Year: 2019
Paper: 2
Question Number: 9
Course: LFM Pure and Mechanics
Section: Projectiles
The Pure questions were again the most popular of the paper, with only one of the questions attempted by fewer than half of the candidates (of the remaining four questions, only question 9 was attempted by more than a quarter of the candidates). In many of the questions candidates were often unable to make good use of the results shown in the earlier parts of the question in order to solve the more complex later parts. Nevertheless, some good solutions were seen to all of the questions. For many of the questions, solutions were seen in which the results were reached, but without sufficient justification of some of the steps.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
A particle $P$ is projected from a point $O$ on horizontal ground with speed $u$ and angle of projection $\alpha$, where $0 < \alpha < \frac{1}{2}\pi$.
\begin{questionparts}
\item Show that if $\sin \alpha < \frac{2\sqrt{2}}{3}$, then the distance $OP$ is increasing throughout the flight.
Show also that if $\sin \alpha > \frac{2\sqrt{2}}{3}$, then $OP$ will be decreasing at some time before the particle lands.
\item At the same time as $P$ is projected, a particle $Q$ is projected horizontally from $O$ with speed $v$ along the ground in the opposite direction from the trajectory of $P$. The ground is smooth. Show that if
$$2\sqrt{2}v > (\sin \alpha - 2\sqrt{2} \cos \alpha)u,$$
then $QP$ is increasing throughout the flight of $P$.
\end{questionparts}\begin{questionparts}
\item Notice that $P = \begin{pmatrix} u \cos \alpha t\\ u \sin \alpha t - \frac12 g t^2 \end{pmatrix}$, so
\begin{align*}
&& |OP|^2 &= u^2 \cos^2 \alpha t^2 + \left (u \sin \alpha t - \frac12 g t^2 \right)^2 \\
&&&= u^2 \cos^2 \alpha t^2 +u^2 \sin^2 \alpha t^2 - u \sin \alpha g t^3 +\frac14 g^2 t^4 \\
&&&= u^2 t^2 -u\sin \alpha g t^3 + \frac14g^2t^4 \\
&& \frac{\d |OP|^2}{\d t} &= 2u^2 t - 3u \sin \alpha g t^2+g^2 t^3 \\
&&&= t \left (2u^2 - 3u \sin \alpha (gt)+(gt)^2\right) \\
&& \Delta &= 9u^2 \sin^2 \alpha -4 \cdot 2u^2 \cdot 1 \\
&&&= u^2 (9\sin^2 \alpha -8) \\
\end{align*}
Therefore if $\sin \alpha < \frac{2\sqrt{2}}3$ the discriminant is negative, the quadratic factor is always positive and the distance $|OP|$ is always increasing.
Similarly, if $\sin \alpha > \frac{2 \sqrt{2}}3$ then the derivative has a root. This means somewhere on its (possibly extended) trajectory $OP$ is decreasing. This must be before it lands, since if it were after it 'landed' then both the $x$ and $y$ distances are increasing, therefore it cannot occur after it 'lands'.
\item Note that $Q = \begin{pmatrix} -v t \\0 \end{pmatrix}$
\begin{align*}
&& |QP|^2 &= (u \cos \alpha t+vt)^2 + \left (u \sin \alpha t - \frac12 g t^2 \right)^2 \\
&&&= u^2 \cos^2 \alpha t^2+2u\cos \alpha v t^2 + v^2 t^2 +u^2 \sin^2 \alpha t^2 - u \sin \alpha g t^3 +\frac14 g^2 t^4 \\
&&&= (u^2+2u v \cos \alpha+v^2) t^2 - u \sin \alpha g t^3 + \frac14 g^2 t^4 \\
\\
\Rightarrow && \frac{\d |QP|^2}{\d t} &= 2(u^2+u v \cos \alpha+v^2) t - 3u \sin \alpha g t^2 + g^2 t^3 \\
&&&= t \left ( 2(u^2+2u v \cos \alpha+v^2) - 3u \sin \alpha (g t) + (g t)^2\right) \\
&& \Delta &= 9u^2 \sin^2 \alpha - 8(u^2+2u v \cos \alpha+v^2) \\
&&&= (9 \sin^2 \alpha -8)u^2 - 16v \cos \alpha u - 8v^2 \\
&&&= \left (( \sin \alpha-2\sqrt{2}\cos \alpha)u-2\sqrt{2} v \right) \left ( ( \sin \alpha+2\sqrt{2}\cos \alpha)u+2\sqrt{2} v \right)
\end{align*}
Since the second bracket is clearly positive, the first bracket must be negative (for $\Delta < 0$ and our derivative to be positive), ie $2\sqrt{2} v > ( \sin \alpha-2\sqrt{2}\cos \alpha)u$
\end{questionparts}
This was the most popular of the Mechanics and Statistics questions, but also one of the questions that attracted a large number of solutions that received no marks. Students seemed relatively good at setting up the kinematics equations in this question and most had the useful idea of differentiating. Somewhat fewer thought about using either completing the square or the quadratic discriminant to decide where the derivative was positive. The logic of the question was very poorly understood, with many students seeing the given inequality as the end point rather than the starting point of the question. In the second part of part (i) it was important that students demonstrated not just that a time existed where the distance is decreasing, but that this time was in the acceptable domain of the question. Part (ii) was conceptually very similar to part (i) but most students found the increased algebraic demand too much.