| 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. \(\frac { m _ { 2 } g } { \sin 53.13 }\) |
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| 3 | (b) | (i) | \(\begin{aligned} | T _ { 2 } \cos \theta + m _ { 1 } g = T _ { 1 } \cos \theta |
| T _ { 1 } = T _ { 2 } + \frac { 9.8 m _ { 1 } } { 0.8 } = |
| \qquad 12.25 m _ { 2 } + 12.25 m _ { 1 } = \frac { 49 } { 4 } \left( m _ { 1 } + m _ { 2 } \right) \end{aligned}\) | | | | 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 |
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| 3 | (b) | (ii) | \(\begin{aligned} | T _ { 1 } \sin \theta + T _ { 2 } \sin \theta = m _ { 1 } a |
| 12.25 \left( m _ { 1 } + m _ { 2 } \right) \times 0.6 + 12.25 m _ { 2 } \times 0.6 = m _ { 1 } \times 0.6 \omega ^ { 2 } |
| \omega ^ { 2 } = \frac { 7.35 m _ { 1 } + 14.7 m _ { 2 } } { 0.6 m _ { 1 } } = \frac { 49 \left( m _ { 1 } + 2 m _ { 2 } \right) } { 4 m _ { 1 } } \end{aligned}\) | | | | 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 \(a = 0.6 \omega ^ { 2 }\) |
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