16 A bungee jumper of mass \(m \mathrm {~kg}\) 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 \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\)
16
- A model for the motion while the rope remains taut assumes that the forces acting on the bungee jumper are
- the weight of the bungee jumper
- a tension in the rope of magnitude \(k x\) newtons
- an air resistance force of magnitude \(R v\) newtons
where \(k\) and \(R\) are constants such that \(4 k m > R ^ { 2 }\)
16 - Show that this model gives the result
$$\left. \left. x = \mathrm { e } ^ { - \frac { R t } { 2 m } } \left( A \cos \frac { \sqrt { 4 k m - R ^ { 2 } } } { 2 m } \right) t + B \sin \frac { \sqrt { 4 k m - R ^ { 2 } } } { 2 m } \right) t \right) + \frac { m g } { k }$$
where \(A\) and \(B\) are constants, and \(g \mathrm {~ms} ^ { - 2 }\) is the acceleration due to gravity.
You do not need to find the value of \(A\) or the value of \(B\)
16 - (ii) It is also given that:
$$\begin{aligned}
k & = 16
R & = 20
m & = 62.5
g & = 9.8 \mathrm {~ms} ^ { - 2 }
\end{aligned}$$
and that the speed of the bungee jumper when the rope becomes taut is \(14 \mathrm {~ms} ^ { - 1 }\) Show that, to the nearest integer, \(A = - 38\) and \(B = 16\)
[0pt]
[6 marks]
16 - A second, simpler model assumes that the air resistance is zero.
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.
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