Year: 2012
Paper: 2
Question Number: 11
Course: UFM Mechanics
Section: Momentum and Collisions 1
There were just over 1000 entries for paper II this year, almost exactly the same number as last year. Overall, the paper was found marginally easier than its predecessor, which means that it was pitched at exactly the level intended and produced the hoped-for outcomes. Almost 50 candidates scored 100 marks or more, with more than 400 gaining at least half marks on the paper. At the lower end of the scale, around a quarter of the entry failed to score more than 40 marks. It was pleasing to note that the advice of recent years, encouraging students not to make attempts at lots of early parts to questions but rather to spend their time getting to grips with the six that can count towards their paper total, was more obviously being heeded in 2012 than I can recall being the case previously. As in previous years, the pure maths questions provided the bulk of candidates' work, with relatively few efforts to be found at the applied ones. Questions 1 and 2 were the most popular questions, although each drew only around 800 "hits" – fewer than usual. Questions 3 – 5 & 8 were almost as popular (around 700), with Q6 attracting the interest of under 450 candidates and Q7 under 200. Q9 was the most popular applied question – and, as it turned out, the most successfully attempted question on the paper – with very little interest shown in the rest of Sections B or C.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1484.9
Banger Comparisons: 1
A small block of mass $km$ is initially at rest on a smooth horizontal surface.
Particles $P_1$, $P_2$, $P_3$, $\ldots$ are fired, in order, along the surface from a fixed point towards the block.
The mass of the $i$th particle is $im$ ($i = 1, 2, \ldots$)and the speed at which it is fired is $u/i\,$. Each particle that collides with the block is embedded in it. Show that, if the $n$th particle collides with the block, the speed of the block after the collision is
\[
\frac{2nu}{2k +n(n+1)}\,.
\]
In the case $2k = N(N+1)$, where $N$ is a positive integer,
determine the number of collisions that occur. Show that the total kinetic energy lost in all the collisions
is
\[
\tfrac12 mu^2\bigg( \sum_{n=2}^{N+1} \frac 1 n \bigg)\,.
\]\begin{align*}
\text{COM}: && \sum_{i=1}^n im \cdot \frac{u}{i} &= \left ( km + \sum_{i=1}^n im \right) v \\
\Rightarrow && nu &= \left ( k + \frac{n(n+1)}{2} \right) v \\
\Rightarrow && v &= \frac{2nu}{2k + n(n+1)}
\end{align*}
If $2k = N(N+1)$, there will be no more collisions when $v_n > \frac{u}{n+1}$, ie
\begin{align*}
&& \frac{u}{n+1} &<\frac{2nu}{2k + n(n+1)} \\
\Leftrightarrow && N(N+1) + n(n+1) &< 2n(n+1) \\
\Leftrightarrow && N(N+1) &< n(n+1) \\
\end{align*}
Therefore $n = N+1$ and there will be $N+1$ collisions.
The loss of kinetic energy is:
\begin{align*}
&& \text{initial k.e.} &= \sum_{k=1}^{N+1} \frac12 im \cdot \frac{u^2}{i^2} \\
&&&= \frac12 m u^2 \left ( \sum_{k=1}^{N+1} \frac{1}{i}\right) \\
&& \text{final k.e.} &= \frac12 \left ( k + \frac{(N+1)(N+2)}{2}\right)m \left ( \frac{2(N+1)u}{N(N+1)+(N+1)(N+2)} \right)^2 \\
&&&= \frac12 m u^2 \frac{2(N+1)^2}{(N+1)(2N+2)} \\
&&&= \frac12 mu^2 \\
\Rightarrow && \Delta \text{ k.e.} &= \frac12 m u^2 \left ( \sum_{k=2}^{N+1} \frac{1}{i}\right)
\end{align*}
There were many very capable attempts at this question, taken by around a quarter of all candidates. At some stage, a general approach was required to the use of the principle of conservation of linear momentum; some resorted to "pattern-spotting" (which lost them a couple of marks) and others to an inductive approach, collision by collision, which worked well though was generally a lengthier bit of work. Some candidates mixed up n with N in the following part, while others incorrectly considered > rather than ≥ and ended up missing the answer by 1. A little bit of care was needed with the summation in the final part, and there was a bit of fiddling going on in order to get the given answer. Nevertheless, it was pleasing to see the principles understood well, even if the details were less carefully attended to.