| Exam Board | OCR |
|---|---|
| Module | FM1 AS (Further Mechanics 1 AS) |
| Year | 2021 |
| Session | June |
| Marks | 14 |
| Topic | Circular Motion 1 |
| Type | Angular speed conversion and basic circular motion quantities |
| Difficulty | Standard +0.3 This is a straightforward conical pendulum problem requiring vertical force resolution (T cos θ = mg) and simple algebraic manipulation. It's a standard FM1 exercise with clear setup and routine application of equilibrium conditions, making it slightly easier than average A-level difficulty. |
| Spec | 3.03r Friction: concept and vector form3.03v Motion on rough surface: including inclined planes6.02a Work done: concept and definition6.02b Calculate work: constant force, resolved component |
| B1 for each of two correct statements about the models. | |||||||||
| If commenting on the accuracy of (a), must emphasise that (a) is very inaccurate or at least quite inaccurate | |||||||||
| Do not allow e.g. | |||||||||
| - model (a) is not very effective | |||||||||
| - Neither model is accurate | |||||||||
| - (a) and (b) are not very accurate | |||||||||
| Clear comparison between the accuracy of the two models (must emphasise that (b) is fairly accurate or considerably more accurate than (a)), or other suitable distinct second comment | |||||||||
| Do not allow e.g. | |||||||||
| - model (b) is more accurate than model (a) | |||||||||
| - (b) is not accurate | |||||||||
| Do not allow statement claiming that resistance is proportional to speed, or to speed \({ } ^ { 2 }\) | |||||||||
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| 3 | (a) | \(T _ { 2 } \cos \theta = m _ { 2 } g\) \(T _ { 2 } = \frac { m _ { 2 } \times 9.8 } { 0.8 } = 12.25 m _ { 2 }\) |
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| 3 | (b) | (i) | \(\begin{aligned} | T _ { 2 } \cos \theta + m _ { 1 } g = T _ { 1 } \cos \theta | |||||||||||||
| T _ { 1 } = T _ { 2 } + \frac { 9.8 m _ { 1 } } { 0.8 } = | |||||||||||||||||
| \qquad 12.25 m _ { 2 } + 12.25 m _ { 1 } = \frac { 49 } { 4 } \left( m _ { 1 } + m _ { 2 } \right) \end{aligned}\) |
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| 3 | (b) | (ii) | \(\begin{aligned} | T _ { 1 } \sin \theta + T _ { 2 } \sin \theta = m _ { 1 } a | |||||||||||||
| 12.25 \left( m _ { 1 } + m _ { 2 } \right) \times 0.6 + 12.25 m _ { 2 } \times 0.6 = m _ { 1 } \times 0.6 \omega ^ { 2 } | |||||||||||||||||
| \omega ^ { 2 } = \frac { 7.35 m _ { 1 } + 14.7 m _ { 2 } } { 0.6 m _ { 1 } } = \frac { 49 \left( m _ { 1 } + 2 m _ { 2 } \right) } { 4 m _ { 1 } } \end{aligned}\) |
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| 3 | (c) | \(\begin{aligned} | \text { E.g } m _ { 1 } \gg m _ { 2 } \Rightarrow \frac { 2 m _ { 2 } } { m _ { 1 } } \approx 0 \text { or } \frac { 49 m _ { 2 } } { 4 m _ { 1 } } \approx 0 | |||||||||||
| \omega \approx \sqrt { \frac { 49 m } { 4 m } } = 3.5 \end{aligned}\) |
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| 3 | \multirow{3}{*}{(d)} |
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| [5] |
4 \\
2.4 \\
& \begin{tabular}{l}
B1 for each of two correct statements about the models. \\
If commenting on the accuracy of (a), must emphasise that (a) is very inaccurate or at least quite inaccurate \\
Do not allow e.g. \\
- model (a) is not very effective \\
- Neither model is accurate \\
- (a) and (b) are not very accurate \\
Clear comparison between the accuracy of the two models (must emphasise that (b) is fairly accurate or considerably more accurate than (a)), or other suitable distinct second comment \\
Do not allow e.g. \\
- model (b) is more accurate than model (a) \\
- (b) is not accurate \\
Do not allow statement claiming that resistance is proportional to speed, or to speed ${ } ^ { 2 }$ \\
& \begin{tabular}{l}
Suitable comments for (a): \\
- is very inaccurate \\
- predicted speed is nearly three times the actual value \\
- constant resistance is not a suitable model \\
- both models underestimate the resistance (as top speed is lower than expected) \\
For the linear model (b) \\
- is fairly accurate (but probably underestimates the resistance at higher speeds) \\
- resistance is not proportional to speed but is a much better model than constant resistance \\
\end{tabular} \\
\hline
\end{tabular}
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\hline
3 & (a) & & \(T _ { 2 } \cos \theta = m _ { 2 } g\) \(T _ { 2 } = \frac { m _ { 2 } \times 9.8 } { 0.8 } = 12.25 m _ { 2 }\) & \begin{tabular}{l}
M1 \\
A1 \\[0pt]
[2] \\
\end{tabular} & \begin{tabular}{l}
1.1a \\
1.1 \\
\end{tabular} & \begin{tabular}{l}
Resolving $T _ { 2 }$ vertically and balancing forces on $R$ \\
Do not allow extra forces present \\
Allow use of g, e.g. $\frac { 5 } { 4 } g m _ { 2 }$ \\
\end{tabular} & \begin{tabular}{l}
In this solution $\theta$ is the angle between $R P$ and $R A$ Sin may be seen instead if $\theta$ is measured horizontally. \\
Do not allow incomplete expressions e.g. $\frac { m _ { 2 } g } { \sin 53.13 }$ \\
\end{tabular} \\
\hline
3 & (b) & (i) & \(\begin{aligned} & T _ { 2 } \cos \theta + m _ { 1 } g = T _ { 1 } \cos \theta \\ & T _ { 1 } = T _ { 2 } + \frac { 9.8 m _ { 1 } } { 0.8 } = \\ & \qquad 12.25 m _ { 2 } + 12.25 m _ { 1 } = \frac { 49 } { 4 } \left( m _ { 1 } + m _ { 2 } \right) \end{aligned}\) & \begin{tabular}{l}
M1 \\
A1 \\[0pt]
[2] \\
\end{tabular} & \begin{tabular}{l}
3.1b \\
2.1 \\
\end{tabular} & \begin{tabular}{l}
Vertical forces on $P$; 3 terms including resolving of $T _ { 1 }$; allow sign error \\
AG Dividing by $\cos \theta ( = 0.8 )$, substituting their $T _ { 2 }$ and rearranging \\
Allow 12.25 instead of $\frac { 49 } { 4 }$ \\
\end{tabular} & \begin{tabular}{l}
Or $T _ { 1 } \cos \theta = m _ { 1 } g + m _ { 2 } g$ (equation for the system as a whole) \\
At least one intermediate step must be seen \\
\end{tabular} \\
\hline
3 & (b) & (ii) & \(\begin{aligned} & T _ { 1 } \sin \theta + T _ { 2 } \sin \theta = m _ { 1 } a \\ & 12.25 \left( m _ { 1 } + m _ { 2 } \right) \times 0.6 + 12.25 m _ { 2 } \times 0.6 = m _ { 1 } \times 0.6 \omega ^ { 2 } \\ & \omega ^ { 2 } = \frac { 7.35 m _ { 1 } + 14.7 m _ { 2 } } { 0.6 m _ { 1 } } = \frac { 49 \left( m _ { 1 } + 2 m _ { 2 } \right) } { 4 m _ { 1 } } \end{aligned}\) & \begin{tabular}{l}
M1 \\
M1 \\
A1 \\[0pt]
[3] \\
\end{tabular} & \begin{tabular}{l}
3.1b \\
1.1 \\
2.1 \\
\end{tabular} & \begin{tabular}{l}
NII horizontally for $P ; 3$ terms including resolving of tensions; allow sign error \\
Substituting for $T _ { 1 }$, their $T _ { 2 } , \sin \theta$ and $\alpha$ \\
AG Must see an intermediate step \\
\end{tabular} & \begin{tabular}{l}
Could see $a$ or $0.6 \omega ^ { 2 }$ or $\frac { v ^ { 2 } } { 0.6 }$ or $\omega ^ { 2 } r$ or $\frac { v ^ { 2 } } { r } \sin \theta = 0.6$ \\
must be $a = 0.6 \omega ^ { 2 }$ \\
\end{tabular} \\
\hline
& & & & & & & \\
\hline
\end{tabular}
\end{center}
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
3 & (c) & \(\begin{aligned} & \text { E.g } m _ { 1 } \gg m _ { 2 } \Rightarrow \frac { 2 m _ { 2 } } { m _ { 1 } } \approx 0 \text { or } \frac { 49 m _ { 2 } } { 4 m _ { 1 } } \approx 0 \\ & \omega \approx \sqrt { \frac { 49 m } { 4 m } } = 3.5 \end{aligned}\) & \begin{tabular}{l}
M1 A1 \\[0pt]
[2] \\
\end{tabular} & \begin{tabular}{l}
1.1 \\
1.1 \\
\end{tabular} & \begin{tabular}{l}
Allow argument such as if $m _ { 1 } \gg m _ { 2 }$ then $m _ { 1 } + 2 m _ { 2 } \approx m _ { 1 }$ \\
AG $m$ may be missing \\
SC1 for result following argument that $m _ { 2 }$ is negligible (by comparison with $m _ { 1 }$ ) without justification, or using trial values of $m _ { 1 }$ and $m _ { 2 }$ with $m _ { 1 } \gg m _ { 2 }$. \\
\end{tabular} & \begin{tabular}{l}
Do not allow the assumption that $m _ { 2 } = 0$ \\
If using trial values, $m _ { 1 }$ must be at least $70 \times m _ { 2 }$ to give $\omega = 3.5$ to 1 dp . \\
\end{tabular} \\
\hline
\end{tabular}
\end{center}
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
3 & \multirow{3}{*}{(d)} & \begin{tabular}{l}
\(v = r \omega = 0.6 \sqrt { \frac { 49 \times 2.5 + 98 \times 2.8 } { 4 \times 2.5 } }\) \\
Final energy $= 2.5 \times g \times 1$ \(\text { Initial } \mathrm { KE } = \frac { 1 } { 2 } \times 2.5 \times 0.6 ^ { 2 } \times \frac { 49 \times 2.5 + 98 \times 2.8 } { 4 \times 2.5 }\) \\
Initial PE $= 2.5 \times g \times 1.2 + 2.8 \times g \times 0.4$ \\
Energy loss $= 17.8605 + 40.376 - 24.5 = 33.7365$ \\
\end{tabular} & \begin{tabular}{l}
M1 \\
B1 \\
M1 \\
M1 \\
A1 \\
\end{tabular} & \begin{tabular}{l}
1.2 \\
1.1 \\
1.1 \\
1.1 \\
3.2a \\
\end{tabular} & \begin{tabular}{l}
Use of $v = r \omega$ with values for $m _ { 1 }$ and $m _ { 2 }$ \\
(Assuming zero PE level at 2 m below $A$; other values possible) \\
Do not allow use of $\omega = 3.5$ \\
oe with different zero PE level awrt 33.7 \\
\end{tabular} & \begin{tabular}{l}
( $v = 3.78 , v ^ { 2 } = 14.2884$ ) \\
NB $\omega = 6.3$ (24.5) \\
(17.8605) \\
(40.376) \\
\end{tabular} \\
\hline
& & \begin{tabular}{l}
Alternate method \(v = r \omega = 0.6 \sqrt { \frac { 49 \times 2.5 + 98 \times 2.8 } { 4 \times 2.5 } }\) \\
Initial KE $= \frac { 1 } { 2 } \times 2.5 \times 0.6 ^ { 2 } \times \frac { 49 \times 2.5 + 98 \times 2.8 } { 4 \times 2.5 }$ \\
$\triangle P E$ for $m _ { 1 } = \pm 2.5 \times 9.8 \times ( 0.8 - 1 )$ \\
$\triangle P E$ for $m _ { 2 } = \pm 2.8 \times 9.8 ( 1.6 - 2 )$ \\
Energy loss $= 17.8605 + 4.9 + 10.976$ \\
\end{tabular} & \begin{tabular}{l}
M1 \\
M1 \\
M1 \\
M1 \\
A1 \\
\end{tabular} & & \begin{tabular}{l}
Use of $v = r \omega$ with values for $m _ { 1 }$ and $m _ { 2 }$ \\
Or $- \triangle P E$ \(= 2.5 \times 9.8 \times 0.2 + 2.8 \times 9.8 \times 0.4\) \\
awrt 33.7 \\
\end{tabular} & \begin{tabular}{l}
( $v = 3.78 , v ^ { 2 } = 14.2884$ ) $\mathrm { NB } \omega = 6.3$ \\
(17.8605) \\
$( \pm 4.9 )$ \\
$( \pm 10.976 )$ \\
$( \pm 15.876 )$ \\
Or 15.876 + 17.8605 \\
\end{tabular} \\
\hline
& & & [5] & & & \\
\hline
\end{tabular}
\end{center}
\hfill \mbox{\textit{OCR FM1 AS 2021 Q4 [14]}}