In a game between two players both players have an equal chance of winning each point. The game continues until one player has scored \(N\) points. Find the probability \(p_r\) that the winning player has a lead of exactly \(r\) points when the game is completed. Deduce that $$(2N - r - 1)p_{r+1} = 2(N - r)p_r \quad (r = 1, 2, \ldots, N),$$ and hence find the expected value of the lead at the end of the game.
A uniform solid circular cylinder of radius \(a\) and mass \(M\) has, rigidly attached to the cylinder, a uniform rod of length \(a\) and mass \(M\). The rod lies in a plane perpendicular to the axis of the cylinder, passes through the axis and has one end on the circumference of the cylinder, as illustrated in the diagram. The cylinder rolls on a rough horizontal table with the rod just overhanging the edge of the table. The system is released from rest with the rod horizontal. By energy considerations, show that, when the rod is inclined at an angle \(\theta\) to the downward vertical, the angular velocity \(\dot{\theta} = d\theta/dt\) is given by $$a\dot{\theta}^2(19 - 8\cos\theta) = 8g\cos\theta.$$ Find the frictional force between the cylinder and the table as a function of \(\theta\) (not involving the time derivatives of \(\theta\)), and show that it vanishes at a value of \(\theta\) between \(\cos^{-1}0.8\) and \(\cos^{-1}0.4\).
A smooth wedge of mass \(M\) rests on a smooth horizontal plane. The sloping face of the wedge makes an acute angle \(\alpha\) with the horizontal. A particle of mass \(m\) is dropped from a point vertically above the centre of mass of the wedge, so that its velocity immediately before impact is \(u\). The coefficient of restitution for impacts between the particle and the wedge is \(e\). Immediately after the \(n\)th impact, the velocity of the wedge is \(U_n\), the component of the velocity of the particle parallel to the sloping face of the wedge is \(V_n\), and the component of the relative velocity of the particle and wedge perpendicular to the sloping face of the wedge is \(W_n\), all measured so as to have positive values. Find \(U_1\), \(V_1\) and \(W_1\), and show that \begin{align} (M + m\sin^2\alpha)U_n &= m(V_n\cos\alpha + W_n\sin\alpha),\\ V_{n+1} &= V_n + 2W_n\tan\alpha\\ \text{and} \quad W_{n+1} &= eW_n. \end{align}
A small ring of mass \(m\) is placed around the midpoint of a rough uniform rod \(AB\) of mass \(M\) and length \(2l\). The coefficient of friction between the ring and the rod is \(\mu\). The rod is pivoted at its end \(A\) so as to be free to swing in a vertical plane, but the ring is such that this plane is constrained to rotate about the vertical through \(A\), pivoted at constant angular velocity \(\omega\). If \(\mu\) is sufficiently large that the ring does not slide along the rod, then a position of equilibrium will exist in which the rod makes an angle \(\theta > 0\) with the downward vertical through \(A\). Assuming the validity of the use of `centrifugal forces', find the moment about \(A\) of the centrifugal forces acting on the rod in this equilibrium situation. Hence show that $$\omega^2 = \frac{3(M + m)g}{4Ml + 3ml}\cos\theta.$$ Show also that, for the ring not to slip in this position, one must have $$\mu\sin 2\theta - \cos 2\theta \geq 7 + (6m/M).$$ Find the minimum value of \(\mu\) for which this has a solution for \(\theta\). What is the value of \(\theta\) in this case? [It may be assumed that the rod has circular cross-section and that the ring sits loosely around it. There is thus only one point of contact between the rod and the ring, and the direction of the reaction at that point lies in the vertical plane through the rod.]
A ship is steaming due east at a constant speed. The ship sends out an SOS call which is received by an aeroplane. The navigator of the aeroplane correctly determines the bearing of the ship as \(\alpha\) radians east of north, and calculates that, allowing for the motion of the ship, if they fly on a bearing \(\beta\) radians east of north at speed \(v\), they should reach the ship after flying a distance \(l\). The pilot accepts this course, but due to errors in his instruments he actually flies on a bearing \((\beta + \phi)\) radians east of north at speed \(v(1 + \epsilon)\), where \(\phi\) and \(\epsilon\) are small. Show that, to first order in \(\phi\) and \(\epsilon\), their closest distance of approach to the ship is $$l[\epsilon \sin(\beta - \alpha) + \phi \cos(\beta - \alpha)].$$
\(P\), \(Q\), \(O\) and \(R\) are four distinct points which are not coplanar. Let \(a\) be the angle between the planes \(QOP\) and \(POR\). Define \(b\), \(c\), \(A\), \(B\) and \(C\) similarly by cyclic permutation of \(P\), \(Q\) and \(R\). Let \(\mathbf{p}\), \(\mathbf{q}\) and \(\mathbf{r}\) be the position vectors of \(P\), \(Q\) and \(R\) respectively with respect to \(O\) as origin, and let \(p\), \(q\), and \(r\) be the corresponding magnitudes of these vectors. By geometrical considerations, evaluate $$|(\mathbf{p} \times \mathbf{q}) \times (\mathbf{p} \times \mathbf{r})|$$ in terms of \(p\), \(q\), \(r\) and the angles \(a\), \(b\), \(c\), \(A\), \(B\) and \(C\). By expanding this repeated vector product, show also that $$\frac{|(\mathbf{p} \times \mathbf{q}) \times (\mathbf{p} \times \mathbf{r})|}{|(\mathbf{q} \times \mathbf{r}) \times (\mathbf{q} \times \mathbf{p})|} = \frac{p}{q}.$$ Deduce that $$\frac{\sin a}{\sin A} = \frac{\sin b}{\sin B} = \frac{\sin c}{\sin C}.$$ [It may be assumed without proof that, for any three vectors \(\mathbf{a}\), \(\mathbf{b}\) and \(\mathbf{c}\), $$(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{b} \cdot \mathbf{c})\mathbf{a}$$ and $$(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = (\mathbf{b} \times \mathbf{c}) \cdot \mathbf{a} = (\mathbf{c} \times \mathbf{a}) \cdot \mathbf{b}.]$$