Year: 2020
Paper: 2
Question Number: 10
Course: UFM Mechanics
Section: Work, energy and Power 1
No solution available for this problem.
There were just over 800 entries for this paper, and good solutions were seen to all of the questions. Candidates should be aware of the need to provide clear explanations of their reasoning throughout the paper (and particularly in questions where the result to be shown is given in the question). Short explanatory comments at key points in solutions can be very helpful in this regard, as can clearly drawn diagrams of the situation described in the question. The paper included a few questions where a statement of the form "A if and only if B" needed to be proven – candidates should be aware of the meaning of such statements and make sure that both directions of the implication are covered clearly. In general, candidates who performed better on the questions in this paper recognised the relationships between the different parts of each question and were able to adapt methods used in earlier parts when working on the later sections of the question.
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
A particle $P$ of mass $m$ moves freely and without friction on a wire circle of radius $a$, whose axis is horizontal. The highest point of the circle is $H$, the lowest point of the circle is $L$ and angle $PHL = \theta$. A light spring of modulus of elasticity $\lambda$ is attached to $P$ and to $H$. The natural length of the spring is $l$, which is less than the diameter of the circle.
\begin{questionparts}
\item Show that, if there is an equilibrium position of the particle at $\theta = \alpha$, where $\alpha > 0$, then $\cos\alpha = \dfrac{\lambda l}{2(a\lambda - mgl)}$.
Show also that there will only be such an equilibrium position if $\lambda > \dfrac{2mgl}{2a - l}$.
When the particle is at the lowest point $L$ of the circular wire, it has speed $u$.
\item Show that, if the particle comes to rest before reaching $H$, it does so when $\theta = \beta$, where $\cos\beta$ satisfies
\[(\cos\alpha - \cos\beta)^2 = (1 - \cos\alpha)^2 + \frac{mu^2}{2a\lambda}\cos\alpha,\]
where $\cos\alpha = \dfrac{\lambda l}{2(a\lambda - mgl)}$.
Show also that this will only occur if $u^2 < \dfrac{2a\lambda}{m}(2 - \sec\alpha)$.
\end{questionparts}
This was the least popular question on the paper and was found to be quite difficult by those who did attempt it. It is very useful in questions of this type to produce a good sketch of the situation described and, unfortunately, many attempts did not do this. The candidates who made good progress did tend to have good diagrams. When resolving forces, it is useful to consider the different directions that could be chosen. In the case of this question, many candidates chose to resolve horizontally and vertically and as a result produced more complicated equations to deal with. While in most cases they were able to simplify the equations that they obtained, this method did result in more work than was necessary. When dividing equations and inequalities by expressions, candidates should be careful to consider whether the expressions are known to be non-zero (and in the case of inequalities that it is known whether it is positive or negative). In the second part of the question, candidates struggled with the more complicated algebraic expressions that had to be manipulated and many candidates gave up before reaching the end of the question.