M3 June 2017 Q2
2.
A particle \(P\) of mass \(m\) is attached to one end of a light inextensible string. The other end of the string is attached to a fixed point \(A\). The particle moves in a horizontal circle with constant angular speed \(\sqrt{58.8}\) rad s\(^{-1}\). The centre \(O\) of the circle is vertically below \(A\) and the string makes a constant angle \(\theta^\circ\) with the downward vertical, as shown in Figure 2.
Given that the tension in the string is \(1.2mg\), find
| Scheme | Marks |
|---|---|
| \(1.2mg\cos\theta = mg\) or \(T\cos\theta = mg\) | M1A1 |
| (i) \(\cos\theta^\circ = \dfrac{1}{1.2} \quad \theta^\circ = \cos^{-1}\dfrac{1}{1.2},\ \ \theta = 33.55\ldots\) (accept 34, 33.6 or better) | A1 |
| \(1.2mg\sin\theta = mr\omega^2\) or \(T\sin\theta = mr\omega^2\) | M1A1 |
| \(1.2mg\sin\theta = m \times l\sin\theta\,\omega^2\) | A1 |
| (ii) \(1.2mg = 58.8lm \ \Rightarrow\ l = \dfrac{1.2 \times 9.8}{58.8} = 0.2\ (\text{m})\) | dM1A1 |
| (8) | |
| (8 marks) |
Notes
M1 Resolve vertically. Tension to be resolved, weight not resolved.
A1 Fully correct equation with substitution for \(T\) made.
(i)A1 Correct value of \(\theta\) Min 2 sf Use of radians scores A0
M1 Attempt NL2 horizontally. Tension must be resolved, acceleration can be in either form.
A1 LHS correct, RHS can be \(mr\omega^2\) or \(m\dfrac{v^2}{r}\) here. \(T\) substituted now or later
A1 RHS correct, acceleration as shown. \(\sin\theta\) may be numerical \(\dfrac{\sqrt{11}}{6}\) or 0.5527... (min 3 sf) or a numerical value for \(r\) (\(\dfrac{\sqrt{11}}{30}\) or 0.110...) may be seen.
dM1 Use the above equation to obtain a numerical value for \(l\). Depends on the second M mark
(ii)A1 Correct value of \(l\). Accept 0.2, 0.20, 0.200. Exceptionally allow \(\dfrac{1}{5}\) here.
(Corrected from the printed mark scheme: the alternative horizontal equation is printed as \(T\cos\theta = mr\omega^2\).)