M3 June 2018 Q2
2. A light elastic string has natural length 1.2 m and modulus of elasticity \(\lambda\) newtons. One end of the string is attached to a fixed point \(O\). A particle of mass 0.5 kg is attached to the other end of the string. The particle is moving with constant angular speed \(\omega\) rad s\(^{-1}\) in a horizontal circle with the string stretched. The circle has radius 0.9 m and its centre is vertically below \(O\). The string is inclined at 60\(^\circ\) to the horizontal.
Find
| Scheme | Marks |
|---|---|
| \(T\cos 30 = 0.5g\) | M1A1 |
| ext \(= \dfrac{0.9}{\cos 60} - 1.2 = 0.6\) m | M1A1 |
| \(T = \dfrac{\lambda x}{l} = \dfrac{\lambda \times \text{"}0.6\text{"}}{1.2}\) | M1 |
| \(\dfrac{\lambda}{2} \times \dfrac{\sqrt{3}}{2} = \dfrac{g}{2} \qquad \lambda = \dfrac{2g}{\sqrt{3}} = 11.31\ldots = 11.3\) or 11 | dM1A1 |
| (7) |
Notes
NB Here and in qu 7 penalise only once for decimal answers with more than 3 sf
M1 Resolve vertically. Tension must be resolved (cos or sin allowed), weight not resolved.
A1 Correct equation
M1 Use trigonometry to calculate the extension. Must not use an erroneous 1.2 m on the vertical (Ignore it on their diagram)
A1 Correct extension
M1 Use Hooke's Law with their extension.
dM1 Eliminate \(T\) and solve to \(\lambda = \ldots\) Depends on first and third M marks above
A1 Correct answer, 2 or 3 significant figures
| Scheme | Marks |
|---|---|
| \(T\cos 60 = 0.9m\omega^2\) | |
| \(T\cos 60 = 0.9 \times 0.5\omega^2\) | M1A1 |
| \(\dfrac{2g}{\sqrt{3}} \times \dfrac{0.6}{1.2} \times \dfrac{1}{2} = 0.9 \times 0.5\omega^2\) or \(\dfrac{0.5g}{\cos 30} \times \cos 60 = 0.9 \times 0.5\omega^2\) \(\omega = 2.507\ldots = 2.5\) or 2.51 | dM1A1 |
| (4) | |
| (11 marks) |
Notes
M1 Equation of motion along the radius. \(T\) must be resolved (cos or sin), acceleration in either form. \(m\) or 0.5
A1 Correct equation, mass to be 0.5 here or later and acceleration \(0.9\omega^2\)
dM1 Eliminate \(T\) by using Hooke's Law with their \(\lambda\) (from (a)) or using their vertical equation from (a) and solve to \(\omega = \ldots\) Depends on the first M in (b)
A1 Correct value for \(\omega\). Must be 2 or 3 sig figs
NB: Full marks can be awarded in (b) if use of their \(T\) (obtained from a correct use of HL) and \(\lambda\) leads to the correct value.