2019 Paper 1 Q10

Year: 2019
Paper: 1
Question Number: 10

Course: LFM Pure and Mechanics
Section: Projectiles

Difficulty: 1500.0 Banger: 1500.0

Problem

In this question, the \(x\)-axis is horizontal and the positive \(y\)-axis is vertically upwards. A particle is projected from the origin with speed \(u\) at an angle \(\alpha\) to the vertical. The particle passes through the fixed point \((h \tan \beta, h)\), where \(0 < \beta < 90^{\circ}\) and \(h > 0\).
  1. Show that $$c^2 - ck \cot \beta + 1 + k \cot^2 \beta = 0, \quad (*)$$ where \(c = \cot \alpha\) and \(k = \frac{2u^2}{gh}\). You are given that there are two distinct values of \(\alpha\) that satisfy equation \((*)\). Let \(\alpha_1\) and \(\alpha_2\) be these values.
    1. Show that $$\cot \alpha_1 + \cot \alpha_2 = k \cot \beta.$$ Show also that $$\alpha_1 + \alpha_2 = \beta.$$
    2. Show that $$k > 2(1 + \sec \beta).$$
  2. By considering the greatest height attained by the particle, show that \(k \geq 4 \sec^2 \alpha\).

Solution

  1. The horizontal position of the particle at time \(t\) is \(u \sin\alpha t\), so \(T = \frac{h \tan \beta}{u \sin \alpha}\) The vertical position of the particle at this time \(T\) satisifes: \begin{align*} && h &= u \cos\alpha \frac{h \tan \beta}{u \sin\alpha} - \frac12 g \left ( \frac{h \tan \beta}{u \sin\alpha} \right)^2 \\ &&&= h\cot \alpha \tan \beta - \frac{gh^2}{2u^2} \tan^2 \beta \cosec^2 \beta \\ \Rightarrow && 1 &= c \tan \beta - \frac{1}{k} \tan^2 \beta (1 + c^2) \\ \Rightarrow && k \cot^2 \beta &= kc\cot \beta -1-c^2 \\ \Rightarrow && 0 &= c^2 -ck \cot \beta + 1 + k \cot^2 \beta \end{align*}
    1. As a quadratic in \(c\) the sum of the roots is \(k \cot \beta\), therefore \(\cot \alpha_1 + \cot \alpha_2 = k \cot \beta\). We also have that \(\cot \alpha_1 \cot \alpha_2 = 1 + k \cot^2 \beta\), so \begin{align*} && \cot (\alpha_1 + \alpha_2) &= \frac{\cot \alpha_1 \cot \alpha_2-1}{\cot \alpha_1 + \cot \alpha_2} \\ &&&= \frac{1 + k \cot^2 \beta - 1}{k \cot \beta} \\ &&&= \cot \beta \\ \Rightarrow && \beta &= \alpha_1 + \alpha_2 \pmod{\pi} \end{align*} but since \(\alpha_i \in (0, \frac{\pi}{2})\) the equation must hold exactly.
    2. Since it has two real roots we must have \begin{align*} && 0 &<\Delta = k^2 \cot^2 \beta - 4 (1 + k \cot^2 \beta) \\ &&&= k^2 \cot^2 \beta-4k \cot^2 \beta -4 \\ &&&= \cot^2 \beta (k^2 - 4k - 4(\sec^2 \beta - 1)) \\ &&&= \cot^2 \beta ( (k-2)^2 -4\sec^2 \beta) \\ \Rightarrow && k &> 2 + 2\sec \beta = 2(1+\sec \beta) \end{align*}
  2. The greatest height will satisfy \(v^2 = u^2 + 2as\) so \(0 = u^2 \cos^2 \alpha - 2gh_{max} \Rightarrow 4\sec^2 \alpha = \frac{2u^2}{gh_{max}} = k_{max}\), but this decreases with \(h\), so the smallest \(k\) can be is \(4\sec^2 \alpha\), ie \(k \geq 4 \sec^2 \alpha\)
Examiner's report
— 2019 STEP 1, Question 10
Mean: ~4.8 / 20 (inferred) Below Average Inferred ~4.8/20: bounded 3.0-10.0 (above Q5=3.0 least successful); 'poorly done' → lower quarter of 3.0-10.0 ≈ 4.8. Applied question with very few hits; no specific popularity figure.

This mechanics question was poorly done, despite being on the topic most frequently occurring on the early STEPs. Many candidates knew a general result for the trajectory of a projectile, but some could not adapt it when the angle given was with the vertical. Candidates did not seem to be sufficiently familiar with the sums and products of roots of a quadratic, which is new to the syllabus. This question introduced two inequalities, and this is always a problem for candidates. The first came from the quadratic discriminant and the second came from the given point being not greater than the maximum height. However, candidates often resorted to algebraic meandering rather than clear thinking about how an inequality might arise. Those candidates who really engaged with the question generally did well, although very high marks were seldom acquired.

In 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; all of which are available from the STEP and Cambridge Examinations Board websites. 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 for several months beforehand. 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 it is that a candidate should never think that they are simply required to 'go through the motions' but must expect, sooner or later, to 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. When you read through the report and look at the solutions (either in the mark-scheme or the Hints & 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's paper produced the usual sorts of outcomes, with far too many candidates wasting valuable time by attempting more than six questions, and with many of these candidates picking up 0-4 marks on several 'false starts' which petered out the moment some understanding was required. Around one candidate in eight failed to hit the 30 mark overall, though this is an improvement on last year. Most candidates were able to produce good attempts at two or more questions. At the top end of the scale, around a hundred candidates scored 100 or more out of 120, with four hitting the maximum of 120 and many others not far behind. 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, although under two-thirds of the entry attempted it this year, and it also turned out to be the most successful question on the paper with a mean score of about 12 out of 20. In order of popularity, Q1 was followed by Qs.3, 4 and 2. Indeed, it was the pure maths questions in Section A that attracted the majority of attention from candidates, with the applied questions combined scoring fewer 'hits' than any one of the first four questions on its own. Though slightly more popular than the applied questions, the least successful question of all was Q5, on vectors. This question was attempted by almost 750 candidates, but 70% of these scored no more than 2 marks, leaving it with a mean score of just over 3 out of 20. Q9 (a statics question) was found only marginally more appetising, with a mean score of almost 3½ out of 20. In general, it was found that explanations were poorly supplied, with many candidates happy to overlook completely any requests for such details.

Source: Cambridge STEP 2019 Examiner's Report · 2019-p1.pdf
Rating Information

Difficulty Rating: 1500.0

Difficulty Comparisons: 0

Banger Rating: 1500.0

Banger Comparisons: 0

Show LaTeX source
Problem source
In this question, the $x$-axis is horizontal and the positive $y$-axis is vertically upwards.
A particle is projected from the origin with speed $u$ at an angle $\alpha$ \textit{to the vertical}. The particle passes through the fixed point $(h \tan \beta, h)$, where $0 < \beta < 90^{\circ}$ and $h > 0$.
\begin{questionparts}
\item Show that
$$c^2 - ck \cot \beta + 1 + k \cot^2 \beta = 0, \quad (*)$$
where $c = \cot \alpha$ and $k = \frac{2u^2}{gh}$.
You are given that there are two distinct values of $\alpha$ that satisfy equation $(*)$. Let $\alpha_1$ and $\alpha_2$ be these values.
\begin{enumerate}
\item Show that
$$\cot \alpha_1 + \cot \alpha_2 = k \cot \beta.$$
Show also that
$$\alpha_1 + \alpha_2 = \beta.$$
\item Show that
$$k > 2(1 + \sec \beta).$$
\end{enumerate}
\item By considering the greatest height attained by the particle, show that $k \geq 4 \sec^2 \alpha$.
\end{questionparts}
Solution source
\begin{questionparts}
\item The horizontal position of the particle at time $t$ is $u \sin\alpha t$, so $T = \frac{h \tan \beta}{u \sin \alpha}$

The vertical position of the particle at this time $T$ satisifes:

\begin{align*}
&& h &= u \cos\alpha \frac{h \tan \beta}{u \sin\alpha} - \frac12 g \left ( \frac{h \tan \beta}{u \sin\alpha} \right)^2 \\
&&&= h\cot \alpha \tan \beta - \frac{gh^2}{2u^2} \tan^2 \beta \cosec^2 \beta \\
\Rightarrow && 1 &= c \tan \beta - \frac{1}{k} \tan^2 \beta (1 + c^2) \\
\Rightarrow && k \cot^2 \beta &= kc\cot \beta -1-c^2 \\
\Rightarrow && 0 &= c^2 -ck \cot \beta + 1 + k \cot^2 \beta
\end{align*}

\begin{enumerate}
\item As a quadratic in $c$ the sum of the roots is $k \cot \beta$, therefore $\cot \alpha_1 + \cot \alpha_2 = k \cot \beta$. We also have that $\cot \alpha_1 \cot \alpha_2 = 1 + k \cot^2 \beta$, so

\begin{align*}
&& \cot (\alpha_1 + \alpha_2) &= \frac{\cot \alpha_1 \cot \alpha_2-1}{\cot \alpha_1 + \cot \alpha_2} \\
&&&= \frac{1 + k \cot^2 \beta - 1}{k \cot \beta} \\
&&&= \cot \beta \\
\Rightarrow && \beta &= \alpha_1 + \alpha_2 \pmod{\pi}
\end{align*}

but since $\alpha_i \in (0, \frac{\pi}{2})$ the equation must hold exactly.

\item Since it has two real roots we must have 
\begin{align*}
&& 0 &<\Delta = k^2 \cot^2 \beta - 4 (1 + k \cot^2 \beta) \\
&&&= k^2 \cot^2 \beta-4k \cot^2 \beta -4 \\
&&&= \cot^2 \beta (k^2 - 4k - 4(\sec^2 \beta - 1)) \\
&&&= \cot^2 \beta ( (k-2)^2 -4\sec^2 \beta) \\
\Rightarrow && k &> 2 + 2\sec \beta = 2(1+\sec \beta)
\end{align*}
\end{enumerate}

\item The greatest height will satisfy $v^2 = u^2 + 2as$ so $0 = u^2 \cos^2 \alpha - 2gh_{max} \Rightarrow 4\sec^2 \alpha = \frac{2u^2}{gh_{max}} = k_{max}$, but this decreases with $h$, so the smallest $k$ can be is $4\sec^2 \alpha$, ie $k \geq 4 \sec^2 \alpha$

\end{questionparts}