1 problem found
In this question, take the value of \(g\) to be \(10\ \mathrm{ms^{-2}.\)} A body of mass \(m\) kg is dropped vertically into a deep pool of liquid. Once in the liquid, it is subject to gravity, an upward buoyancy force of \(\frac{6}{5}\) times its weight, and a resistive force of \(2mv^{2}\mathrm{N}\) opposite to its direction of travel when it is travelling at speed \(v\) \(\mathrm{ms}^{-1}.\) Show that the body stops sinking less than \(\frac{1}{4}\pi\) seconds after it enters the pool. Suppose now that the body enters the liquid with speed \(1\ \mathrm{ms}^{-1}.\) Show that the body descends to a depth of \(\frac{1}{4}\ln2\) metres and that it returns to the surface with speed \(\frac{1}{\sqrt{2}}\ \mathrm{ms}^{-1},\) at a time \[ \frac{\pi}{8}+\frac{1}{4}\ln\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right) \] seconds after entering the pool.
Solution: While descending, the body experiences the force \(-\frac15mg - 2mv^2\). \begin{align*} \text{N2:} && m \dot{v} &= -\frac15 mg - 2mv^2 \\ \Rightarrow && \frac{\dot{v}}{\frac15g + 2v^2} &= -1 \\ \Rightarrow && \frac{1}{2}\tan^{-1} v_1 - \frac{1}{2}\tan^{-1} {v_0} &= -T \end{align*} We care about when \(v_1 = 0\), ie \(\displaystyle T = \frac{1}{2}\tan^{-1} {v_0} < \frac12 \frac{\pi}2 = \frac{\pi}4\) seconds. If the body enters at speed \(1\ \mathrm{ms}^{-1}.\) then for the first part of it's journey it will experience forces \(-\frac15mg - 2mv^2\) and so: \begin{align*} \text{N2:} && m v \frac{\d v}{\d x} &= -\frac15 mg - 2mv^2 \\ \Rightarrow && \int \frac{v}{2(1 + v^2)} \d v &= \int -1 \d x \\ \Rightarrow && \frac14 \ln (1 + v^2) &= -x \end{align*} Therefore the depth is \(\frac14 \ln 2\) metres. When the body is rising, it experiences forces of: \(\frac15mg - 2mv^2\) and so: \begin{align*} \text{N2:} && m v \frac{\d v}{\d x} &= \frac15mg - 2mv^2 \\ \Rightarrow && \int \frac{v}{2(1 - v^2)} \d v &= \int -1 \d x \\ \Rightarrow && -\frac14 \ln (1 - v^2) &= \frac14 \ln 2 \\ \Rightarrow && 1-v^2 &= \frac12 \\ \Rightarrow && v &= \frac{1}{\sqrt{2}} \ \mathrm{ms}^{-1} \end{align*} This will take \begin{align*} \text{N2:} && m \dot{v} &= \frac15mg - 2mv^2 \\ \Rightarrow && \frac{\dot{v}}{2(1-v^2)} &= -1 \\ \Rightarrow && \dot{v} \frac{1}{4}\l \frac{1}{1 - v} + \frac{1}{1+v} \r &= -1 \\ \Rightarrow && \frac{1}{4} \l -\ln(1 - v) + \ln(1 + v)\r &= -T \end{align*} Since \(v = \frac{1}{\sqrt{2}}\) \begin{align*} T &= \frac{1}{4} \ln \l \frac{1+ \frac1{\sqrt{2}}}{1 - \frac1{\sqrt{2}}}\r \\ &= \frac14 \ln \l \frac{\sqrt{2} + 1}{\sqrt{2}-1} \r \end{align*} and therefore the total time will be: \begin{align*} & \frac12 \tan^{-1} 1 + \frac14 \ln \l \frac{\sqrt{2} + 1}{\sqrt{2}-1} \r \\ =& \frac{\pi}{8} + \frac14 \ln \l \frac{\sqrt{2} + 1}{\sqrt{2}-1} \r \end{align*}