Edexcel M4 (Mechanics 4) 2004 June

1. \hspace{0pt} [In this question \(\mathbf { i }\) and \(\mathbf { j }\) are horizontal unit vectors due east and due north respectively.]

An aeroplane makes a journey from a point \(P\) to a point \(Q\) which is due east of \(P\). The wind velocity is \(w ( \cos \theta \mathbf { i } + \sin \theta \mathbf { j } )\), where \(w\) is a positive constant. The velocity of the aeroplane relative to the wind is \(v ( \cos \phi \mathbf { i } - \sin \phi \mathbf { j } )\), where \(v\) is a constant and \(v > w\). Given that \(\theta\) and \(\phi\) are both acute angles,

1. show that \(v \sin \phi = w \sin \theta\),
2. find, in terms of \(v , w\) and \(\theta\), the speed of the aeroplane relative to the ground.

2. \begin{figure}[h] \end{figure} A smooth uniform sphere \(P\) is at rest on a smooth horizontal plane, when it is struck by an identical sphere \(Q\) moving on the plane. Immediately before the impact, the line of motion of the centre of \(Q\) is tangential to the sphere \(P\), as shown in Fig. 1. The direction of motion of \(Q\) is turned through \(30 ^ { \circ }\) by the impact. Find the coefficient of restitution between the spheres.

3. At noon, two boats \(A\) and \(B\) are 6 km apart with \(A\) due east of \(B\). Boat \(B\) is moving due north at a constant speed of \(13 \mathrm {~km} \mathrm {~h} ^ { - 1 }\). Boat \(A\) is moving with constant speed \(12 \mathrm {~km} \mathrm {~h} ^ { - 1 }\) and sets a course so as to pass as close as possible to boat \(B\). Find

1. the direction of motion of \(A\), giving your answer as a bearing,
2. the time when the boats are closest,
3. the shortest distance between the boats.

4. \begin{figure}[h] \end{figure} A uniform rod \(P Q\), of length \(2 a\) and mass \(m\), is free to rotate in a vertical plane about a fixed smooth horizontal axis through the end \(P\). The end \(Q\) is attached to one end of a light elastic string, of natural length \(a\) and modulus of elasticity \(\frac { m g } { 2 \sqrt { 3 } }\). The other end of the string is attached to a fixed point \(O\), where \(O P\) is horizontal and \(O P = 2 a\), as shown in Fig. 2. \(\angle O P Q\) is denoted by \(2 \theta\).

1. Show that, when the string is taut, the potential energy of the system is $$- \frac { m g a } { \sqrt { 3 } } ( 2 \cos 2 \theta + \sqrt { 3 } \sin 2 \theta + 2 \sin \theta ) + \text { constant } .$$
2. Verify that there is a position of equilibrium at \(\theta = \frac { \pi } { 6 }\).
3. Determine whether this is a position of stable equilibrium.

5. A particle \(P\) of mass \(m\) is attached to one end of a light elastic string, of natural length \(a\) and modulus of elasticity \(2 m a k ^ { 2 }\), where \(k\) is a positive constant. The other end of the string is attached to a fixed point \(A\). At time \(t = 0 , P\) is released from rest from a point which is a distance \(2 a\) vertically below \(A\). When \(P\) is moving with speed \(v\), the air resistance has magnitude \(2 m k v\). At time \(t\), the extension of the string is \(x\).

1. Show that, while the string is taut, $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 2 k \frac { \mathrm {~d} x } { \mathrm {~d} t } + 2 k ^ { 2 } x = g$$ You are given that the general solution of this differential equation is $$x = \mathrm { e } ^ { - k t } ( C \sin k t + D \cos k t ) + \frac { g } { 2 k ^ { 2 } } , \quad \text { where } C \text { and } D \text { are constants. }$$
2. Find the value of \(C\) and the value of \(D\). Assuming that the string remains taut,
3. find the value of \(t\) when \(P\) first comes to rest,
4. show that \(2 k ^ { 2 } a < g \left( 1 + \mathrm { e } ^ { \pi } \right)\).

6. A particle \(P\) of mass \(m\) is attached to one end of a light inextensible string and hangs at rest at time \(t = 0\). The other end of the string is then raised vertically by an engine which is working at a constant rate \(k m g\), where \(k > 0\). At time \(t\), the distance of \(P\) above its initial position is \(x\), and \(P\) is moving upwards with speed \(v\).

1. Show that \(v ^ { 2 } \frac { \mathrm {~d} v } { \mathrm {~d} x } = ( k - v ) g\).
2. Show that \(g x = k ^ { 2 } \ln \left( \frac { k } { k - v } \right) - k v - \frac { 1 } { 2 } v ^ { 2 }\).
3. Hence, or otherwise, find \(t\) in terms of \(k , v\) and \(g\).