M3 June 2018 Q4
4. One end of a light elastic string, of modulus of elasticity \(2mg\) and natural length \(l\), is fixed to a point \(O\) on a rough plane. The plane is inclined at angle \(\alpha\) to the horizontal, where \(\sin\alpha = \dfrac{3}{5}\). The other end of the string is attached to a particle \(P\) of mass \(m\) which is held at rest on the plane at the point \(O\). The coefficient of friction between \(P\) and the plane is \(\dfrac{1}{4}\). The particle is released from rest and slides down the plane, coming to instantaneous rest at the point \(A\), where \(OA = kl\).
Given that \(k > 1\), find, to 3 significant figures, the value of \(k\). (7)
| Scheme | Marks |
|---|---|
| Elastic energy \(= \dfrac{1}{2} \times 2mg\dfrac{x^2}{l}\) | B1 |
| Work done by friction = \((l + x)\mu mg\cos\alpha\) | B1 |
| Energy from release: \((l + x)\mu mg\cos\alpha + \dfrac{1}{2} \times 2mg\dfrac{x^2}{l} = (l + x)mg\sin\alpha\) | M1A1ft work and EE |
| \(\dfrac{1}{4} \times \dfrac{4}{5}(l + x) + \dfrac{x^2}{l} = \dfrac{3}{5}(l + x)\) | |
| \(l^2 + lx + 5x^2 = 3l^2 + 3lx\) | |
| \(5x^2 - 2lx - 2l^2 = 0\) | |
| \(x = 0.863\ldots l\) | dM1A1 |
| \(k = 1.86\) | A1 |
| (7) | |
| (7 marks) |
Notes
B1 Correct elastic energy
B1 Correct work done by friction
M1 Attempt a work-energy equation. Must have 3 terms: work done by friction, elastic energy, GPE.
EPE term must be of the form \(= k\lambda\dfrac{x^2}{l}\) \(k = 1, 2\) or \(\dfrac{1}{2}\)
Work done term must be of the form distance \(\times\ \mu mg\cos\) or \(\sin\alpha\)
A1ft Correct equation, ft their EPE and work terms
dM1 Solve their 3 term quadratic to obtain a value for the extension as a multiple of \(l\). Award if correct answer follows a correct quadratic. If the quadratic is incorrect award only if working shown (ie general formula shown explicitly and used or by implication through substitution correct for their equation, pos root only needed)
A1 Correct extension decimal or exact
A1 Complete by adding 1 to the numerical multiple of \(l\) Must be 3 significant figures.
(Corrected from the printed mark scheme: the fifth line is printed as \(l^2 + 4lx + 5x^2 = 3l^2 + 3lx\).)
ALT
Using dist moved: \(kl\) or \(x\)
EPE \(= \dfrac{1}{2} \times 2mg\dfrac{(kl - l)^2}{l}\), WD \(= kl\mu mg\cos\alpha\) B1, B1
\(kl\mu mg\cos\alpha + \dfrac{1}{2} \times 2mg\dfrac{(kl - l)^2}{l} = klmg\sin\alpha\) M1A1ft
\(5k^2 - 12k + 5 = 0\)
\(k = 1.86\) M1A2 (Give A1 for correct ans with more than 3 sf or exact)