OCR FM1 AS 2018 March — Question 6 9 marks

Exam BoardOCR
ModuleFM1 AS (Further Mechanics 1 AS)
Year2018
SessionMarch
Marks9
TopicCircular Motion 2
TypeVertical circle with peg/obstacle
DifficultyHard +2.3 This is a challenging Further Mechanics problem requiring energy conservation across multiple stages (initial motion to barrier, collision with restitution, motion through lowest point, and ascent to bell), careful tracking of the coefficient of restitution through the analysis, and optimization to find the critical condition. The 'show that' format with parameters and the need to prove an inequality for all barrier positions requires sophisticated mathematical reasoning beyond standard circular motion exercises.
Spec6.02i Conservation of energy: mechanical energy principle6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration

6 A fairground game involves a player kicking a ball, \(B\), from rest so as to project it with a horizontal velocity of magnitude \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The ball is attached to one end of a light rod of length \(l \mathrm {~m}\). The other end of the rod is smoothly hinged at a fixed point \(O\) so that \(B\) can only move in the vertical plane which contains \(O\), a fixed barrier and a bell which is fixed \(l \mathrm {~m}\) vertically above \(O\). Initially \(B\) is vertically below \(O\). The barrier is positioned so that when \(B\) collides directly with the barrier, \(O B\) makes an angle \(\theta\) with the downwards vertical through \(O\) (see diagram). \includegraphics[max width=\textwidth, alt={}, center]{bf86ac88-0fd1-4d49-a705-9b8d06fbac2a-4_643_659_584_724} The coefficient of restitution between \(B\) and the barrier is \(e . B\) rebounds from the barrier, passes through its original position and continues on a circular path towards the bell. The bell will only ring if the ball strikes it with a speed of at least \(V \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The player wins the game if the player causes the bell to ring having kicked \(B\) so that it first collides with the barrier. You may assume that \(B\) and the bell are small and that the barrier has negligible thickness. Show that, whatever the position of the barrier, the player cannot win the game if \(u ^ { 2 } < 4 g l + \frac { V ^ { 2 } } { e ^ { 2 } }\). \section*{END OF QUESTION PAPER}

AnswerMarks Guidance
AnswerMarks Guidance
Initial energy \(= \frac{1}{2}mu^2\)B1
Energy just before collision \(= \frac{1}{2}mv^2 + mgl(1 - \cos\theta)\)B1
\(\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgl(1 - \cos\theta)\)B1ft Conservation of energy
Speed after collision \(= ev\)B1 Condone sign errors
Energy \(= e^2\left(\frac{1}{2}mu^2 - mgl(1 - \cos\theta)\right) + mgl(1 - \cos\theta)\)M1
This must be at least \(2mgl + \frac{1}{2}mV^2\) for player to winM1 Comparison with energy to ring bell
Player fails when \(u^2 < \frac{4gl + V^2 - 2gl(1 - \cos\theta)(1 - e^2)}{e^2}\)A1
Least value of RHS is when \(\cos\theta = -1\)M1
So must fail if \(u^2 < 4gl + \frac{V^2}{e^2}\)A1 [9] AG working must be fully correct
Alternative solution
Initial energy \(= \frac{1}{2}mu^2\)B1
Speed before impact \(= v \Rightarrow\) speed after impact \(= ev\)M1
Effect of impact is loss of some kinetic energy onlyE1
Minimum \(u\) requires loss to be least, i.e. smallest \(v\)E1 Explanations must be complete, correct and convincing
This occurs when barrier is at the highest pointE1
Condition for failure in this case is \(ev < V\)M1
\(\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mg \times 2l\)M1 Conservation of energy
\(u^2 = 4gl + v^2\)A1
\(\Rightarrow u^2 < 4gl + \frac{V^2}{e^2}\)A1 AG working must be fully correct
| Answer | Marks | Guidance |
|--------|-------|----------|
| Initial energy $= \frac{1}{2}mu^2$ | B1 | |
| Energy just before collision $= \frac{1}{2}mv^2 + mgl(1 - \cos\theta)$ | B1 | |
| $\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mgl(1 - \cos\theta)$ | B1ft | Conservation of energy |
| Speed after collision $= ev$ | B1 | Condone sign errors |
| Energy $= e^2\left(\frac{1}{2}mu^2 - mgl(1 - \cos\theta)\right) + mgl(1 - \cos\theta)$ | M1 | |
| This must be at least $2mgl + \frac{1}{2}mV^2$ for player to win | M1 | Comparison with energy to ring bell |
| Player fails when $u^2 < \frac{4gl + V^2 - 2gl(1 - \cos\theta)(1 - e^2)}{e^2}$ | A1 | |
| Least value of RHS is when $\cos\theta = -1$ | M1 | |
| So must fail if $u^2 < 4gl + \frac{V^2}{e^2}$ | A1 [9] | AG working must be fully correct |
| **Alternative solution** | | |
| Initial energy $= \frac{1}{2}mu^2$ | B1 | |
| Speed before impact $= v \Rightarrow$ speed after impact $= ev$ | M1 | |
| Effect of impact is loss of some kinetic energy only | E1 | |
| Minimum $u$ requires loss to be least, i.e. smallest $v$ | E1 | Explanations must be complete, correct and convincing |
| This occurs when barrier is at the highest point | E1 | |
| Condition for failure in this case is $ev < V$ | M1 | |
| $\frac{1}{2}mu^2 = \frac{1}{2}mv^2 + mg \times 2l$ | M1 | Conservation of energy |
| $u^2 = 4gl + v^2$ | A1 | |
| $\Rightarrow u^2 < 4gl + \frac{V^2}{e^2}$ | A1 | AG working must be fully correct |
6 A fairground game involves a player kicking a ball, $B$, from rest so as to project it with a horizontal velocity of magnitude $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The ball is attached to one end of a light rod of length $l \mathrm {~m}$. The other end of the rod is smoothly hinged at a fixed point $O$ so that $B$ can only move in the vertical plane which contains $O$, a fixed barrier and a bell which is fixed $l \mathrm {~m}$ vertically above $O$.

Initially $B$ is vertically below $O$. The barrier is positioned so that when $B$ collides directly with the barrier, $O B$ makes an angle $\theta$ with the downwards vertical through $O$ (see diagram).\\
\includegraphics[max width=\textwidth, alt={}, center]{bf86ac88-0fd1-4d49-a705-9b8d06fbac2a-4_643_659_584_724}

The coefficient of restitution between $B$ and the barrier is $e . B$ rebounds from the barrier, passes through its original position and continues on a circular path towards the bell. The bell will only ring if the ball strikes it with a speed of at least $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The player wins the game if the player causes the bell to ring having kicked $B$ so that it first collides with the barrier. You may assume that $B$ and the bell are small and that the barrier has negligible thickness.

Show that, whatever the position of the barrier, the player cannot win the game if $u ^ { 2 } < 4 g l + \frac { V ^ { 2 } } { e ^ { 2 } }$.

\section*{END OF QUESTION PAPER}

\hfill \mbox{\textit{OCR FM1 AS 2018 Q6 [9]}}