Year: 2016
Paper: 2
Question Number: 11
Course: LFM Pure and Mechanics
Section: Projectiles
As in previous years the Pure questions were the most popular of the paper with questions 1, 3 and 7 the most popular of these. The least popular questions of the paper were questions 10, 11, 12 and 13 with fewer than 400 attempts for each of them. There were many examples of solutions in this paper that were insufficiently well explained, given that the answer to be reached had been provided in the question.
Difficulty Rating: 1600.0
Difficulty Comparisons: 0
Banger Rating: 1484.0
Banger Comparisons: 1
\begin{questionparts}
\item
Two particles move on a smooth horizontal surface.
The positions, in Cartesian coordinates, of the
particles at time $t$
are $(a+ut\cos\alpha \,,\, ut\sin\alpha)$ and
$(vt\cos\beta\,,\, b+vt\sin\beta )$, where $a$, $b$, $u$ and $v$ are positive
constants, $\alpha$ and $\beta$ are constant acute angles, and $t\ge0$.
Given that the two particles collide, show that
\[
u \sin(\theta+\alpha) = v\sin(\theta +\beta)\,,
\]
where $\theta $ is the acute angle satisfying $\tan\theta = \dfrac b a$.
\item A gun is placed on the top of a vertical tower of height $b$ which stands on horizontal ground. The gun fires a bullet with speed $v$ and (acute) angle of elevation $\beta$. Simultaneously, a target is projected from a point on the ground a horizontal distance $a$ from the foot of the tower. The target is projected with speed $u$ and (acute) angle of elevation $\alpha$, in a direction directly away from the tower.
Given that the target is hit before it reaches the ground, show that
\[
2u\sin\alpha (u\sin\alpha - v\sin\beta) > bg\,.
\]
Explain, with reference to part (i), why the target can only be hit if $\alpha > \beta$.
\end{questionparts}\begin{questionparts}
\item The particles collide if there exists a time when
\begin{align*}
&& a + ut \cos \alpha &= vt \cos \beta \\
\Rightarrow && t (v \cos \beta-u \cos \alpha) &= a\\
&& ut \sin \alpha &= b + vt \sin \beta \\
\Rightarrow && t(u \sin \alpha - v \sin \beta) &= b\\
\Rightarrow && a(u\sin \alpha - v \sin \beta) &= b(v \cos \beta - u \cos \alpha) \\
\Rightarrow && u(a \sin \alpha + b \cos \alpha) &= v (b \cos \beta + a \sin \beta) \\
\Rightarrow && u \sin (\alpha + \theta) &= v \sin (\beta + \theta)
\end{align*}
\item The path of the bullet is $(vt \cos \beta, b + vt \sin \beta -\frac12 g t^2)$. The path of the target is $(a+ut \cos \alpha, ut \sin \alpha - \frac12 g t^2)$.
By comparing components as in part (i) and noting the acceleration doesn't change the story, we can see that $t(u \sin \alpha - v \sin \beta) = b$ and we also need $u t \sin \alpha - \frac12 gt^2 >0$ or $u \sin \alpha - \frac12 gt > 0$
\begin{align*}
&& u \sin \alpha & > \frac12 gt \\
&& 2u \sin \alpha & > g \frac{b}{(u \sin \alpha - v \sin \beta)} \\
\Rightarrow && 2u \sin \alpha( u \sin \alpha - v \sin \beta) & > gb
\end{align*}
Notice we must have $u \sin \alpha > v \sin \beta$ and $u \sin (\alpha + \theta) = v \sin (\beta + \theta)$ so $ \frac{\sin \alpha}{\sin (\alpha + \theta)} > \frac{\sin \beta}{\sin (\beta + \theta)}$, but if we consider $f(t) = \frac{\sin t}{\sin(t+x)}$ we can see $f'(t) = \frac{\cos t \sin(t + x) - \sin t \cos(t+x)}{\sin^2(t+x)} = \frac{\sin x}{\sin^2(t+x)} > 0$ is increasing, therefore $\alpha > \beta$.
\end{questionparts}
Although not a popular question, this was one of the better-answered questions in terms of the average number of marks achieved per candidate. Many candidates who attempted this question were able to gain many of the marks for part (i), often by substituting λ = b/a into the simultaneous equations and then eliminating. Some candidates, however, lost some marks for assuming that x = aλ and y = bλ. Part (ii) was well answered by many candidates, but very few solutions successfully explained the link between the two parts of the question.