Year: 2013
Paper: 1
Question Number: 9
Course: LFM Pure and Mechanics
Section: Projectiles
Around 1500 candidates sat this paper, a significant increase on last year. Overall, responses were good with candidates finding much to occupy them profitably during the three hours of the examination. In hindsight, two or three of the questions lacked sufficient 'punch' in their later parts, but at least most candidates showed sufficient skill to identify them and work on them as part of their chosen selection of questions. On the whole, nearly all candidates managed to attempt 4-6 questions – although there is always a significant minority who attempt 7, 8, 9, … bit and pieces of questions – and most scored well on at least two. Indeed, there were many scripts with 6 question-attempts, most or all of which were fantastically accomplished mathematically, and such excellence is very heart-warming.
Difficulty Rating: 1516.0
Difficulty Comparisons: 1
Banger Rating: 1516.0
Banger Comparisons: 1
Two particles, $A$ and $B$, are projected simultaneously towards each other from two points which are a distance $d$ apart in a horizontal plane. Particle $A$ has mass $m$ and is projected at speed $u$ at angle $\alpha$ above the horizontal. Particle $B$ has mass $M$ and is projected at speed $v$ at angle $\beta$ above the horizontal. The trajectories of the two particles lie in the same vertical plane. The particles collide directly when each is at its point of greatest height above the plane. Given that both $A$ and $B$ return to their starting points, and that momentum is conserved in the collision, show that
\[
m\cot \alpha = M \cot \beta\,.
\]
Show further that the collision occurs at a point which is a horizontal distance $b$ from the point of projection of $A$ where
\[
b= \frac{Md}{m+M}\, ,
\]
and find, in terms of $b$ and $\alpha$, the height above the horizontal plane at which the collision occurs.Since $A$ and $B$ return to their starting points, and at their highest points there is no vertical component to their velocities, their horizontal must perfectly reverse, ie
\begin{align*}
&& m u \cos \alpha - M v \cos \beta &= -m u \cos \alpha + M v \cos \beta \\
\Rightarrow && mu \cos \alpha &= Mv \cos \beta
\end{align*}
Since they reach their highest points at the same time, they must have the same initial vertical speed, ie $u \sin \alpha = v \sin \beta$, so
\begin{align*}
&& m v \frac{\sin \beta}{\sin \alpha} \cos \alpha &= M v \cos \beta \\
\Rightarrow && m \cot \alpha &= M \cot \beta
\end{align*}
The horizontal distance travelled by $A$ & $B$ will be:
\begin{align*}
&& d_A &= u \cos \alpha t \\
&& d_B &= v \cos \beta t \\
\Rightarrow && \frac{d_A}{d_A+d_B} &= \frac{u \cos \alpha}{u \cos \alpha + v \cos \beta} \\
&&&= \frac{\frac{M}{m}v \cos \beta}{\frac{M}{m}v \cos \beta + v \cos \beta} \\
&&&= \frac{M}{M+m} \\
\Rightarrow && d_A = b &= \frac{Md}{m+M}
\end{align*}
Applying $v^2 = u^2 + 2as$ we see that
\begin{align*}
&& 0 &= u \sin \alpha - gt \\
\Rightarrow && t &= \frac{u \sin \alpha}{g} \\
&& b &=u \cos \alpha \frac{u \sin \alpha}{g} \\
\Rightarrow && u^2 &= \frac{2bg}{\sin 2 \alpha} \\
&& 0 &= u^2 \sin^2 \alpha - 2g h \\
\Rightarrow && h &= \frac{u^2 \sin^2 \alpha}{2g} \\
&&&= \frac{2bg}{\sin 2 \alpha} \frac{ \sin^2 \alpha}{2g} \\
&&&= \frac12 b \tan \alpha
\end{align*}
This question was the most popular of the applied questions, drawing well over 500 responses, and the most successful of the mechanics questions, with an average score of 8.8/20. It proved to be a surprisingly good discriminator, giving a good range of marks. The use of constant-acceleration formulae for the projectile motion provided a routine and straightforward start to the question, but this was followed by the momentum equation for the collision, which proved trickier, with quite a few candidates getting to mucosθ – Mvcosθ = MwB – mwA but no further. A lot of candidates resorted to writing down the result mucosθ = Mvcosθ without any attempt to justify it. The second result then found many candidates going round in algebraic circles, and very few indeed managed to find the answer (not given) to the very final part of the question.