A2 October 2020 Q7
7. A light elastic spring has natural length \(l\) and modulus of elasticity \(4mg\). A particle \(P\) of mass \(m\) is attached to one end of the spring. The other end of the spring is attached to a fixed point \(A\). The point \(B\) is vertically below \(A\) with \(AB = \dfrac{7}{4}l\). The particle \(P\) is released from rest at \(B\).
| Scheme | Marks | AO |
|---|---|---|
| Equation of motion about equilibrium position: | M1 | 3.1a |
| \(\dfrac{4mg(x + e)}{l} - mg = -m\ddot{x}\) | A1 | 1.1b |
| Extension \(e\) at equilibrium: \(\dfrac{4mge}{l} = mg,\quad \left(e = \dfrac{l}{4}\right)\) | B1 | 1.1b |
| \(\Rightarrow \dfrac{4gx}{l} = -\ddot{x},\ \left(\ddot{x} = -\dfrac{4g}{l}x\right)\) | M1 | 3.1a |
| This is of the form \(\ddot{x} = -\omega^2 x\), so SHM * | A1* | 3.2a |
| Period \(= \dfrac{2\pi}{\omega}\) | M1 | 3.4 |
| \(= 2\pi\sqrt{\dfrac{l}{4g}} = \pi\sqrt{\dfrac{l}{g}}\) * | A1* | 2.2a |
| (7) |
Notes
M1: Equation of motion about equilibrium position. Need all terms. Dimensionally correct. Allow with their \(e \ne 0\). Condone sign errors.
A1ft: Correct unsimplified equation with \(e\) or their \(e \ne 0\)
B1: Correct \(e\)
M1: Complete strategy e.g. use equation of motion and equilibrium position to form equation in \(x\).
A1*: Reach given conclusion from correct working
M1: Use the model to find periodic time (their \(\omega\))
A1*: Obtain given answer from correct working
Alternative (a)
| Scheme | Marks | AO |
|---|---|---|
| Equation of motion for extension \(x\): | M1 | 3.1a |
| \(\dfrac{4mgx}{l} - mg = -m\ddot{x},\quad \ddot{x} = -\dfrac{4g}{l}\left(x - \dfrac{l}{4}\right)\) | A1 | 1.1b |
| Use substitution \(X = x - \dfrac{l}{4}\) | B1 | 1.1b |
| \(\Rightarrow \dfrac{4gX}{l} = -\ddot{X},\ \left(\ddot{X} = -\dfrac{4g}{l}X\right)\) | M1 | 3.1a |
| This is of the form \(\ddot{X} = -\omega^2 X\), so SHM * | A1* | 3.2a |
| Period \(= \dfrac{2\pi}{\omega}\) | M1 | 3.4 |
| \(= 2\pi\sqrt{\dfrac{l}{4g}} = \pi\sqrt{\dfrac{l}{g}}\) * | A1* | 2.2a |
| (7) |
| Scheme | Marks | AO |
|---|---|---|
| Max speed \(= a\omega\ \left(= \dfrac{l}{2}\sqrt{\dfrac{4g}{l}}\right)\) | M1 | 3.4 |
| Max KE \(= \dfrac{1}{2}m\left(\dfrac{l}{2}\sqrt{\dfrac{4g}{l}}\right)^2\) | M1 | 1.2 |
| \(= \dfrac{1}{2}m\dfrac{l^2}{4}\times\dfrac{4g}{l} = \dfrac{1}{2}mlg\) | A1 | 1.1b |
| (3) |
Notes
M1: Use the model to find the max speed. Follow their \(\omega\)
M1: Follow their \(a\), \(\omega\)
A1: Correct simplified
| Scheme | Marks | AO |
|---|---|---|
| \(x = a\cos\omega t = \dfrac{l}{2}\cos\sqrt{\dfrac{4g}{l}}t\) | B1ft | 2.2a |
| Length of spring \(\lt l \Rightarrow x = -\dfrac{l}{4},\quad -\dfrac{l}{4} = \dfrac{l}{2}\cos\sqrt{\dfrac{4g}{l}}t\) | M1 | 1.1b |
| \(\Rightarrow \sqrt{\dfrac{4g}{l}}t = \dfrac{2\pi}{3}\) or \(\dfrac{4\pi}{3}\), \(\quad t = \dfrac{\pi}{3}\sqrt{\dfrac{l}{g}}\) or \(t = \dfrac{2\pi}{3}\sqrt{\dfrac{l}{g}}\) | A1 | 1.1b |
| Correct strategy | M1 | 3.1a |
| Length of time \(= \dfrac{2\pi}{3}\sqrt{\dfrac{l}{g}} - \dfrac{\pi}{3}\sqrt{\dfrac{l}{g}} = \dfrac{\pi}{3}\sqrt{\dfrac{l}{g}}\) | A1 | 2.2a |
| (5) | ||
| (15 marks) |
Notes
B1: Or equivalent. Follow their \(a\), \(\omega\)
M1: Follow their \(e\) and solve for \(t\)
A1: One correct solution. Accept \(t = \dfrac{2\pi}{3\omega}\), or \(t = \dfrac{4\pi}{3\omega}\)
M1: Complete strategy to find the required interval: select formula for displacement as function of time and use symmetry of motion to find the time interval.
A1: Correct answer from correct working