A2 June 2023 Paper 2 Q16
16 A bungee jumper of mass \(m\) kg is attached to an elastic rope.
The other end of the rope is attached to a fixed point.
The bungee jumper falls vertically from the fixed point.
At time \(t\) seconds after the rope first becomes taut, the extension of the rope is \(x\) metres and the speed of the bungee jumper is \(v\) m s\(^{-1}\)
- the weight of the bungee jumper
- a tension in the rope of magnitude \(kx\) newtons
- an air resistance force of magnitude \(Rv\) newtons
where \(k\) and \(R\) are constants such that \(4km \gt R^2\)
where \(A\) and \(B\) are constants, and \(g\) m s\(^{-2}\) is the acceleration due to gravity.
You do not need to find the value of \(A\) or the value of \(B\) [6 marks]
and that the speed of the bungee jumper when the rope becomes taut is 14 m s\(^{-1}\)
Show that, to the nearest integer, \(A = -38\) and \(B = 16\) [6 marks]
The values of \(k\), \(m\) and \(g\) remain the same.
Find an expression for \(x\) in terms of \(t\) according to this simpler model, giving the values of all constants to two significant figures. [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| (i) Forms an equation of motion with at least three terms correct. Accept \(a\) and/or \(v\). | M1 | 3.3 |
| (i) Obtains fully correct differential equation. | A1 | 1.1b |
| (i) Obtains solutions of their auxiliary equation. | M1 | 1.1a |
| (i) Obtains a complementary function consistent with their solutions to their auxiliary equation. Must justify the choice of complementary function either from the roots of their auxiliary equation or by a clear explanation. | A1 | 1.1b |
| (i) Uses a correct method to obtain the correct particular integral. | B1 | 1.1b |
| (i) Completes a fully correct reasoned argument to obtain the required result. Condone incorrect brackets. | R1 | 2.1 |
| (6) | ||
| (ii) Correctly substitutes all relevant values into the general equation. | B1 | 3.1b |
| (ii) Substitutes \(t = 0\) and \(x = 0\) into the general equation, either with or without other values substituted. | M1 | 3.4 |
| (ii) Obtains the correct value for \(A\) Must have \(-38.3\) or better. | A1 | 1.1b |
| (ii) Uses the product rule to differentiate the expression for displacement. | M1 | 1.1a |
| (ii) Substitutes \(t = 0\) and \(v = 14\) into their equation for \(v\), either with or without other values substituted. | M1 | 3.4 |
| (ii) Completes a fully correct argument to show that to the nearest integer, \(A = -38\) and \(B = 16\) | A1 | 1.1b |
| (6) |
Typical solution
(i)
\[m\ddot{x} = mg - kx - R\dot{x}\]\[m\ddot{x} + R\dot{x} + kx = mg\]CF:
\[m\lambda^2 + R\lambda + k = 0\]\[\lambda = \frac{-R \pm \sqrt{R^2 - 4km}}{2m} = -\frac{R}{2m} \pm \mathrm{i}\left(\frac{\sqrt{4km - R^2}}{2m}\right)\]CF:
\[x = \mathrm{e}^{-\frac{Rt}{2m}}\left(A\cos\left(\frac{\sqrt{4km - R^2}}{2m}\right)t + B\sin\left(\frac{\sqrt{4km - R^2}}{2m}\right)t\right)\]PI:
\[x = p,\ \dot{x} = 0,\ \ddot{x} = 0 \Rightarrow kp = mg \Rightarrow p = \frac{mg}{k}\]So
\[x = \mathrm{e}^{-\frac{Rt}{2m}}\left(A\cos\left(\frac{\sqrt{4km - R^2}}{2m}\right)t + B\sin\left(\frac{\sqrt{4km - R^2}}{2m}\right)t\right) + \frac{mg}{k}\](ii)
\[x = \mathrm{e}^{-0.16t}\big(A\cos(0.48t) + B\sin(0.48t)\big) + 38.3\]When \(t = 0\), \(x = 0\)
\[0 = A + 38.3\]\[A = -38.3\]\[x = \mathrm{e}^{-0.16t}\big({-}38.3\cos(0.48t) + B\sin(0.48t)\big) + 38.3\]\[\begin{aligned} \dot{x} &= -0.16\mathrm{e}^{-0.16t}\big({-}38.3\cos(0.48t) + B\sin(0.48t)\big) \\ &\quad + \mathrm{e}^{-0.16t}\big(18.4\sin(0.48t) + 0.48B\cos(0.48t)\big) \end{aligned}\]When \(t = 0\), \(\dot{x} = 14\)
\[14 = -0.16(-38.3) + 0.48B\]\[B = 16.4\]To the nearest integer, \(A = -38\) and \(B = 16\)
| Scheme | Marks | AO |
|---|---|---|
| Deduces that \(R = 0\) in the equation from part (a)(i) OR Obtains a correct solution to the differential equation formed from an equation of motion using \(R = 0\) | B1 | 2.2a |
| Differentiates displacement | M1 | 1.1a |
| Substitutes \(t = 0\), \(x = 0\) and \(v = 14\) into their equations for \(x\) and \(v\) | M1 | 3.4 |
| Obtains a completely correct expression for \(x\), with values to 2 significant figures or better. | A1 | 1.1b |
| (4) | ||
| (16 marks) |
Typical solution
Setting \(R = 0\)
\[x = \left(A\cos\left(\frac{\sqrt{4km}}{2m}\right)t + B\sin\left(\frac{\sqrt{4km}}{2m}\right)t\right) + \frac{mg}{k}\]\[x = \big(A\cos(0.506t) + B\sin(0.506t)\big) + 38.3\]When \(t = 0\), \(x = 0 \Rightarrow A = -38.3\)
\[x = \big({-}38.3\cos(0.506t) + B\sin(0.506t)\big) + 38.3\]\[\dot{x} = 19.4\sin(0.506t) + 0.506B\cos(0.506t)\]When \(t = 0\), \(\dot{x} = 14\)
\[B = \frac{14}{0.506} = 27.7\]\[x = \big({-}38\cos(0.51t) + 28\sin(0.51t)\big) + 38\]