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1997 Paper 2 Q10
D: 1600.0 B: 1500.0

\noindent{\it In this question the effect of gravity is to be neglected.} A small body of mass \(M\) is moving with velocity \(v\) along the axis of a long, smooth, fixed, circular cylinder of radius \(L\). An internal explosion splits the body into two spherical fragments, with masses \(qM\) and \((1-q)M\), where \(q\le\frac{1}{2}\). After bouncing perfectly elastically off the cylinder (one bounce each) the fragments collide and coalesce at a point \(\frac{1}{2}L\) from the axis. Show that \(q=\frac{3}{ 8}\). The collision occurs at a time \(5L/v\) after the explosion. Find the energy imparted to the fragments by the explosion, and find the velocity after coalescence.

1994 Paper 2 Q10
D: 1600.0 B: 1486.7

A truck is towing a trailer of mass \(m\) across level ground by means of an elastic rope of natural length \(l\) whose modulus of elasticity is \(\lambda.\) At first the rope is slack and the trailer stationary. The truck then accelerates until the rope becomes taut and thereafter the truck travels in a straight line at a constant speed \(u\). Assuming that the effect of friction on the trailer is negligible, show that the trailer will collide with the truck at a time \[ \pi\left(\frac{lm}{\lambda}\right)^{\frac{1}{2}}+\frac{l}{u} \] after the rope first becomes taut.

1990 Paper 1 Q11
D: 1500.0 B: 1484.0

A shell of mass \(m\) is fired at elevation \(\pi/3\) and speed \(v\). Superman, of mass \(2m\), catches the shell at the top of its flight, by gliding up behind it in the same horizontal direction with speed \(3v\). As soon as Superman catches the shell, he instantaneously clasps it in his cloak, and immediately pushes it vertically downwards, without further changing its horizontal component of velocity, but giving it a downward vertical component of velocity of magnitude \(3v/2\). Calculate the total time of flight of the shell in terms of \(v\) and \(g\). Calculate also, to the nearest degree, the angle Superman's flight trajectory initially makes with the horizontal after releasing the shell, as he soars upwards like a bird. {[}Superman and the shell may be regarded as particles.{]}


Solution: The particle has initial velocity \(\displaystyle \binom{v \cos \frac{\pi}{3}}{v \sin \frac{\pi}{3}}\) and acceleration \(\displaystyle \binom{0}{-g}\). It will have zero vertical speed (ie be at the top of its trajectory) when \(t = \frac{\sqrt{3}v}{2g}\). Since \(0 = v^2-u^2 + 2as\) the height achieved will be \(\frac{3v^2}{8g}\) At this point it will need to travel the same distance again, but this time the initial speed is \(\frac{3v}{2}\) so: \begin{align*} && \frac{3v^2}{8g} &= \frac{3v}{2} t + \frac12 g t^2 \\ \Rightarrow && 0 &= 4g^2t^2+12vgt - 3v^2 \\ \Rightarrow && t &= \l \frac{-3+2\sqrt{3}}{2} \r \frac{v}{g} \end{align*} Therefore the total time is: \begin{align*} \l \frac{\sqrt{3}}{2} - \frac32 + \sqrt{3} \r \frac{v}{g} &= \frac{3\sqrt{3}-3}{2}\frac{v}{g} \end{align*} \begin{align*} COM(\uparrow): && 0 &= 2m v_y - m \frac{3}{2}v \\ \Rightarrow &&v_y &= \frac34 v \\ COM(\rightarrow): && 3mV &= 2m (3v) +m \frac{v}{2} \\ \Rightarrow && V &= \frac{13}6\\ \end{align*} Therefore superman is now travelling at a vector of \(\displaystyle \binom{\frac{13}6}{\frac34}v\) ie an angle of \(\tan^{-1} \frac 9{26}\) to the horizontal, approximately \(19^\circ\)

1987 Paper 2 Q15
D: 1500.0 B: 1500.0

A train of length \(l_{1}\) and a lorry of length \(l_{2}\) are heading for a level crossing at speeds \(u_{1}\) and \(u_{2}\) respectively. Initially the front of the train and the front of the lorry are at distances \(d_{1}\) and \(d_{2}\) from the crossing. Find conditions on \(u_{1}\) and \(u_{2}\) under which a collision will occur. On a diagram with \(u_{1}\) and \(u_{2}\) measured along the \(x\) and \(y\) axes respectively, shade in the region which represents collision. Hence show that if \(u_{1}\) and \(u_{2}\) are two independent random variables, both uniformly distributed on \((0,V)\), then the probability of a collision in the case when initially the back of the train is nearer to the crossing than the front of the lorry is \[ \frac{l_{1}l_{2}+l_{2}d_{1}+l_{1}d_{2}}{2d_{2}\left(l_{2}+d_{2}\right)}. \] Find the probability of a collision in each of the other two possible cases.