M3 June 2013 Q6
6.
The points \(A\) and \(B\) are 3.75 m apart on a smooth horizontal floor. A particle \(P\) has mass 0.8 kg. One end of a light elastic spring, of natural length 1.5 m and modulus of elasticity 24 N, is attached to \(P\) and the other end is attached to \(A\). The ends of another light elastic spring, of natural length 0.75 m and modulus of elasticity 18 N, are attached to \(P\) and \(B\). The particle \(P\) rests in equilibrium at the point \(O\), where \(AOB\) is a straight line, as shown in Figure 5.
The point \(C\) lies on the straight line \(AOB\) between \(O\) and \(B\). The particle \(P\) is held at \(C\) and released from rest.
The maximum speed of \(P\) is \(\sqrt{2}\) m s\(^{-1}\).
| Scheme | Marks |
|---|---|
| \(\dfrac{24e}{1.5} = \dfrac{18(1.5 - e)}{0.75}\) | M1A1 |
| \(16e = 36 - 24e\) | |
| \(e = 0.9\) | A1 |
| \(AO = 2.4\) m* | A1ft |
| (4) |
Notes
M1 for using Hooke's Law for each string, equating the two tensions and solving to find the extension in either string. The extensions should add to 1.5. The formula for Hooke's law must be correct, either shown explicitly in its general form or implicitly by the substitution.
A1 for a correct equation
A1 for \(e = 0.9\)
A1cso for 2.4 (m) *
Alternative: Find the ratio of the two extensions and divide 1.5 m in that ratio.
M1 complete method A1 correct ratio A1 extension in \(AO\) A1 2.4 (m)
| Scheme | Marks |
|---|---|
| \(\dfrac{18(0.6 - x)}{0.75} - \dfrac{24(0.9 + x)}{1.5} = m\ddot{x}\) or \(0.8\ddot{x}\) | M1A1A1 |
| \(14.4 - 24x - 14.4 - 16x = m\ddot{x}\) or \(0.8\ddot{x}\) | |
| \(\ddot{x} = -\dfrac{40x}{0.8\text{ or }m}\ \ \ (= -50x)\ \ \ \therefore\) SHM | M1depA1 |
| (5) |
Notes
M1 for an equation of motion for \(P\). There must be a difference of two tensions. The acceleration can be \(a\) or \(\ddot{x}\) here and \(x\) should be measured from the equilibrium position (\(O\)) unless a suitable substitution is made later. Mass can be \(m\) or 0.8
A1,A1 for \(\dfrac{18(0.6 - x)}{0.75} - \dfrac{24(0.9 + x)}{1.5} = m\ddot{x}\) or \(0.8\ddot{x}\) or \(a\) instead of \(\ddot{x}\) Give A1A1 if the equation is completely correct and A1 if only one error. Note that if the difference of the tensions is the wrong way round, this is one error
M1dep for simplifying to \(\ddot{x} = \mathrm{f}(x)\) Must be \(\ddot{x}\) now.
A1 for \(\ddot{x} = -\dfrac{40x}{0.8\text{ or }m}\ \ (= -50x)\) and the conclusion (ie \(\therefore\) SHM)
| Scheme | Marks |
|---|---|
| \(\ddot{x} = -50x \Rightarrow \omega = \sqrt{50}\) or \(5\sqrt{2}\) | B1 |
| max. speed \(= \sqrt{2} \Rightarrow a \times 5\sqrt{2} = \sqrt{2}\) | M1 |
| \(a = \dfrac{1}{5}\) | A1 |
| \(-0.1 = 0.2\cos\left(5\sqrt{2}\right)t\) | M1 |
| \(t = \dfrac{1}{5\sqrt{2}}\cos^{-1}\left(-\dfrac{1}{2}\right)\) | |
| \(t = \dfrac{1}{5\sqrt{2}} \times \dfrac{2\pi}{3} = \dfrac{\pi\sqrt{2}}{15}\) or 0.296 s (0.2961...) Accept 0.30, or better | A1 |
| (5) | |
| (14 marks) |
Notes
B1 for \(\omega = \sqrt{50}\) or \(5\sqrt{2}\) need not be shown explicitly
M1 for using max speed \(= a\omega = \sqrt{2}\) with their \(\omega\)
A1 for \(a = \dfrac{1}{5}\)
M1 for using \(x = a\cos\omega t\) with their \(\omega\) and \(a\) and \(x = \pm(0.3 - a)\) or \(x = a\sin\omega t\) provided the work is completed by adding a quarter of their period is added to the time to complete the method.
A1cao for \(t = \dfrac{\pi\sqrt{2}}{15}\) or 0.296 s (0.2961...) Accept 0.30 or better