Year: 2012
Paper: 2
Question Number: 9
Course: LFM Pure and Mechanics
Section: Projectiles
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: 1500.0
Banger Comparisons: 0
A tennis ball is projected from a height of $2h$ above horizontal ground with speed $u$ and at an angle of $\alpha$ below the horizontal. It travels in a plane perpendicular to a vertical net of height $h$ which is a horizontal distance of $a$ from the point of projection. Given that the ball passes over the net, show that
\[
\frac 1{u^2}< \frac {2(h-a\tan\alpha)}{ga^2\sec^2\alpha}\,.
\]
The ball lands before it has travelled a horizontal distance
of $b$ from the point of projection. Show that
\[
\sqrt{u^2\sin^2\alpha +4gh \ } < \frac{bg}{u\cos\alpha} + u \sin\alpha\,.
\]
Hence show that
\[
\tan\alpha < \frac{h(b^2-2a^2)}{ab(b-a)}\,.
\]\begin{align*}
&& s &= ut \\
\Rightarrow && a &= u \cos \alpha t\\
\Rightarrow && t &= \frac{a}{u \cos \alpha}\\
&& s &= ut+ \frac12at^2 \\
\Rightarrow && -h &< -u\sin \alpha \frac{a}{u \cos \alpha}-\frac12 g \left (\frac{a}{u \cos \alpha} \right)^2 \\
&&&= -a \tan \alpha-\frac12 g a^2 \frac{1}{u^2} \sec^2 \alpha \\
\Rightarrow && \frac12 g a^2 \frac{1}{u^2} \sec^2 \alpha &< h -a\tan \alpha \\
\Rightarrow &&\frac{1}{u^2} &< \frac{2(h-a\tan \alpha)}{ga^2 \sec^2 \alpha}
\end{align*}
\begin{align*}
&& s &= ut + \frac12a t^2 \\
\Rightarrow && 2h &= u\sin \alpha t + \frac12 gt^2 \\
\Rightarrow && t &= \frac{-u\sin \alpha \pm \sqrt{u^2 \sin^2 \alpha+4hg}}{g}\\
&& t &= \frac{-u\sin \alpha +\sqrt{u^2 \sin^2 \alpha+4hg}}{g}\\
&& s &= ut \\
\Rightarrow && b &> u \cos \alpha t \\
\Rightarrow && \frac{b}{u \cos \alpha} &> \frac{-u\sin \alpha +\sqrt{u^2 \sin^2 \alpha+4hg}}{g} \\
\Rightarrow && \sqrt{u^2 \sin^2 \alpha+4hg} &< \frac{bg}{u \cos \alpha} + u \sin \alpha \\
\end{align*}
\begin{align*}
\Rightarrow && u^2 \sin^2 \alpha+4hg &< \frac{b^2g^2}{u^2 \cos^2 \alpha} +u^2 \sin^2 \alpha + 2bg \tan \alpha \\
\Rightarrow && 4hg - 2bg \tan \alpha &< \frac{b^2g^2}{u^2 \cos^2 \alpha} \\
&&&< \frac{b^2g^2}{\cos^2 \alpha} \frac{2(h-a\tan \alpha)}{ga^2 \sec^2 \alpha} \\
&&&= \frac{2b^2g(h-a\tan \alpha)}{a^2} \\
\Rightarrow && \tan \alpha \left (\frac{2b^2g}{a} - 2bg \right) &< \frac{2b^2gh}{a^2} - 4hg \\
\Leftrightarrow && \tan \alpha \left (\frac{2b^2g- 2abg}{a} \right) &< \frac{2b^2gh- 4hga^2}{a^2} \\
\Leftrightarrow && \tan \alpha \left (\frac{2bg(b- a)}{a} \right) &< \frac{2hg(b^2- 2a^2)}{a^2} \\
\Rightarrow && \tan \alpha &< \frac{h(b^2-2a^2)}{ab(b-a)}
\end{align*}
Despite its obviously algebraic nature, and incorporating inequalities, this was a remarkably popular question with candidates, more than 400 of whom chose to do it. Moreover, it also proved to be the most successful question on the paper, with the average mark exceeding a score of 12. Marks that were lost generally arose from a lack of care with signs (directions) or a failure to justify the direction of the inequality from the physical nature of the situation.