2 problems found
A projectile of mass \(m\) is fired horizontally from a toy cannon of mass \(M\) which slides freely on a horizontal floor. The length of the barrel is \(l\) and the force exerted on the projectile has the constant value \(P\) for so long as the projectile remains in the barrel. Initially the cannon is at rest. Show that the projectile emerges from the barrel at a speed relative to the ground of \[ \sqrt{\frac{2PMl}{m(M+m)}}. \]
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.