A2 June 2022 Q8
8. Throughout this question, use \(g = 10\ \text{m s}^{-2}\)
A light elastic string has natural length 1.25 m and modulus of elasticity 25 N.
A particle \(P\) of mass 0.5 kg is attached to one end of the string. The other end of the string is attached to a fixed point \(A\). Particle \(P\) hangs freely in equilibrium with \(P\) vertically below \(A\)
The particle is then pulled vertically down to a point \(B\) and released from rest.
The maximum kinetic energy of \(P\) during the subsequent motion is 2.5 J.
The particle returns to \(B\) for the first time \(T\) seconds after it was released from rest at \(B\)
| Scheme | Marks | AO |
|---|---|---|
| At equilibrium: \(0.5g = \dfrac{25e}{1.25},\quad e = \dfrac{0.5\times 10\times 1.25}{25} = \dfrac{1}{4}\) | B1 | 3.3 |
| For taut string, when distance \(x\) from equilibrium, equation of motion | M1 | 2.1 |
| Alternative for M1: Conservation of energy using a known point (\(E\) or \(B\)) and a general position: From \(E\): \(\dfrac{25e^2}{2\times 1.25} + KE(\text{constant} \ne 0) + 0.5gx = \dfrac{25(e + x)^2}{2\times 1.25} + \dfrac{1}{2}0.5v^2 + 0(\text{GPE})\) and differentiate wrt \(x\) for M1 \(\Rightarrow 0.5g = \dfrac{25(e + x)}{1.25} + \dfrac{1}{2}v\dfrac{\mathrm{d}v}{\mathrm{d}x}\) | ||
| \(\dfrac{25(e + x)}{1.25} - 0.5g = -0.5\ddot{x}\) | A1ft | 1.1b |
| \(\ddot{x} = -40x\) hence SHM* | A1* | 2.2a |
| Periodic time: | M1 | 3.4 |
| \(T = \dfrac{2\pi}{\sqrt{40}} = \dfrac{\pi}{\sqrt{10}}\) * | A1* | 2.2a |
| (6) |
Notes
B1: Correct only Award if see \(mg = \dfrac{\lambda e}{l}\) used in their equation of motion
M1: Equation of motion with \(x\) measured from the equilibrium position. Need all terms and dimensionally correct. Allow with their \(e \ne 0\). Condone sign errors. Allow with \(a\) in place of \(\ddot{x}\)
A1ft: Correct unsimplified equation with their \(e\) or \(e \ne 0\).
Could have the negative of the whole equation
or \(x\) replaced with \(-x\) throughout
A1*: Reach given conclusion from correct working.
Condone correct conclusion without explanation
M1: Use the model to find the periodic time: \(T = \dfrac{2\pi}{\omega}\)
From an equation of the form \(\ddot{x} = -\omega^2 x\)
A1*: Obtain given answer from correct working throughout.
Available if only error is not to conclude SHM
| Scheme | Marks | AO |
|---|---|---|
| Max KE \(= 2.5 = \dfrac{1}{2}\times\dfrac{1}{2}\times\max v^2 \qquad \Rightarrow \max v^2 = 10\) | B1 | 1.2 |
| Max speed \(= a\omega\): \(\sqrt{10} = a\sqrt{40}\) | M1 | 3.4 |
| \(AB = 1.25 + \dfrac{1}{4} + \dfrac{1}{2} = 2\,(\text{m})\) * | A1* | 1.1b |
| (3) |
Notes
B1: Use the KE to find \(\max v\) or \(\max v^2\)
M1: Use the model to find the amplitude of the motion
A1*: Obtain the given answer from correct working.
Alternative (b)
| Scheme | Marks | AO |
|---|---|---|
| Energy: \(\dfrac{25e^2}{2.5} + 2.5 + 0.5ga = \dfrac{25(e + a)^2}{2.5}\) | B1 | |
| Solve for \(a\) | M1 | |
| \(AB = 1.25 + \dfrac{1}{4} + \dfrac{1}{2} = 2\,(\text{m})\) * | A1* | 1.1b |
| (3) |
B1: Using correct \(\lambda\) and \(l\) and \(e\) or their \(e\)
M1: Requires an energy equation with all the right terms
A1*: Obtain the given answer from correct working.
| Scheme | Marks | AO |
|---|---|---|
| \(a = 0.5,\ x = 0.5\cos\sqrt{40}t\) | B1 | 2.2a |
| \(-0.25 = 0.5\cos\sqrt{40}t \quad \Rightarrow t = 0.3311\ldots\) | M1 | 3.1a |
| \(v^2 = 40\left(0.5^2 - 0.25^2\right) = \dfrac{15}{2}\) | M1 | 3.4 |
| Total time \(= 2\times 0.3311\ldots + \dfrac{2\times\sqrt{7.5}}{10}\) | DM1 | 3.1a |
| \(= 1.2\,(\text{s})\) or better | A1 | 2.2a |
| (5) | ||
| (14 marks) |
Notes
B1: Correct equation for SHM seen or implied
M1: Find the time until the string goes slack
If working from \(x = 0.5\sin\sqrt{40}t\) need \(\dfrac{T}{4} + \dfrac{1}{\sqrt{40}}\sin^{-1}\dfrac{1}{2}\)
M1: Use the model to find \(v\) or \(v^2\) at the instant the string goes slack \((v = 2.738\ldots)\)
Using SHM formula or conservation of energy.
M1: Complete method to find the total time until return to \(B\)
Requires the preceding M marks
If they use suvat to find the time as a projectile it must be a complete method e.g. \(\sqrt{\dfrac{15}{2}} = -\sqrt{\dfrac{15}{2}} + gt\) or a combination of \(v^2 = u^2 + 2as\) and \(s = ut + \dfrac{1}{2}at^2\)
A1: \(= 1.2\,(\text{s})\) or better Condone an answer to \(\gt\) 2 s.f.
Not scored if they have used 9.8.