June 2022 Paper 2 Q18
18 An object, \(O\), of mass \(m\) kilograms is hanging from a ceiling by two light, inelastic strings of different lengths.
The shorter string, of length 0.8 metres, is fixed to the ceiling at \(A\).
The longer string, of length 1.2 metres, is fixed to the ceiling at \(B\).
This object hangs 0.6 metres directly below the ceiling as shown in the diagram.
(a) Show that the tension in the shorter string is over 30% more than the tension in the longer string. [4 marks]
(b) The tension in the longer string is known to be \(2g\) newtons.
Find the value of \(m\). [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains both \(\sin^{-1}\left(\dfrac{0.6}{0.8}\right)\) and \(\sin^{-1}\left(\dfrac{0.6}{1.2}\right)\) OE Accept complementary angles or exact values \(\dfrac{\sqrt{3}}{2}\) and \(\dfrac{\sqrt{7}}{4}\) | B1 | 1.1b |
| Resolves forces horizontally to form equilibrium equation, one component correct Or Uses a triangle of forces and applies the sine rule | M1 | 3.3 |
| Obtains correct equation with angles substituted | A1 | 1.1b |
| Rearranges the correct equation to show that \(T_{OA} = k\,T_{OB}\) OE where \(1.305 \leqslant k \leqslant 1.325\) Completes argument to conclude that the tension in the shorter string is over 30% more than the tension in the longer string | R1 | 2.1 |
| (4) |
Typical solution
Angle for \(OA = \sin^{-1}\left(\dfrac{0.6}{0.8}\right) = 48.59^\circ\)
Angle for \(OB = \sin^{-1}\left(\dfrac{0.6}{1.2}\right) = 30^\circ\)
\[T_{OA}\cos A = T_{OB}\cos B\]\[T_{OA} = T_{OB}\frac{\cos B}{\cos A}\]\[T_{OA} = T_{OB}\frac{\cos 30}{\cos 48.59}\]\[\therefore T_{OA} = 1.309T_{OB}\]\[\therefore T_{OA} \gt 1.3T_{OB}\]So, the tension in the shorter string is more than 30% greater than the tension in the longer string
| Scheme | Marks | AO |
|---|---|---|
| Obtains \(T_{OA} = 2g \times\) their ratio | B1 | 1.1b |
| Resolves forces vertically to form a three-term equilibrium equation, with at least two terms correct Or Uses a triangle of forces and applies the sine rule | M1 | 3.3 |
| Forms fully correct equation of forces in equilibrium This mark can be awarded for \(mg = T_{OA}\sin A + T_{OB}\sin B\) | A1 | 1.1b |
| Substitutes their \(T_{OA}\) and \(T_{OB}\) and correct values for angles into the correct equation and obtains AWRT \(m\) = 3 Might come from 2.96.. FT their ratio from part (a) provided their \(m\) = AWRT 3 | A1F | 3.4 |
| (4) | ||
| (8 marks) |