Year: 2017
Paper: 1
Question Number: 9
Course: LFM Pure and Mechanics
Section: Projectiles
The pure questions were again the most popular of the paper with questions 1 and 3 being attempted by almost all candidates. The least popular questions on the paper were questions 10, 11, 12 and 13 and a significant proportion of attempts at these were brief, attracting few or no marks. Candidates generally demonstrated a high level of competence when completing the standard processes and there were many good attempts made when questions required explanations to be given, particularly within the pure questions. A common feature of the stronger responses to questions was the inclusion of diagrams.
Difficulty Rating: 1484.0
Difficulty Comparisons: 1
Banger Rating: 1484.0
Banger Comparisons: 1
A particle is projected at speed $u$ from a point $O$ on a horizontal plane.
It passes through a fixed point $P$ which is at a horizontal distance $d$ from $O$ and at a height $d \tan \beta$ above the plane, where $d>0$ and $\beta $ is an acute angle.
The angle of projection $\alpha$ is chosen so that $u$ is as small as
possible.
\begin{questionparts}
\item Show that $u^2 = gd \tan \alpha$ and $2\alpha = \beta + 90^\circ\,$.
\item At what angle to the horizontal is the particle travelling when it passes through $P$? Express your answer in terms of $\alpha$ in its simplest form.
\end{questionparts}\begin{questionparts}
\item $\,$
\begin{align*}
&& d &= u \cos \alpha t \\
&& d \tan \beta &= u \sin \alpha t - \frac12 gt^2 \\
&& &= d\tan \alpha - \frac1{2u^2} g d^2 \sec^2 \alpha \\
\Rightarrow && u^2 &= \frac{gd \sec^2 \alpha}{2(\tan \alpha + \tan \beta)} \\
&&&= \frac{gd t^2}{2(t+\tan \beta)} \\
&& \frac{\d}{\d t} \left (u^2 \right) &= \frac{2gdt\cdot 2(t+\tan \beta) - gdt^2 \cdot 2}{4(t+\tan \beta)^2} \\
&&&= \frac{2gdt(2t-t+2\tan \beta)}{4(t+\tan \beta)^2} \\
&&&= \frac{gdt(t+2\tan \beta)}{2(t+\tan \beta)^2} \\
\end{align*}
So either $t = 0$ or $t = -2 \tan \beta$
\begin{align*}
&& u^2 &= \frac{gd\cdot 4 \tan^2 \beta}{2(-2\tan \beta + \tan \beta)} \\
&&&= \frac{2gd \tan \beta}{-1} \\
&&&= gd (-2\tan \beta) \\
&&&= gd \tan \alpha
\end{align*}
\begin{align*}
&& d \tan \beta &= d \tan \alpha - \frac12 \frac{gd^2}{gd \tan \alpha \cdot \cos^2 \alpha } \\
\Rightarrow && \tan \beta&= \tan \alpha - \frac{1}{2\sin \alpha \cos \alpha} \\
&&&= \frac{2 \sin^2 \alpha - 1}{2 \sin \alpha \cos \alpha} \\
&&&= \frac{-\cos 2 \alpha}{\sin 2 \alpha} \\
&&&= -\cot 2 \alpha \\
&&&= \tan (2\alpha - 90^\circ) \\
\Rightarrow && \beta &= 2\alpha - 90^\circ \\
\Rightarrow && 2\alpha &= \beta + 90^\circ
\end{align*}
\item Suppose the angle to the horizontal is $\theta$, then $\tan \theta = \frac{v_y}{v_x}$ so
\begin{align*}
&& \tan \theta &= \frac{u \sin \alpha - gt}{u \cos \alpha} \\
&&&= \frac{u \sin \alpha - \frac12 g \frac{d}{u \cos \alpha}}{u \cos \alpha} \\
&&&= \frac{u^2\sin \alpha \cos \alpha - gd}{u^2 \cos^2 \alpha} \\
&&&= \frac{gd \tan \alpha \sin \alpha \cos \alpha- gd}{ gd \tan \alpha \cdot \cos^2 \alpha} \\
&&&= \frac{\tan \alpha \sin \alpha \cos \alpha - 1}{ \sin \alpha \cos \alpha} \\
&&&= \frac{\sin^2 \alpha - 1}{\sin \alpha \cos \alpha} \\
&&&= -\frac{\cos \alpha}{\sin \alpha}\\
&&&= - \cot \alpha = \tan (\alpha - 90^\circ)\\
\Rightarrow && \theta &= \alpha - 90^\circ
\end{align*}
\end{questionparts}
This was the most popular of the Mechanics questions, but still less popular than half of the Pure questions. Of all of the questions on the paper, this is the question that received the lowest average mark. Many attempts were able to produce the correct equations for the horizontal and vertical components of the motion. The differentiation required to then establish the result in part (i) proved quite complicated for many candidates and so many did not reach the required result. Those who got as far as part (ii) were able to draw some conclusions about the relationships between the two angles, but struggled to reach the simplest form. Only a few candidates were able to achieve full marks for this question.