6. A light elastic spring \(A B\) has natural length \(2 a\) and modulus of elasticity \(2 m n ^ { 2 } a\), where \(n\) is a constant. A particle \(P\) of mass \(m\) is attached to the end \(A\) of the spring. At time \(t = 0\), the spring, with \(P\) attached, lies at rest and unstretched on a smooth horizontal plane. The other end \(B\) of the spring is then pulled along the plane in the direction \(A B\) with constant acceleration \(f\). At time \(t\) the extension of the spring is \(x\).
- Show that
$$\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + n ^ { 2 } x = f .$$
- Find \(x\) in terms of \(n , f\) and \(t\).
Hence find
- the maximum extension of the spring,
- the speed of \(P\) when the spring first reaches its maximum extension.
Turn over
- \hspace{0pt} [In this question \(\mathbf { i }\) and \(\mathbf { j }\) are unit vectors due east and due north respectively]
A man cycles at a constant speed \(u \mathrm {~m} \mathrm {~s} ^ { - 1 }\) on level ground and 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 positive constant.
When the man cycles with velocity \(\frac { 1 } { 5 } u ( - 3 \mathbf { i } + 4 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\), the velocity of the wind appears to be \(w \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 1 }\), where \(w\) is a positive constant.
Find, in terms of \(u\), the true velocity of the wind.
2. Two smooth uniform spheres \(S\) and \(T\) have equal radii. The mass of \(S\) is 0.3 kg and the mass of \(T\) is 0.6 kg . The spheres are moving on a smooth horizontal plane and collide obliquely. Immediately before the collision the velocity of \(S\) is \(\mathbf { u } _ { 1 } \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the velocity of \(T\) is \(\mathbf { u } _ { 2 } \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The coefficient of restitution between the spheres is 0.5 . Immediately after the collision the velocity of \(S\) is \(( - \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\) and the velocity of \(T\) is \(( \mathbf { i } + \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\). Given that when the spheres collide the line joining their centres is parallel to \(\mathbf { i }\), - find
- \(\mathbf { u } _ { 1 }\),
- \(\mathbf { u } _ { 2 }\).
After the collision, \(T\) goes on to collide with a smooth vertical wall which is parallel to \(\mathbf { j }\). Given that the coefficient of restitution between \(T\) and the wall is also 0.5 , find
- the angle through which the direction of motion of \(T\) is deflected as a result of the collision with the wall,
- the loss in kinetic energy of \(T\) caused by the collision with the wall.
- At 12 noon, \(\operatorname { ship } A\) is 8 km due west of \(\operatorname { ship } B\). Ship \(A\) is moving due north at a constant speed of \(10 \mathrm {~km} \mathrm {~h} ^ { - 1 }\). Ship \(B\) is moving at a constant speed of \(6 \mathrm {~km} \mathrm {~h} ^ { - 1 }\) on a bearing so that it passes as close to \(A\) as possible.
- Find the bearing on which ship \(B\) moves.
- Find the shortest distance between the two ships.
- Find the time when the two ships are closest.
- A particle of mass \(m\) is projected vertically upwards, at time \(t = 0\), with speed \(U\). The particle is subject to air resistance of magnitude \(\frac { m g v ^ { 2 } } { k ^ { 2 } }\), where \(v\) is the speed of the particle at time \(t\) and \(k\) is a positive constant.
- Show that the particle reaches its greatest height above the point of projection at time
$$\frac { k } { g } \tan ^ { - 1 } \left( \frac { U } { k } \right)$$ - Find the greatest height above the point of projection attained by the particle.
5.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{cf941854-3a33-4d9d-9fa0-ce9a63227599-22_419_1212_260_365}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{figure}
The end \(A\) of a uniform rod \(A B\), of length \(2 a\) and mass \(4 m\), is smoothly hinged to a fixed point. The end \(B\) is attached to one end of a light inextensible string which passes over a small smooth pulley, fixed at the same level as \(A\). The distance from \(A\) to the pulley is \(4 a\). The other end of the string carries a particle of mass \(m\) which hangs freely, vertically below the pulley, with the string taut. The angle between the rod and the downward vertical is \(\theta\), where \(0 < \theta < \frac { \pi } { 2 }\), as shown in Figure 1. - Show that the potential energy of the system is
$$2 m g a ( \sqrt { } ( 5 - 4 \sin \theta ) - 2 \cos \theta ) + \text { constant }$$
- Hence, or otherwise, show that any value of \(\theta\) which corresponds to a position of equilibrium of the system satisfies the equation
$$4 \sin ^ { 3 } \theta - 6 \sin ^ { 2 } \theta + 1 = 0$$
- Given that \(\theta = \frac { \pi } { 6 }\) corresponds to a position of equilibrium, determine its stability.
- Two points \(A\) and \(B\) lie on a smooth horizontal table with \(A B = 4 a\). One end of a light elastic spring, of natural length \(a\) and modulus of elasticity \(2 m g\), is attached to \(A\). The other end of the spring is attached to a particle \(P\) of mass \(m\). Another light elastic spring, of natural length \(a\) and modulus of elasticity \(m g\), has one end attached to \(B\) and the other end attached to \(P\). The particle \(P\) is on the table at rest and in equilibrium.
- Show that \(A P = \frac { 5 a } { 3 }\).
The particle \(P\) is now moved along the table from its equilibrium position through a distance \(0.5 a\) towards \(B\) and released from rest at time \(t = 0\). At time \(t , P\) is moving with speed \(v\) and has displacement \(x\) from its equilibrium position. There is a resistance to motion of magnitude \(4 m \omega v\) where \(\omega = \sqrt { } \left( \frac { g } { a } \right)\). - Show that \(\frac { \mathrm { d } ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 4 \omega \frac { \mathrm {~d} x } { \mathrm {~d} t } + 3 \omega ^ { 2 } x = 0\).
- Find the velocity, \(\frac { \mathrm { d } x } { \mathrm {~d} t }\), of \(P\) in terms of \(a , \omega\) and \(t\).
Turn over
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1.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{cf941854-3a33-4d9d-9fa0-ce9a63227599-27_780_1022_228_488}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{figure}
Two smooth uniform spheres \(A\) and \(B\) have masses \(2 m \mathrm {~kg}\) and \(3 m \mathrm {~kg}\) respectively and equal radii. The spheres are moving on a smooth horizontal surface. Initially, sphere \(A\) has velocity \(( 3 \mathbf { i } - 4 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\) and sphere \(B\) has velocity \(( 2 \mathbf { i } - 3 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\). When the spheres collide, the line joining their centres is parallel to \(\mathbf { j }\), as shown in Figure 1. The coefficient of restitution between the spheres is \(\frac { 3 } { 7 }\). Find, in terms of \(m\), the total kinetic energy lost in the collision.
2.
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{cf941854-3a33-4d9d-9fa0-ce9a63227599-29_680_853_285_543}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{figure}
Figure 2 represents part of the smooth rectangular floor of a sports hall. A ball is at \(B\), 4 m from one wall of the hall and 5 m from an adjacent wall. These two walls are smooth and meet at the corner \(C\). The ball is kicked so that it travels along the floor, bounces off the first wall at the point \(X\) and hits the second wall at the point \(Y\). The point \(Y\) is 7.5 m from the corner \(C\).
The coefficient of restitution between the ball and the first wall is \(\frac { 3 } { 4 }\).
Modelling the ball as a particle, find the distance \(C X\).
- \hspace{0pt} [In this question the unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are due east and due north respectively.]
A coastguard patrol boat \(C\) is moving with constant velocity \(( 8 \mathbf { i } + u \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }\). Another ship \(S\) is moving with constant velocity \(( 12 \mathbf { i } + 16 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }\). - Find, in terms of \(u\), the velocity of \(C\) relative to \(S\).
At noon, \(S\) is 10 km due west of \(C\).
If \(C\) is to intercept \(S\), - find the value of \(u\).
- Using this value of \(u\), find the time at which \(C\) would intercept \(S\).
If instead, at noon, \(C\) is moving with velocity \(( 8 \mathbf { i } + 8 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }\) and continues at this constant velocity,
- find the distance of closest approach of \(C\) to \(S\).
- A hiker walking due east at a steady speed of \(5 \mathrm {~km} \mathrm {~h} ^ { - 1 }\) notices that the wind appears to come from a direction with bearing 050. At the same time, another hiker moving on a bearing of 320 , and also walking at \(5 \mathrm {~km} \mathrm {~h} ^ { - 1 }\), notices that the wind appears to come from due north.
Find - the direction from which the wind is blowing,
- the wind speed.
5. A particle \(Q\) of mass 6 kg is moving along the \(x\)-axis. At time \(t\) seconds the displacement of \(Q\) from the origin \(O\) is \(x\) metres and the speed of \(Q\) is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The particle moves under the action of a retarding force of magnitude ( \(a + b v ^ { 2 }\) ) N, where \(a\) and \(b\) are positive constants. At time \(t = 0 , Q\) is at \(O\) and moving with speed \(U \mathrm {~m} \mathrm {~s} ^ { - 1 }\) in the positive \(x\)-direction. The particle \(Q\) comes to instantaneous rest at the point \(X\). - Show that the distance \(O X\) is
$$\frac { 3 } { b } \ln \left( 1 + \frac { b U ^ { 2 } } { a } \right) \mathrm { m }$$
Given that \(a = 12\) and \(b = 3\),
- find, in terms of \(U\), the time taken to move from \(O\) to \(X\).
6. A particle \(P\) of mass 4 kg moves along a horizontal straight line under the action of a force directed towards a fixed point \(O\) on the line. At time \(t\) seconds, \(P\) is \(x\) metres from \(O\) and the force towards \(O\) has magnitude \(9 x\) newtons. The particle \(P\) is also subject to air resistance, which has magnitude \(12 v\) newtons when \(P\) is moving with speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
- Show that the equation of motion of \(P\) is
$$4 \frac { \mathrm {~d} ^ { 2 } x } { \mathrm {~d} t ^ { 2 } } + 12 \frac { \mathrm {~d} x } { \mathrm {~d} t } + 9 x = 0$$
It is given that the solution of this differential equation is of the form
$$x = \mathrm { e } ^ { - \lambda t } ( A t + B )$$
When \(t = 0\) the particle is released from rest at the point \(R\), where \(O R = 4 \mathrm {~m}\).
Find,
- the values of the constants \(\lambda , A\) and \(B\),
- the greatest speed of \(P\) in the subsequent motion.