Edexcel M4 (Mechanics 4) 2005 January

2. [In this question \(\mathbf { i }\) and \(\mathbf { j }\) are horizontal unit vectors due east and due north respectively.] A man cycling at a constant speed \(u\) on horizontal ground finds that, when his velocity is \(u \mathbf { j } \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the velocity of the wind appears to be \(v ( 3 \mathbf { i } - 4 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\), where \(v\) is a constant. When the velocity of the man is \(\frac { u } { 5 } ( - 3 \mathbf { i } + 4 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\), he finds that the velocity of the wind appears to be \(w \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 1 }\), where \(w\) is a constant.

1. Show that \(v = \frac { u } { 20 }\), and find \(w\) in terms of \(u\).
2. Find, in terms of \(u\), the true velocity of the wind.

3. Two ships \(A\) and \(B\) are sailing in the same direction at constant speeds of \(12 \mathrm {~km} \mathrm {~h} ^ { - 1 }\) and \(16 \mathrm {~km} \mathrm {~h} ^ { - 1 }\) respectively. They are sailing along parallel lines which are 4 km apart. When the distance between the ships is \(4 \mathrm {~km} , B\) turns through \(30 ^ { \circ }\) towards \(A\). Find the shortest distance between the ships in the subsequent motion.

4. A car of mass \(M\) moves along a straight horizontal road. The total resistance to motion of the car is modelled as having constant magnitude \(R\). The engine of the car works at a constant rate \(R U\). Find the time taken for the car to accelerate from a speed of \(\frac { 1 } { 4 } U\) to a speed of \(\frac { 1 } { 2 } U\). \section*{5. [In this question \(\mathbf { i }\) and \(\mathbf { j }\) are horizontal perpendicular unit vectors.]} The vector \(\mathbf { n } = \left( - \frac { 3 } { 5 } \mathbf { i } + \frac { 4 } { 5 } \mathbf { j } \right)\) and the vector \(\mathbf { p } = \left( - \frac { 4 } { 5 } \mathbf { i } + \frac { 3 } { 5 } \mathbf { j } \right)\) are perpendicular unit vectors.

1. Verify that \(\frac { 9 } { 5 } \mathbf { n } + \frac { 13 } { 5 } \mathbf { p } = ( \mathbf { i } + 3 \mathbf { j } )\). A smooth uniform sphere \(S\) of mass 0.5 kg is moving on a smooth horizontal plane when it collides with a fixed smooth vertical wall which is parallel to \(\mathbf { p }\). Immediately after the collision the velocity of \(S\) is \(( \mathbf { i } + 3 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\). The coefficient of restitution between \(S\) and the wall is \(\frac { 9 } { 16 }\).
2. Find, in terms of \(\mathbf { i }\) and \(\mathbf { j }\), the velocity of \(S\) immediately before the collision.
3. Find the energy lost in the collision. \section*{6.} \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Figure 1} \includegraphics[alt={},max width=\textwidth]{d6e5bd56-0a01-44a2-b439-f80cb356d46d-3_681_747_1121_679}
   \end{figure} A smooth wire \(P M Q\) is in the shape of a semicircle with centre \(O\) and radius \(a\). The wire is fixed in a vertical plane with \(P Q\) horizontal and the mid-point \(M\) of the wire vertically below \(O\). A smooth bead \(B\) of mass \(m\) is threaded on the wire and is attached to one end of a light elastic string. The string has modulus of elasticity \(4 m g\) and natural length \(\frac { 5 } { 4 } a\). The other end of the string is attached to a fixed point \(F\) which is a distance \(a\) vertically above \(O\), as shown in Fig. 1.

1. Show that, when \(\angle B F O = \theta\), the potential energy of the system is $$\frac { 1 } { 10 } m g a ( 8 \cos \theta - 5 ) ^ { 2 } - 2 m g a \cos ^ { 2 } \theta + \text { constant } .$$
2. Hence find the values of \(\theta\) for which the system is in equilibrium.
3. Determine the nature of the equilibrium at each of these positions.

7. A particle of mass \(m\) is attached to one end \(P\) of a light elastic spring \(P Q\), of natural length \(a\) and modulus of elasticity \(m a n ^ { 2 }\). At time \(t = 0\), the particle and the spring are at rest on a smooth horizontal table, with the spring straight but unstretched and uncompressed. The end \(Q\) of the spring is then moved in a straight line, in the direction \(P Q\), with constant acceleration \(f\). At time \(t\), the displacement of the particle in the direction \(P Q\) from its initial position is \(x\) and the length of the spring is \(( a + y )\).

1. Show that \(x + y = \frac { 1 } { 2 } f t ^ { 2 }\).
2. Hence show that $$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + n ^ { 2 } x = \frac { 1 } { 2 } n ^ { 2 } f t ^ { 2 }$$ You are given that the general solution of this differential equation is $$x = A \cos n t + B \sin n t + \frac { 1 } { 2 } f t ^ { 2 } - \frac { f } { n ^ { 2 } }$$ where \(A\) and \(B\) are constants.
3. Find the values of \(A\) and \(B\).
4. Find the maximum tension in the spring. END