<s>[INST] A squash bridge has a length \(l\), and a bolt is positioned at a distance \(d\) from one end. The bridge is then compressed between two rigid surfaces (a squash racket and a court surface) so that its length is reduced to \(l/2\). Neglect friction, and assume that the bridge is initially in equilibrium. Initially, the squash bridge is of length \(l\). When the racket compresses it to length \(l/2\), the work done by the bolt can be modeled by the internal energy of stretching of the bridge's material: \[U\,=\frac{1}{2}\,k(\Delta x)^{2},\] where \(\Delta x\) equals one-half the change in the bridge's length. When the squash racket is quickly removed, the ball flies off at an angle \(\theta\) with velocity \(v\). **(a)**: Find the component of the momentum of the squash ball perpendicular to the court surface. (b) Find the speed at which the squash ball moves over the court surface. **(c)**: Show that the ball's kinetic energy equals the work done by the bolt. **(d)**: Suppose that the squash racket does work only on the top half of the first squash ball's projected path. What is the angle \(\theta\) at which the ball can be projected so that the ball can just clear the net?[/INST]