Year: 2022
Paper: 2
Question Number: 10
Course: LFM Pure and Mechanics
Section: Projectiles
No solution available for this problem.
Candidates appeared to be generally well prepared for most topics within the examination, but there were a few situations in questions where some did not appear to be as proficient in standard techniques as needed. In particular, the method for finding invariant lines required in question 8 and the manipulation of trigonometric functions that were needed in question 10 caused considerable difficulties for some candidates. An additional issue that occurred at numerous points in the paper relates to the direction in which a deduction is required. It is important that candidates make sure that they know which statement is the one that they should start from as they deduce the other and that it is clear in their solution that the logic has gone in the correct direction. Clarity of solution is also an issue that candidates should be aware of, especially in the situations where the result to be reached has been given. It is important to check that there are no special cases that need to be considered separately, and when dividing by a function it is necessary to confirm that the function cannot be equal to 0 (and in the case of inequalities that the function always has the same sign). When drawing diagrams and sketching graphs it is useful if significant points that need to be clear are not drawn over the lines on the page as these can be difficult to interpret during the marking process.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
\begin{questionparts}
\item Show that, if a particle is projected at an angle $\alpha$ above the horizontal with speed $u$, it will reach height $h$ at a horizontal distance $s$ from the point of projection where
\[h = s\tan\alpha - \frac{gs^2}{2u^2\cos^2\alpha}\,.\]
\end{questionparts}
The remainder of this question uses axes with the $x$- and $y$-axes horizontal and the $z$-axis vertically upwards. The ground is a sloping plane with equation $z = y\tan\theta$ and a road runs along the $x$-axis. A cannon, which may have any angle of inclination and be pointed in any direction, fires projectiles from ground level with speed $u$. Initially, the cannon is placed at the origin.
\begin{questionparts}\setcounter{enumi}{1}
\item Let a point $P$ on the plane have coordinates $(x,\, y,\, y\tan\theta)$. Show that the condition for it to be possible for a projectile from the cannon to land at point $P$ is
\[x^2 + \left(y + \frac{u^2\tan\theta}{g}\right)^2 \leqslant \frac{u^4\sec^2\theta}{g^2}\,.\]
\item Show that the furthest point directly up the plane that can be reached by a projectile from the cannon is a distance
\[\frac{u^2}{g(1+\sin\theta)}\]
from the cannon.
How far from the cannon is the furthest point directly down the plane that can be reached by a projectile from it?
\item Find the length of road which can be reached by projectiles from the cannon.
The cannon is now moved to a point on the plane vertically above the $y$-axis, and a distance $r$ from the road. Find the value of $r$ which maximises the length of road which can be reached by projectiles from the cannon. What is this maximum length?
\end{questionparts}
While the first part of the question was answered well in general, candidates did not perform well on the question as a whole. Many candidates only answered the first part of the question, although some misunderstood and attempted to consider the maximum height reached. Others assigned the sign to acceleration incorrectly in the vertical motion equation. Many candidates also struggled with the trigonometric functions and failed to identify that trigonometric identities could be applied; this often meant that part (ii) was not possible. The lack of familiarity with trigonometric manipulations caused further marks to be lost later in the question also.