4
2.4
\end{tabular} &
| 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 \({ } ^ { 2 }\) |
&
| 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 |
\hline
\end{tabular}
\end{center}
| 3 | (a) | | \(T _ { 2 } \cos \theta = m _ { 2 } g\) \(T _ { 2 } = \frac { m _ { 2 } \times 9.8 } { 0.8 } = 12.25 m _ { 2 }\) | | | | Resolving \(T _ { 2 }\) vertically and balancing forces on \(R\) | | Do not allow extra forces present | | Allow use of g, e.g. \(\frac { 5 } { 4 } g m _ { 2 }\) |
| | 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. \(\frac { m _ { 2 } g } { \sin 53.13 }\) |
|
| 3 | (b) | (i) | \(\begin{aligned} | T _ { 2 } \cos \theta + m _ { 1 } g = T _ { 1 } \cos \theta |
| T _ { 1 } = T _ { 2 } + \frac { 9.8 m _ { 1 } } { 0.8 } = |
| \qquad 12.25 m _ { 2 } + 12.25 m _ { 1 } = \frac { 49 } { 4 } \left( m _ { 1 } + m _ { 2 } \right) \end{aligned}\) | | | | 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 \(\frac { 49 } { 4 }\) |
| | 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 |
|
| 3 | (b) | (ii) | \(\begin{aligned} | T _ { 1 } \sin \theta + T _ { 2 } \sin \theta = m _ { 1 } a |
| 12.25 \left( m _ { 1 } + m _ { 2 } \right) \times 0.6 + 12.25 m _ { 2 } \times 0.6 = m _ { 1 } \times 0.6 \omega ^ { 2 } |
| \omega ^ { 2 } = \frac { 7.35 m _ { 1 } + 14.7 m _ { 2 } } { 0.6 m _ { 1 } } = \frac { 49 \left( m _ { 1 } + 2 m _ { 2 } \right) } { 4 m _ { 1 } } \end{aligned}\) | | | | 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 |
| | 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 \(a = 0.6 \omega ^ { 2 }\) |
|
| | | | | | | |
| 3 | (c) | \(\begin{aligned} | \text { E.g } m _ { 1 } \gg m _ { 2 } \Rightarrow \frac { 2 m _ { 2 } } { m _ { 1 } } \approx 0 \text { or } \frac { 49 m _ { 2 } } { 4 m _ { 1 } } \approx 0 |
| \omega \approx \sqrt { \frac { 49 m } { 4 m } } = 3.5 \end{aligned}\) | | | | 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 \(m _ { 1 } \gg m _ { 2 }\). |
| | 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 . |
|
| 3 | \multirow{3}{*}{(d)} | | \(v = r \omega = 0.6 \sqrt { \frac { 49 \times 2.5 + 98 \times 2.8 } { 4 \times 2.5 } }\) | | 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 \(= 17.8605 + 40.376 - 24.5 = 33.7365\) |
| | | | 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 |
| | ( \(v = 3.78 , v ^ { 2 } = 14.2884\) ) | | NB \(\omega = 6.3\) (24.5) | | (17.8605) | | (40.376) |
|
| | | 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 \(= 17.8605 + 4.9 + 10.976\) |
| | | | 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 |
| | ( \(v = 3.78 , v ^ { 2 } = 14.2884\) ) \(\mathrm { NB } \omega = 6.3\) | | (17.8605) | | \(( \pm 4.9 )\) | | \(( \pm 10.976 )\) | | \(( \pm 15.876 )\) | | Or 15.876 + 17.8605 |
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| | | [5] | | | |