A2 June 2019 Q7
7. A particle \(P\), of mass \(m\), is attached to one end of a light elastic spring of natural length \(a\) and modulus of elasticity \(kmg\).
The other end of the spring is attached to a fixed point \(O\) on a ceiling.
The point \(A\) is vertically below \(O\) such that \(OA = 3a\)
The point \(B\) is vertically below \(O\) such that \(OB = \dfrac{1}{2}a\)
The particle is held at rest at \(A\), then released and first comes to instantaneous rest at the point \(B\).
| Scheme | Marks | AO |
|---|---|---|
| From \(A\) to \(B\) EPE lost = GPE gained | M1 | 2.1 |
| \(\dfrac{kmg \times 4a^2}{2a} - \dfrac{kmg \times \frac{a^2}{4}}{2a} = mg \times \dfrac{5a}{2}\) | A1 | 1.1b |
| \(k = \dfrac{4}{3}\) * | A1* | 2.2a |
| (3) |
Notes
M1: Use conservation of energy with EPE \(= \dfrac{\lambda x^2}{2a}\). (Condone EPE \(= \dfrac{\lambda x^2}{a}\) here). All three terms required. Must be dimensionally correct. Condone sign errors.
A1: Correct unsimplified equation in \(k\)
A1*: Derive given result from correct working.
| Scheme | Marks | AO |
|---|---|---|
| At \(A\), equation of motion: | M1 | 3.1a |
| \((T - mg =)\ \dfrac{4mg \times 2a}{3a} - mg = m \times \text{acceleration}\) | A1 | 1.1b |
| \(\Rightarrow \text{acceleration} = \dfrac{5g}{3}\) | A1 | 1.1b |
| (3) |
Notes
M1: Use \(T = \dfrac{\lambda x}{a}\) and N2L to form equation of motion. All terms required. Dimensionally correct. Condone sign errors
A1: Correct unsimplified equation
A1: Correct only ISW. Condone \(1.7g\) or better. Accept + / -
| Scheme | Marks | AO |
|---|---|---|
| Max speed at equilibrium position | M1 | 3.1a |
| \(\dfrac{4mge}{3a} = mg\), \(e = \dfrac{3a}{4}\) | A1 | 1.1b |
| Forms equation using conservation of energy | M1 | 3.1a |
| \(\dfrac{4mg \times 4a^2}{3 \times 2a} = \dfrac{4mg \times \frac{9a^2}{16}}{3 \times 2a} + \dfrac{1}{2}mv^2 + mg \times \dfrac{5a}{4}\) | A1ft A1ft | 1.1b 1.1b |
| \(v = \dfrac{5}{2}\sqrt{\dfrac{ga}{3}}\) | A1 | 1.1b |
| (6) | ||
| (12 marks) |
Notes
M1: Maximum speed at equilibrium seen or implied, and correct method to find \(e\)
A1: Correct \(e\)
Alternative: form energy equation for movement through a height of \(h\) and differentiate \(v^2\) wrt \(h\) to find \(h\) for max \(v\) M1
\(h = \dfrac{5a}{4}\) A1
M1: Form energy equation for movement from \(A\) to equilibrium position. Need all 4 terms. Correct form for EPE. Dimensionally correct. Condone sign errors. Allow in \(a\), \(g\) and \(e\) (with \(e\) defined)
A1ft A1ft: Unsimplified equation in their \(e\) with at most one error
Correct unsimplified equation (using their \(e\)) for \(v\)
A1: Any equivalent form. Accept \(1.44\sqrt{ag}\) or \(1.4\sqrt{ag}\)
SHM is not on this specification, but you might see some candidates using it. See below for SHM alternative for parts (b) and (c)
SHM alternative for parts (b) and (c)
| Scheme | Marks | AO |
|---|---|---|
| At equilibrium, \(\dfrac{4mge}{3a} = mg\), \(e = \dfrac{3a}{4}\) Equation of motion: \(mg - \dfrac{4mg}{3a}(e + x) = m\ddot{x}\), so \(\ddot{x} = -\dfrac{4g}{3a}x\) Hence SHM They need to start by showing that they have SHM in order to justify using the standard results. No marks scored for this at this stage. | ||
| (b) Use of \(x = \dfrac{5a}{4}\) and their \(\omega^2\) Substitute to find acceleration | M1 | |
| \(\ddot{x} = -\dfrac{4g}{3a} \times \dfrac{5a}{4} = -\dfrac{5g}{3}\), \(|\ddot{x}| = \dfrac{5g}{3}\) Correct only ISW. Condone \(1.7g\) or better | A1 | |
| (2) | ||
| (c) \(\dfrac{4mge}{3a} = mg\), | M1 | |
| \(e = \dfrac{3a}{4}\) This work now scores the two marks provided it is used in part (c) | A1 | |
| Use of \(v_{\max} = \omega a\) Correct method to find max \(v\) | M1 | |
| \(v_{\max} = \sqrt{\dfrac{4g}{3a}} \times \dfrac{5a}{4}\) Follow their \(e\) and \(\omega\) | A2ft | |
| \(v_{\max} = \dfrac{5}{2}\sqrt{\dfrac{ga}{3}}\) Any equivalent form. Accept \(1.44\sqrt{ag}\) or \(1.4\sqrt{ag}\) | A1 | |
| (6) |