \\
&&&= \frac12 m(u-p) \left (u-v+(p-q)\right) \\
&&&= \frac12 m(u-p) \left (u-v-e(u-v)\right) \\
&&&= \frac12m(u-p)(u-v)(1-e)
\end{align*}
Since the loss in energy is positive, and $m$, $u-v$ and $1-e$ are all positive, so is $u-p$, ie $u \geq p$
\item \begin{align*}
&& \frac12 m u^2 - \frac12mp^2 &= \frac12Mv^2 - \frac12Mq^2 \\
&& \frac12 m(u-p)(u+p) &= \frac12 M (v-q)(v+q) \\
&& \frac12 m (u-p)(u+p) &= -\frac12 m(u-p)(v+q) \\
\Rightarrow && u+p+v+q &= 0
\end{align*}
\begin{align*}
&& p+q &= -(u+v)\\
&&mp+Mq &= mu+Mv \\
\Rightarrow && (M-m)q &= mu+Mv+mu+mv\\
\Rightarrow && q &= \frac{(M+m)v+2mu}{M-m} \\
\Rightarrow && (m-M)p &= mu+Mv+Mu+Mv \\
\Rightarrow && p &= -\frac{(M+m)u+2Mv}{M-m} \\
\\
&& e &= \frac{q-p}{u-v} \\
&&&= \frac{(M+m)v+2mu+(M+m)u+2Mv}{(u-v)(M-m)} \\
&&&= \frac{(3M+m)v+(3m+M)u}{(u-v)(M-m)}
\end{align*}
\end{questionparts}
This was found to be another very hard question, with most attempts being fragmentary; almost half of students gained 2 or less, though the handful who made it beyond half way achieved either 19 or 20 marks. The first part of the question required correctly stating the laws of conservation of momentum and restitution, together with the formula for kinetic energy. Almost all candidates realised this and attempted to write down the appropriate formulæ. It was particularly disappointing, though, that so many candidates struggled to do even this correctly. Most were comfortable with the conservation of momentum, but having learnt the law of restitution in the form: e = speed of separation / speed of approach, they were at a loss when all of the velocities were in the same direction, and consequently introduced various sign errors. A number of candidates remembered the formula upside-down (as speed of approach over speed of separation), and this was invariably fatal. The next stumbling block was the formula for loss of kinetic energy. A sizeable minority of students wrote that the loss of KE in the first particle is ½m(u−p)² instead of ½mu² − ½mp² = ½m(u² − p²) and similarly for the second particle; this is a disturbing error, although it is not clear whether it arose because of a belief that a² − b² = (a − b)² or a lack of thought about what loss of KE means. A number of students compounded their problems by confusing M and m, often due to careless handwriting. It is a cardinal principle of written mathematics that distinct symbols should look obviously distinct! Those students who made one of these errors generally failed to make any significant further progress with the question. The key step in part (i) was to manipulate the formula for loss of KE to eliminate M and q, making use of the other two equations to achieve this. Most candidates who had reached this point succeeded in doing this and reaching the required formula. Some made their lives more difficult by expanding every bracket fully; it is a valuable skill to recognise when this would be helpful and when not. For the final deduction in part (i), the instruction to "deduce" the result u ⩾ p meant that no credit was given to those students who simply stated "it's obvious"; some justification on the basis of the formula for loss of KE was required. The majority of candidates did not even attempt part (ii), which is a shame, as the ideas were essentially the same as in part (i), albeit with a little more sophistication: the task was to eliminate some of the variables. The first half required writing down the word statement in symbols and eliminating either M or M, while the second half required elimination of p and q followed by some rearrangement. Any attempt to perform these eliminations would have led to the correct results.