A2 June 2025 Q6
6.
A smooth plane is inclined at an angle \(\theta\) to horizontal ground, where \(\sin\theta = \dfrac{1}{3}\)
The points \(A\), \(B\) and \(C\) lie on a line of greatest slope of the plane, with \(B\) between \(A\) and \(C\), with \(A\) above \(B\) and \(AC = 3a\), as shown in Figure 3.
A light elastic string of natural length \(2a\) has one end attached to the fixed point \(A\).
The other end of the string is attached to a parcel \(P\) of mass \(m\).
The modulus of elasticity of the string is \(\dfrac{8}{3}mg\)
The parcel rests in equilibrium on the plane at the point \(B\).
The parcel is modelled as a particle.
The parcel is now held at \(C\) and released from rest.
| Scheme | Marks | AO |
|---|---|---|
| Correct use of Hooke’s law \(T = \dfrac{\frac{8}{3}mge}{2a}\) or \(T = \dfrac{\frac{8}{3}mg(AB - 2a)}{2a}\) | M1 | 3.4 |
| Resolve parallel to slope | M1 | 3.1b |
| \(T = mg\sin\theta \;\left(= \dfrac{1}{3}mg\right)\) | A1 | 1.1b |
| \(\Rightarrow e = \dfrac{2a}{8}\left(= \dfrac{a}{4}\right)\), \(AB = \dfrac{9}{4}a\) * OR \(\Rightarrow AB - 2a = \dfrac{2a}{8}\left(= \dfrac{a}{4}\right)\), \(AB = \dfrac{9}{4}a\) * | A1* | 2.2a |
| (4) |
Notes
M1: Correct use of Hooke’s law with \(\dfrac{8mg}{3}\) and \(2a\) substituted.
M1: Resolve parallel to the slope with all required terms and no extras. Dimensionally correct. Weight must be resolved but condone sin/cos confusion. \(T\) does not need to be replaced.
A1: Correct unsimplified equilibrium equation, \(T\) does not need to be replaced.
A1*: A complete and correct method using Hooke’s law with the parallel equilibrium equation to obtain the given answer. There must be at least one line of working between the initial equations and the given answer. Must see ‘\(AB = \ldots\)’ Accept fractions \(\dfrac{9a}{4}\) or \(\dfrac{9}{4}a\).
| Scheme | Marks | AO |
|---|---|---|
| Loss in EPE | M1 | 3.3 |
| \(\dfrac{\lambda(3a - 2a)^2}{2(2a)} - \dfrac{\lambda\left(\dfrac{9a}{4} - 2a\right)^2}{2(2a)}\) | A1 | 1.1b |
| \(= \dfrac{5mga}{8}\) | A1 | 1.1b |
| (3) |
Notes
M1: Correct method for difference in EPE at \(B\) and \(C\), allow either way round. Dimensionally correct and of the correct structure. No need to substitute \(\lambda\). For M mark, condone denominator of \(2a\). Must use extensions \((3a - 2a)\) and \(\left(\dfrac{9a}{4} - 2a\right).\)
A1: Correct unsimplified expression for change in EPE. Allow \(\pm\)
A1: Correct answer, o.e. ISW. Accept \(0.625mga\) and \(0.63mga\). Must be positive but allow a negative expression to change to a positive expression without justification.
| Scheme | Marks | AO |
|---|---|---|
| Energy equation | M1 | 3.1b |
| \(\text{“}\dfrac{5mga}{8}\text{”} = mg\left(3a - \dfrac{9a}{4}\right)\sin\theta + \dfrac{1}{2}mv^2\) | A1ft | 1.1b |
| \(v = \sqrt{\dfrac{3ga}{4}}\) | A1 | 1.1b |
| (3) | ||
| (10 marks) |
Notes
M1: Use of conservation of energy principle from \(C\) to \(B\) to form an equation with all terms of the correct structure and dimensionally correct. Condone sign errors. Condone sin/cos confusion on vertical height.
For GPE, \(mg\left(3a - \dfrac{9a}{4}\right)\sin\theta\) o.e for example \(mg\dfrac{3a}{4}\sin\theta,\ \ mg\dfrac{3a}{4}\left(\dfrac{1}{3}\right),\ \ mg\dfrac{a}{4}\)
For EPE change, may use their answer from (b) if dimensionally correct (of the form \(kma\) where \(k\) is a constant) or start again with 2 EPE terms:
\(\dfrac{\frac{8mg}{3}(3a - 2a)^2}{2(2a)}\) and \(\dfrac{\frac{8mg}{3}\left(\frac{9}{4}a - 2a\right)^2}{2(2a)}\).
For M mark, condone denominator of \(2a\).
(Corrected from the printed mark scheme: the second EPE term is printed with \(\left(3a - \frac{9}{4}a\right)^2\); the extension at \(B\) is \(\frac{9}{4}a - 2a\).)
A1ft: Correct unsimplified equation. Terms must be correct when the equation is formed but follow their answer to (b) for EPE if used.
A1: Correct answer in terms of \(\sqrt{ag}\) ISW. Accept eg \(\dfrac{1}{2}\sqrt{3ag}\), \(0.87\sqrt{ag}\) or better.
N.B. If the final mark in (b) is A0 due to substituting \(g = 9.8\ \text{m s}^{-2}\), do not penalise again for the same reason here.