Year: 2023
Paper: 3
Question Number: 9
Course: LFM Pure and Mechanics
Section: Newton's laws and connected particles
No solution available for this problem.
The total entry was a marginal increase on that of 2022 (by just over 1%). Two questions were attempted by more than 90% of candidates, another two by 80%, and another two by about two thirds. The least popular questions were attempted by more than a sixth of candidates. All the questions were perfectly answered by at least three candidates (but mostly more than this), with one being perfectly answered by eighty candidates. Very nearly 90% of candidates attempted no more than 7 questions. One general comment regarding all the questions is that candidates need to make sure that they read the question carefully, paying particular attention to command words such as "hence" and "show that".
Difficulty Rating: 1500.0
Difficulty Comparisons: 0
Banger Rating: 1500.0
Banger Comparisons: 0
Two particles, $A$ of mass $m$ and $B$ of mass $M$, are fixed to the ends of a light inextensible string $AB$ of length $r$ and lie on a smooth horizontal plane. The origin of coordinates and the $x$- and $y$-axes are in the plane.
Initially, $A$ is at $(0,\,0)$ and $B$ is at $(r,\,0)$. $B$ is at rest and $A$ is given an instantaneous velocity of magnitude $u$ in the positive $y$ direction.
At a time $t$ after this, $A$ has position $(x,\,y)$ and $B$ has position $(X,\,Y)$. You may assume that, in the subsequent motion, the string remains taut.
\begin{questionparts}
\item Explain by means of a diagram why
\[X = x + r\cos\theta\]
\[Y = y - r\sin\theta\]
where $\theta$ is the angle clockwise from the positive $x$-axis of the vector $\overrightarrow{AB}$.
\item Find expressions for $\dot{X}$, $\dot{Y}$, $\ddot{X}$ and $\ddot{Y}$ in terms of $\ddot{x}$, $\ddot{y}$, $\dot{x}$, $\dot{y}$, $r$, $\ddot{\theta}$, $\dot{\theta}$ and $\theta$, as appropriate.
Assume that the tension $T$ in the string is the only force acting on either particle.
\item Show that
\[\ddot{x}\sin\theta + \ddot{y}\cos\theta = 0\]
\[\ddot{X}\sin\theta + \ddot{Y}\cos\theta = 0\]
and hence that $\theta = \dfrac{ut}{r}$.
\item Show that
\[m\ddot{x} + M\ddot{X} = 0\]
\[m\ddot{y} + M\ddot{Y} = 0\]
and find $my + MY$ in terms of $t$ and $m, M, u, r$ as appropriate.
\item Show that
\[y = \frac{1}{m+M}\left(mut + Mr\sin\!\left(\frac{ut}{r}\right)\right).\]
\item Show that, if $M > m$, then the $y$ component of the velocity of particle $A$ will be negative at some time in the subsequent motion.
\end{questionparts}
This question only just beat question 11 to be the least popular question on the paper. Although its mean score was only 6.9/20, it was more successfully attempted than two of the Pure questions and the other Mechanics question. Parts (i) and (ii) were well answered by a good number of candidates, if candidates once set up the problem and then got going. The diagrams were done well, and the derivatives didn't pose much of a problem for most, although there were some errors in the second derivatives and applying the chain rule, with candidates forgetting to multiply by θ̇ to produce θ̇². Part (iii) was more mixed in terms of good responses. Those who did this by resolving forces horizontally and vertically set up the remainder of the question well, but some seemed to struggle with the first part and just assumed the equations were true. The biggest problem for a large number of candidates here was applying boundary conditions when integrating and justifying the choice of boundary conditions. Parts (iv) and (v) were answered fairly well. Part (vi) lead to a lot of marks lost, as they were required to justify the velocity being negative by finding a suitable time to ensure it happens, which most did not do.