2023 Paper 2 Q10

Year: 2023
Paper: 2
Question Number: 10

Course: LFM Pure and Mechanics
Section: Projectiles

Difficulty: 1500.0 Banger: 1500.0

Problem

In this question, the \(x\)- and \(y\)-axes are horizontal and the \(z\)-axis is vertically upwards.
  1. A particle \(P_\alpha\) is projected from the origin with speed \(u\) at an acute angle \(\alpha\) above the positive \(x\)-axis. The curve \(E\) is given by \(z = A - Bx^2\) and \(y = 0\). If \(E\) and the trajectory of \(P_\alpha\) touch exactly once, show that \[u^2 - 2gA = u^2(1 - 4AB)\cos^2\alpha\,.\] \(E\) and the trajectory of \(P_\alpha\) touch exactly once for all \(\alpha\) with \(0 < \alpha < \frac{1}{2}\pi\). Write down the values of \(A\) and \(B\) in terms of \(u\) and \(g\).
An explosion takes place at the origin and results in a large number of particles being simultaneously projected with speed \(u\) in different directions. You may assume that all the particles move freely under gravity for \(t \geqslant 0\).
  1. Describe the set of points which can be hit by particles from the explosion, explaining your answer.
  2. Show that, at a time \(t\) after the explosion, the particles lie on a sphere whose centre and radius you should find.
  3. Another particle \(Q\) is projected horizontally from the point \((0, 0, A)\) with speed \(u\) in the positive \(x\) direction. Show that, at all times, \(Q\) lies on the curve \(E\).
  4. Show that for particles \(Q\) and \(P_\alpha\) to collide, \(Q\) must be projected a time \(\dfrac{u(1-\cos\alpha)}{g\sin\alpha}\) after the explosion.

No solution available for this problem.

Examiner's report
— 2023 STEP 2, Question 10

Part (i) was answered well. Almost all candidates used the discriminant condition correctly, even if their quadratic contained an error. The final two marks in this part were trickier to achieve. Many candidates substituted two values for alpha and then solved for A, B. Of these, it was fairly common for them to use α = 0, ½π which were excluded from the range being considered. Many missed that one can simply set the coefficients of each side to zero. Some treated α as a variable to be solved for rather than varied. Part (ii) was found very tricky. Some candidates were able to guess or intuit that the safe zone should be the parabola considered in part (i), but almost no candidate gave a proper explanation as to why it was only these points that could be reached. A few considered the 2D parabola correctly but did not consider 3D. Some wondered about the presence of a "floor" at z = 0. This only caused an issue if the candidate thought we were only interested in points of intersection with this plane, possibly caused by imagining the problem in the context of artillery as it is often presented in schools. In this case they answered with a circle and received no credit. Part (iii) was fairly tough for most candidates. Many candidates who could not find the equation of a circle/sphere would attempt an explanation in words, which was almost never sufficient to earn credit. It was possible for candidates to guess the correct centre and radius with no mathematical justification, but this earned no credit. Many candidates who attempted part (iv) were able to complete it successfully. Candidates who attempted part (v) were generally able to pick up at least one mark by appealing to earlier calculations. Most appreciated the need to introduce separate times for Q and P, or a time delay between the two.

Many candidates were able to express their reasoning clearly and presented good solutions to the questions that they attempted. There were excellent solutions seen for all of the questions. An area where candidates struggled in several questions was in the direction of the logic that was required in a solution. Some candidates failed to appreciate that separate arguments may be needed for the "if" and "only if" parts of a question and, in some cases, candidates produced correct arguments, but for the wrong direction. In several questions it was clear that candidates who used sketches or diagrams generally performed much better that those who did not. Sketches often also helped to make the solution clearer and easier to understand. Several questions on the STEP papers ask candidates to show a given result. Candidates should be aware that there is a need to present sufficient detail in their solutions so that it is clear that the reasoning is well understood.

Source: Cambridge STEP 2023 Examiner's Report · 2023-p2.pdf
Rating Information

Difficulty Rating: 1500.0

Difficulty Comparisons: 0

Banger Rating: 1500.0

Banger Comparisons: 0

Show LaTeX source
Problem source
In this question, the $x$- and $y$-axes are horizontal and the $z$-axis is vertically upwards.
\begin{questionparts}
\item A particle $P_\alpha$ is projected from the origin with speed $u$ at an acute angle $\alpha$ above the positive $x$-axis.
The curve $E$ is given by $z = A - Bx^2$ and $y = 0$. If $E$ and the trajectory of $P_\alpha$ touch exactly once, show that
\[u^2 - 2gA = u^2(1 - 4AB)\cos^2\alpha\,.\]
$E$ and the trajectory of $P_\alpha$ touch exactly once for all $\alpha$ with $0 < \alpha < \frac{1}{2}\pi$. Write down the values of $A$ and $B$ in terms of $u$ and $g$.
\end{questionparts}
An explosion takes place at the origin and results in a large number of particles being simultaneously projected with speed $u$ in different directions. You may assume that all the particles move freely under gravity for $t \geqslant 0$.
\begin{questionparts}
\setcounter{enumi}{1}
\item Describe the set of points which can be hit by particles from the explosion, explaining your answer.
\item Show that, at a time $t$ after the explosion, the particles lie on a sphere whose centre and radius you should find.
\item Another particle $Q$ is projected horizontally from the point $(0, 0, A)$ with speed $u$ in the positive $x$ direction.
Show that, at all times, $Q$ lies on the curve $E$.
\item Show that for particles $Q$ and $P_\alpha$ to collide, $Q$ must be projected a time $\dfrac{u(1-\cos\alpha)}{g\sin\alpha}$ after the explosion.
\end{questionparts}