Year: 2017
Paper: 3
Question Number: 9
Course: UFM Mechanics
Section: Work, energy and Power 2
The total entry was only very slightly smaller than that of 2016, which was a record entry, but was still over 10% more than 2015. No question was attempted by in excess of 90%, although two were very popular and also five others were attempted by 60% or more. No question was generally avoided with even the least popular one attracting more than 10% of the entry. Less than 10% of candidates attempted more than 7 questions, and, apart from 18 exceptions, those doing so did not achieve very good totals and seemed to be 'casting around' to find things they could do: the 18 exceptions were very strong candidates who were generally achieving close to full marks on all the questions they attempted. The general trend was that those with six attempts fared better than those with more than six.
Difficulty Rating: 1700.0
Difficulty Comparisons: 0
Banger Rating: 1500.9
Banger Comparisons: 2
Two particles $A$ and $B$ of masses $m$ and $2 m$, respectively, are connected by a light spring of natural length $a$ and modulus of elasticity $\lambda$. They are placed on a smooth horizontal table with $AB$ perpendicular to the edge of the table, and $A$ is held on the edge of the table. Initially the spring is at its natural length.
Particle $A$ is released. At a time $t$ later, particle $A$ has dropped a distance $y$ and particle $ B$ has moved a distance $x$ from its initial position (where $x < a$).
Show that $ y + 2x= \frac12 gt^2$.
The value of $\lambda$ is such that particle $B$ reaches the edge of the table at a time $T$ given by $T= \sqrt{6a/g\,}\,$. By considering the total energy of the system (without solving any differential equations), show that the speed of particle $B$ at this time is $\sqrt{2ag/3\,}\,$.\begin{align*}
\text{N2}(\downarrow): && mg -T &= m\ddot{y} \\
\text{N2}(\rightarrow): && T &= 2m\ddot{x} \\
\Rightarrow && g &= \ddot{y}+2\ddot{x} \\
\Rightarrow && \tfrac12gt^2 &= y + 2x
\end{align*}
At time $T = \sqrt{6a/g}$, we have $y + 2x = 3a$, note also that $\dot{y}+2\dot{x} = gt$
\begin{array}{ccc}
& \text{KE} & \text{GPE} & \text{EPE} \\
\text{Initial} & 0 & 0 & 0 \\
\text{Final} & \frac12m\dot{y}^2 + \frac12(2m)\dot{x}^2 & -mgy & \frac{\lambda (y-x)^2}{2a}
\end{array}
Also note when we head over the table, $x = a$ and $y = a$
\begin{align*}
\text{COE}: && 0 &= \frac12m(gT-2\dot{x})^2+m\dot{x}^2-mga+\frac{\lambda(0)^2}{2a} \\
\Rightarrow && 0 &= (gT-2\dot{x})^2+2\dot{x}^2-2ga \\
&&&= (\sqrt{6ag}-2\dot{x})^2+2\dot{x}^2-2ga \\
&&&= 6\dot{x}^2-4\sqrt{6ag}+4ag \\
\Rightarrow &&&= (\sqrt{6}\dot{x} - 2\sqrt{ag})^2 \\
\Rightarrow && \dot{x} &= \sqrt{2ag/3}
\end{align*}
as required.
The most popular of the applied questions, there were a handful of attempts more for this question than for question 2. However, it was the least successfully attempted applied question with about one third marks scored. Common errors were to assume constant acceleration which does not apply, or to consider the motion of the centre of mass, but ignoring the normal force at the edge of the table, and the fact that the centre of mass does not lie along the string once motion commences. The two constants of integration for the first result were in fact zero but needed to be shown to be so. In considering the energy of the system, many assumed the speeds of and were equal, which they do work out to be, but this could not be known before calculating correctly, and likewise the elastic energy being zero, which again needed to be shown. Numerous attempts resulted in the given correct speed from specious working. Scoring largely occurred in the first section of the solution, though rarely earning all the marks for that first result and then earning little attempting to conserve energy.