2 problems found
A spaceship of mass \(M\) is travelling at constant speed \(V\) in a straight line when it enters a force field which applies a resistive force acting directly backwards and of magnitude \(M\omega(v^{2}+V^{2})/v\), where \(v\) is the instantaneous speed of the spaceship, and \(\omega\) is a positive constant. No other forces act on the spaceship. Find the distance travelled from the edge of the force field until the speed is reduced to \(\frac{1}{2}V\). As soon as the spaceship has travelled this distance within the force field, the field is altered to a constant resistive force, acting directly backwards, whose magnitude is within 10% of that of the force acting on the spaceship immediately before the change. If \(z\) is the extra distance travelled by the spaceship before coming instantaneously to rest, determine limits between which \(z\) must lie.
Solution: Using Newton's second law, we have: \begin{align*} && -M\omega(v^2+V^2)/v &= M v \frac{\d v}{\d x} \\ \Rightarrow && \frac{v^2}{v^2+V^2} \frac{\d v}{\d x} &= -\omega \\ \Rightarrow && \omega X &= \int_{V/2}^V \frac{v^2}{v^2+V^2} \d v \\ &&&= \int_{V/2}^V \l 1 - \frac{V^2}{v^2+V^2} \r \d v \\ &&&= \left [v - V\tan^{-1} \frac{v}{V} \right]_{V/2}^V \\ &&&= V \l \frac12 - \tan^{-1} 1 + \tan^{-1} \frac12 \r \\ \Rightarrow X &= \frac{V}{\omega} \l \tan^{-1} \frac12 + \frac12 - \frac{\pi}{4} \r \end{align*}. The resistive force just before the field changes is \(M \omega (\frac{V^2}{4} + V^2)/\frac{V}{2} = \frac52MV\omega\). Therefor the constant resistive force is between \(\frac{11}4MV\omega\) and \(\frac{9}{4}MV \omega\) and acceleration is \(\frac{11}{4}V\omega, \frac{9}{4}V\omega\). Since \(v^2 = u^2 + 2as \Rightarrow s = \frac{v^2-u^2}{2a} = \frac{\frac{V^2}{4}}{2kV\omega} = \frac{V}{8k\omega}\) therefore \(z \in \left [ \frac{V}{22\omega},\frac{V}{18 \omega} \right]\)
A skater of mass \(M\) is skating inattentively on a smooth frozen canal. She suddenly realises that she is heading perpendicularly towards the straight canal bank at speed \(V\). She is at a distance \(d\) from the bank and can choose one of two methods of trying to avoid it; either she can apply a force of constant magnitude \(F\), acting at right-angles to her velocity, so that she travels in a circle; or she can apply a force of magnitude \(\frac{1}{2}F(V^{2}+v^{2})/V^{2}\) directly backwards, where \(v\) is her instantaneous speed. Treating the skater as a particle, find the set of values of \(d\) for which she can avoid hitting the bank. Comment briefly on the assumption that the skater is a particle.
Solution: Suppose she applies a force of magnitude \(\frac{1}{2}F(V^{2}+v^{2})/V^{2}\) backwards, then \begin{align*} && M v \frac{dv}{dx} &= -\frac{1}{2}F(V^{2}+v^{2})/V^{2} \\ \Rightarrow && M\int_{V}^0 \frac{2v}{V^2+ v^2} \d v &= - \frac{F}{V^2} x \\ \Rightarrow && M \left [ -\log(V^2+v^2) \right]_0^V &= -\frac{Fx}{V^2} \\ \Rightarrow && -M \ln 2&= -\frac{Fx}{V^2} \end{align*} Therefore she will stop quickly enough if \(d > \frac{V^2M \ln 2}{F}\) If she attempts to use the right-angled method, then she will travel a distance at most \(r\) where \(r\) is the radius of her circle: \begin{align*} && F &= M \frac{V^2}{r} \\ \Rightarrow && r &= \frac{MV^2}{F} \end{align*} Therefore she can always avoid the wall if \(d > \frac{MV^2}{F}\). There are two potential issues with being a particle. Firstly we would need to account for any variation in the distance to the wall (which could be accounted for by changing \(d\)). Secondly when she enters circular motion she will rotate and therefore we might need to consider her inertia as well as just her velocity when modelling.