6.02e Calculate KE and PE: using formulae

7 \includegraphics[max width=\textwidth, alt={}, center]{6e3d5f5e-7ffa-4111-903d-468fb4d20192-5_584_686_264_678} \(A B C D\) is a uniform rectangular lamina with mass \(m\) and sides \(A B = 6 a\) and \(A D = 8 a\). The lamina rotates freely in a vertical plane about a fixed horizontal axis passing through \(A\), and it is released from rest in the position with \(D\) vertically above \(A\). When the diagonal \(A C\) makes an angle \(\theta\) below the horizontal, the force acting on the lamina at \(A\) has components \(R\) parallel to \(C A\) and \(S\) perpendicular to \(C A\) (see diagram).

1. Find the moment of inertia of the lamina about the axis through \(A\), in terms of \(m\) and \(a\).
2. Show that the angular speed of the lamina is \(\sqrt { \frac { 3 g ( 4 + 5 \sin \theta ) } { 50 a } }\).
3. Find the angular acceleration of the lamina, in terms of \(a , g\) and \(\theta\).
4. Find \(R\) and \(S\), in terms of \(m , g\) and \(\theta\).

4 A uniform square lamina has mass \(m\) and sides of length \(2 a\).

1. Calculate the moment of inertia of the lamina about an axis through one of its corners perpendicular to its plane. \includegraphics[max width=\textwidth, alt={}, center]{639c658e-0aca-4161-9e77-0f4c494b0b55-3_693_640_434_715} The uniform square lamina has centre \(C\) and is free to rotate in a vertical plane about a fixed horizontal axis passing through one of its corners \(A\). The lamina is initially held such that \(A C\) is vertical with \(C\) above \(A\). The lamina is slightly disturbed from rest from this initial position. When \(A C\) makes an angle \(\theta\) with the upward vertical, the force exerted by the axis on the lamina has components \(X\) parallel to \(A C\) and \(Y\) perpendicular to \(A C\) (see diagram).
2. Show that the angular speed, \(\omega\), of the lamina satisfies \(a \omega ^ { 2 } = \frac { 3 } { 4 } g \sqrt { 2 } ( 1 - \cos \theta )\).
3. Find \(X\) and \(Y\) in terms of \(m , g\) and \(\theta\). \section*{Question 5 begins on page 4.}
   \includegraphics[max width=\textwidth, alt={}]{639c658e-0aca-4161-9e77-0f4c494b0b55-4_767_337_248_863}
   A pendulum consists of a uniform rod \(A B\) of length \(4 a\) and mass \(4 m\) and a spherical shell of radius \(a\), mass \(m\) and centre \(C\). The end \(B\) of the rod is rigidly attached to a point on the surface of the shell in such a way that \(A B C\) is a straight line. The pendulum is initially at rest with \(B\) vertically below \(A\) and it is free to rotate in a vertical plane about a smooth fixed horizontal axis passing through \(A\) (see diagram).
4. Show that the moment of inertia of the pendulum about the axis of rotation is \(47 m a ^ { 2 }\). A particle of mass \(m\) is moving horizontally in the plane in which the pendulum is free to rotate. The particle has speed \(\sqrt { k g a }\), where \(k\) is a positive constant, and strikes the rod at a distance \(3 a\) from \(A\). In the subsequent motion the particle adheres to the rod and the combined rigid body \(P\) starts to rotate.
5. Show that the initial angular speed of \(P\) is \(\frac { 3 } { 56 } \sqrt { \frac { k g } { a } }\).
6. For the case \(k = 4\), find the angle that \(P\) has turned through when \(P\) first comes to instantaneous rest.
7. Find the least value of \(k\) such that the rod reaches the horizontal. \includegraphics[max width=\textwidth, alt={}, center]{639c658e-0aca-4161-9e77-0f4c494b0b55-5_437_903_269_573} A uniform rod \(A B\) has mass \(m\) and length \(2 a\). The rod can rotate in a vertical plane about a smooth fixed horizontal axis passing through \(A\). One end of a light elastic string of natural length \(a\) and modulus of elasticity \(\sqrt { 3 } m g\) is attached to \(A\). The string passes over a small smooth fixed pulley \(C\), where \(A C\) is horizontal and \(A C = a\). The other end of the string is attached to the rod at its mid-point \(D\). The rod makes an angle \(\theta\) below the horizontal (see diagram).
8. Taking \(A\) as the reference level for gravitational potential energy, show that the total potential energy \(V\) of the system is given by $$V = m g a ( \sqrt { 3 } - \sin \theta - \sqrt { 3 } \cos \theta ) .$$
9. Show that \(\theta = \frac { 1 } { 6 } \pi\) is a position of stable equilibrium for the system. The system is making small oscillations about the equilibrium position.
10. By differentiating the energy equation with respect to time, show that $$\frac { 4 } { 3 } a \ddot { \theta } = g ( \cos \theta - \sqrt { 3 } \sin \theta ) .$$
11. Using the substitution \(\theta = \phi + \frac { 1 } { 6 } \pi\), show that the motion is approximately simple harmonic, and find the approximate period of the oscillations. \section*{END OF QUESTION PAPER}

6 A pendulum consists of a uniform rod \(A B\) of length \(2 a\) and mass \(2 m\) and a particle of mass \(m\) that is attached to the end \(B\). The pendulum can rotate in a vertical plane about a smooth fixed horizontal axis passing through \(A\).

1. Show that the moment of inertia of this pendulum about the axis of rotation is \(\frac { 20 } { 3 } m a ^ { 2 }\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{4b50b084-081f-48d2-ad5b-95b2c9e55dfc-4_572_86_852_575} \captionsetup{labelformat=empty} \caption{Fig. 1}
   \end{figure} \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{4b50b084-081f-48d2-ad5b-95b2c9e55dfc-4_582_456_842_1050} \captionsetup{labelformat=empty} \caption{Fig. 2}
   \end{figure} The pendulum is initially held with \(B\) vertically above \(A\) (see Fig.1) and it is slightly disturbed from this position. When the angle between the pendulum and the upward vertical is \(\theta\) radians the pendulum has angular speed \(\omega \mathrm { rads } ^ { - 1 }\) (see Fig. 2).
2. Show that $$\omega ^ { 2 } = \frac { 6 g } { 5 a } ( 1 - \cos \theta ) .$$
3. Find the angular acceleration of the pendulum in terms of \(g , a\) and \(\theta\). At an instant when \(\theta = \frac { 1 } { 3 } \pi\), the force acting on the pendulum at \(A\) has magnitude \(F\).
4. Find \(F\) in terms of \(m\) and \(g\). It is given that \(a = 0.735 \mathrm {~m}\).
5. Show that the time taken for the pendulum to move from the position \(\theta = \frac { 1 } { 6 } \pi\) to the position \(\theta = \frac { 1 } { 3 } \pi\) is given by $$k \int _ { \frac { 1 } { 6 } \pi } ^ { \frac { 1 } { 3 } \pi } \operatorname { cosec } \left( \frac { 1 } { 2 } \theta \right) \mathrm { d } \theta ,$$ stating the value of the constant \(k\). Hence find the time taken for the pendulum to rotate between these two points. (You may quote an appropriate result given in the List of Formulae (MF1).) \section*{END OF QUESTION PAPER}

3 \includegraphics[max width=\textwidth, alt={}, center]{57323af2-8cf3-4721-b2c8-a968264be343-2_439_444_1318_822} A uniform rod \(A B\) has mass \(m\) and length \(4 a\). The rod can rotate in a vertical plane about a smooth fixed horizontal axis passing through \(A\). One end of a light elastic string of natural length \(a\) and modulus of elasticity \(\lambda m g\) is attached to \(B\). The other end of the string is attached to a small light ring which slides on a fixed smooth horizontal rail which is in the same vertical plane as the rod. The rail is a vertical distance \(3 a\) above \(A\). The string is always vertical and the rod makes an angle \(\theta\) radians with the horizontal, where \(0 \leqslant \theta \leqslant \frac { 1 } { 2 } \pi\) (see diagram).

1. Taking \(A\) as the reference level for gravitational potential energy, find an expression for the total potential energy \(V\) of the system, and show that $$\frac { \mathrm { d } V } { \mathrm {~d} \theta } = 2 m g a \cos \theta ( 4 \lambda ( 1 + 2 \sin \theta ) - 1 ) .$$ Determine the positions of equilibrium and the nature of their stability in the cases
2. \(\lambda > \frac { 1 } { 12 }\),
3. \(\lambda < \frac { 1 } { 12 }\). \includegraphics[max width=\textwidth, alt={}, center]{57323af2-8cf3-4721-b2c8-a968264be343-3_392_689_269_671} The diagram shows the curve with equation \(y = \frac { 1 } { 2 } \ln x\). The region \(R\), shaded in the diagram, is bounded by the curve, the \(x\)-axis and the line \(x = 4\). A uniform solid of revolution is formed by rotating \(R\) completely about the \(y\)-axis to form a solid of volume \(V\).
4. Show that \(V = \frac { 1 } { 4 } \pi ( 64 \ln 2 - 15 )\).
5. Find the exact \(y\)-coordinate of the centre of mass of the solid. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{57323af2-8cf3-4721-b2c8-a968264be343-4_385_741_269_646} \captionsetup{labelformat=empty} \caption{Fig. 1}
   \end{figure} Fig. 1 shows part of the line \(y = \frac { a } { h } x\), where \(a\) and \(h\) are constants. The shaded region bounded by the line, the \(x\)-axis and the line \(x = h\) is rotated about the \(x\)-axis to form a uniform solid cone of base radius \(a\), height \(h\) and volume \(\frac { 1 } { 3 } \pi a ^ { 2 } h\). The mass of the cone is \(M\).
6. Show by integration that the moment of inertia of the cone about the \(y\)-axis is \(\frac { 3 } { 20 } M \left( a ^ { 2 } + 4 h ^ { 2 } \right)\). (You may assume the standard formula \(\frac { 1 } { 4 } m r ^ { 2 }\) for the moment of inertia of a uniform disc about a diameter.) \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{57323af2-8cf3-4721-b2c8-a968264be343-4_501_556_1238_726} \captionsetup{labelformat=empty} \caption{Fig. 2}
   \end{figure} A uniform solid cone has mass 3 kg , base radius 0.4 m and height 1.2 m . The cone can rotate about a fixed vertical axis passing through its centre of mass with the axis of the cone moving in a horizontal plane. The cone is rotating about this vertical axis at an angular speed of \(9.6 \mathrm { rad } \mathrm { s } ^ { - 1 }\). A stationary particle of mass \(m \mathrm {~kg}\) becomes attached to the vertex of the cone (see Fig. 2). The particle being attached to the cone causes the angular speed to change instantaneously from \(9.6 \mathrm { rad } \mathrm { s } ^ { - 1 }\) to \(7.8 \mathrm { rad } \mathrm { s } ^ { - 1 }\).
7. Find the value of \(m\). \includegraphics[max width=\textwidth, alt={}, center]{57323af2-8cf3-4721-b2c8-a968264be343-5_534_501_255_767} A triangular frame \(A B C\) consists of three uniform rods \(A B , B C\) and \(C A\), rigidly joined at \(A , B\) and \(C\). Each rod has mass \(m\) and length \(2 a\). The frame is free to rotate in a vertical plane about a fixed horizontal axis passing through \(A\). The frame is initially held such that the axis of symmetry through \(A\) is vertical and \(B C\) is below the level of \(A\). The frame starts to rotate with an initial angular speed of \(\omega\) and at time \(t\) the angle between the axis of symmetry through \(A\) and the vertical is \(\theta\) (see diagram).
8. Show that the moment of inertia of the frame about the axis through \(A\) is \(6 m a ^ { 2 }\).
9. Show that the angular speed \(\dot { \theta }\) of the frame when it has turned through an angle \(\theta\) satisfies $$a \dot { \theta } ^ { 2 } = a \omega ^ { 2 } - k g \sqrt { 3 } ( 1 - \cos \theta ) ,$$ stating the exact value of the constant \(k\).
   Hence find, in terms of \(a\) and \(g\), the set of values of \(\omega ^ { 2 }\) for which the frame makes complete revolutions. At an instant when \(\theta = \frac { 1 } { 6 } \pi\), the force acting on the frame at \(A\) has magnitude \(F\).
10. Given that \(\omega ^ { 2 } = \frac { 2 g } { a \sqrt { 3 } }\), find \(F\) in terms of \(m\) and \(g\). \section*{END OF QUESTION PAPER}

2 A light elastic string AB has stiffness \(k\). The end A is attached to a fixed point and a particle of mass \(m\) is attached at the end B . With the string vertical, the particle is released from rest from a point at a distance \(a\) below its equilibrium position. At time \(t\), the displacement of the particle below the equilibrium position is \(x\) and the velocity of the particle is \(v\).

1. Show that $$m v \frac { \mathrm {~d} v } { \mathrm {~d} x } = - k x$$
2. Show that, while the particle is moving upwards and the string is taut, $$v = - \sqrt { \frac { k } { m } \left( a ^ { 2 } - x ^ { 2 } \right) }$$
3. Hence use integration to find an expression for \(x\) at time \(t\) while the particle is moving upwards and the string is taut.

3 A uniform rigid rod AB of length \(2 a\) and mass \(m\) is smoothly hinged to a fixed point at A so that it can rotate freely in a vertical plane. A light elastic string of modulus \(\lambda\) and natural length \(a\) connects the midpoint of AB to a fixed point C which is vertically above A with \(\mathrm { AC } = a\). The rod makes an angle \(2 \theta\) with the upward vertical, where \(\frac { 1 } { 3 } \pi \leqslant 2 \theta \leqslant \pi\). This is shown in Fig. 3. \begin{figure}[h] \end{figure}

1. Find the potential energy, \(V\), of the system relative to A in terms of \(m , \lambda , a\) and \(\theta\). Show that $$\frac { \mathrm { d } V } { \mathrm {~d} \theta } = 2 a \cos \theta ( 2 \lambda \sin \theta - 2 m g \sin \theta - \lambda ) .$$ Assume now that the system is set up so that the result (*) continues to hold when \(\pi < 2 \theta \leqslant \frac { 5 } { 3 } \pi\).
2. In the case \(\lambda < 2 m g\), show that there is a stable position of equilibrium at \(\theta = \frac { 1 } { 2 } \pi\). Show that there are no other positions of equilibrium in this case.
3. In the case \(\lambda > 2 m g\), find the positions of equilibrium for \(\frac { 1 } { 3 } \pi \leqslant 2 \theta \leqslant \frac { 5 } { 3 } \pi\) and determine for each whether the equilibrium is stable or unstable, justifying your conclusions.

2 A uniform rod AB of length 0.5 m and mass 0.5 kg is freely hinged at A so that it can rotate in a vertical plane. Attached at B are two identical light elastic strings BC and BD each of natural length 0.5 m and stiffness \(2 \mathrm {~N} \mathrm {~m} ^ { - 1 }\). The ends C and D are fixed at the same horizontal level as A and with \(\mathrm { AC } = \mathrm { CD } = 0.5 \mathrm {~m}\). The system is shown in Fig. 2.1 with the angle \(\mathrm { BAC } = \theta\). You may assume that \(\frac { 1 } { 3 } \pi \leqslant \theta \leqslant \frac { 5 } { 3 } \pi\) so that both strings are taut. \begin{figure}[h] \end{figure}

1. Show that the length of BC in metres is \(\sin \frac { 1 } { 2 } \theta\).
2. Find the potential energy, \(V \mathrm {~J}\), of the system relative to AD in terms of \(\theta\). Hence show that $$\frac { \mathrm { d } V } { \mathrm {~d} \theta } = 1.5 \sin \theta - 1.225 \cos \theta - \frac { 0.5 \sin \theta } { \sqrt { 1.25 - \cos \theta } } - 0.5 \cos \frac { 1 } { 2 } \theta .$$
3. Fig. 2.2 shows a graph of the function \(\mathrm { f } ( \theta ) = 1.5 \sin \theta - 1.225 \cos \theta - \frac { 0.5 \sin \theta } { \sqrt { 1.25 - \cos \theta } } - 0.5 \cos \frac { 1 } { 2 } \theta\) for \(0 \leqslant \theta \leqslant 2 \pi\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{bc637a95-b469-493b-8fd4-d3b12049878b-2_453_1264_2021_397} \captionsetup{labelformat=empty} \caption{Fig. 2.2}
   \end{figure} Use the graph both to estimate, correct to 1 decimal place, the values of \(\theta\) for which the system is in equilibrium and also to determine their stability.

4 A uniform lamina of mass \(m\) is in the shape of a sector of a circle of radius \(a\) and angle \(\frac { 1 } { 3 } \pi\). It can rotate freely in a vertical plane about a horizontal axis perpendicular to the lamina through its vertex O .

1. Show by integration that the moment of inertia of the lamina about the axis is \(\frac { 1 } { 2 } m a ^ { 2 }\).
2. State the distance of the centre of mass of the lamina from the axis. The lamina is released from rest when one of the straight edges is horizontal as shown in Fig. 4.1. After time \(t\), the line of symmetry of the lamina makes an angle \(\theta\) with the downward vertical. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{bc637a95-b469-493b-8fd4-d3b12049878b-3_257_441_1475_322} \captionsetup{labelformat=empty} \caption{Fig. 4.1}
   \end{figure} \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{bc637a95-b469-493b-8fd4-d3b12049878b-3_380_732_1635_1014} \captionsetup{labelformat=empty} \caption{Fig. 4.2}
   \end{figure}
3. Show that \(\dot { \theta } ^ { 2 } = \frac { 4 g } { \pi a } ( 2 \cos \theta + 1 )\).
4. Find the greatest speed attained by any point on the lamina.
5. Find an expression for \(\ddot { \theta }\) in terms of \(\theta , a\) and \(g\). The lamina strikes a fixed peg at A where \(\mathrm { AO } = \frac { 3 } { 4 } a\) and is horizontal, as shown in Fig. 4.2. The collision reverses the direction of motion of the lamina and halves its angular speed.
6. Find the magnitude of the impulse that the peg gives to the lamina.
7. Determine the maximum value of \(\theta\) in the subsequent motion.

3 A uniform rigid rod AB of mass \(m\) and length \(2 a\) is freely hinged to a horizontal floor at A . The end B is attached to a light elastic string of modulus \(\lambda\) and natural length \(5 a\). The other end of the string is attached to a small, light, smooth ring C which can slide along a horizontal rail. The rail is a distance \(7 a\) above the floor and C is always vertically above B . The angle that AB makes with the floor is \(\theta\). The system is shown in Fig. 3. \begin{figure}[h] \end{figure}

1. Find the potential energy, \(V\), of the system and hence show that $$\frac { \mathrm { d } V } { \mathrm {~d} \theta } = a \cos \theta \left( m g - \frac { 4 \lambda } { 5 } ( 1 - \sin \theta ) \right) .$$
2. Show that there is a position of equilibrium when \(\theta = \frac { 1 } { 2 } \pi\) and determine whether or not it is stable. There are two further positions of equilibrium when \(0 < \theta < \pi\).
3. Find the magnitude of the tension in the string and the vertical force of the hinge on the rod in these positions.
4. Show that \(\lambda > \frac { 5 m g } { 4 }\).
5. Show that these positions of equilibrium are stable.

2 Fig. 2 shows a system in a vertical plane. A uniform rod AB of length \(2 a\) and mass \(m\) is freely hinged at A . The angle that AB makes with the horizontal is \(\theta\), where \(- \frac { 2 } { 3 } \pi < \theta < \frac { 2 } { 3 } \pi\). Attached at B is a light spring BC of natural length \(a\) and stiffness \(\frac { m g } { a }\). The other end of the spring is attached to a small light smooth ring C which can slide freely along a vertical rail. The rail is at a distance of \(a\) from A and the spring is always horizontal. \begin{figure}[h] \end{figure}

1. Find the potential energy, \(V\), of the system and hence show that \(\frac { \mathrm { d } V } { \mathrm {~d} \theta } = m g a \cos \theta ( 1 - 4 \sin \theta )\).
2. Hence find the positions of equilibrium of the system and investigate their stability.

3 Fig. 3 shows a uniform rigid rod AB of length \(2 a\) and mass \(2 m\). The rod is freely hinged at A so that it can rotate in a vertical plane. One end of a light inextensible string of length \(l\) is attached to B . The string passes over a small smooth fixed pulley at C , where C is vertically above A and \(\mathrm { AC } = 6 a\). A particle of mass \(\lambda m\), where \(\lambda\) is a positive constant, is attached to the other end of the string and hangs freely, vertically below C . The rod makes an angle \(\theta\) with the upward vertical, where \(0 \leqslant \theta \leqslant \pi\). You may assume that the particle does not interfere with the rod AB or the section of the string BC . \begin{figure}[h] \end{figure}

1. Find the potential energy, \(V\), of the system relative to a situation in which the rod AB is horizontal, and hence show that $$\frac { \mathrm { d } V } { \mathrm {~d} \theta } = 2 m g a \sin \theta \left( \frac { 3 \lambda } { \sqrt { 10 - 6 \cos \theta } } - 1 \right) .$$
2. Show that \(\theta = 0\) and \(\theta = \pi\) are the only values of \(\theta\) for which the system is in equilibrium whatever the value of \(\lambda\).
3. Show that, if there is a third value of \(\theta\) for which the system is in equilibrium, then \(\frac { 2 } { 3 } < \lambda < \frac { 4 } { 3 }\).
4. Given that there are three positions of equilibrium, establish whether each of these positions is stable or unstable. It is given that, for small values of \(\theta\), $$\frac { \mathrm { d } V } { \mathrm {~d} \theta } \approx 2 m g a \left[ \left( \frac { 3 } { 2 } \lambda - 1 \right) \theta - \left( \frac { 13 } { 16 } \lambda - \frac { 1 } { 6 } \right) \theta ^ { 3 } \right] .$$
5. Investigate the stability of the equilibrium position given by \(\theta = 0\) in the case when \(\lambda = \frac { 2 } { 3 }\).

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

A small smooth ring of mass 0.5 kg moves along a smooth horizontal wire. The only forces acting on the ring are its weight, the normal reaction from the wire, and a constant force \(( 5 \mathbf { i } + \mathbf { j } - 3 \mathbf { k } ) \mathrm { N }\). The ring is initially at rest at the point with position vector \(( \mathbf { i } + \mathbf { j } + \mathbf { k } ) \mathrm { m }\), relative to a fixed origin. Find the speed of the ring as it passes through the point with position vector \(( 3 \mathbf { i } + \mathbf { k } ) \mathrm { m }\).

6. A particle \(P\) of mass 2 kg moves in the \(x - y\) plane. At time \(t\) seconds its position vector is \(\mathbf { r }\) metres. When \(t = 0\), the position vector of \(P\) is \(\mathbf { i }\) metres and the velocity of \(P\) is ( \(- \mathbf { i } + \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\). The vector \(\mathbf { r }\) satisfies the differential equation $$\frac { \mathrm { d } ^ { 2 } \mathbf { r } } { \mathrm {~d} t ^ { 2 } } + 2 \frac { \mathrm {~d} \mathbf { r } } { \mathrm {~d} t } + 2 \mathbf { r } = \mathbf { 0 }$$

1. Find \(\mathbf { r }\) in terms of \(t\).
2. Show that the speed of \(P\) at time \(t\) is \(\mathrm { e } ^ { - t } \sqrt { 2 } \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
3. Find, in terms of e, the loss of kinetic energy of \(P\) in the interval \(t = 0\) to \(t = 1\).

4 The diagram shows two points A and B on a snowy slope. A is a vertical distance of 25 m above B. \includegraphics[max width=\textwidth, alt={}, center]{d1ec7861-dc8b-450b-8e05-c70479ab0dc2-5_220_1376_306_244} A rider and snowmobile, with a combined mass of 240 kg , start at the top of the slope, heading in the direction of \(B\). As the snowmobile passes \(A\), with a speed of \(3 \mathrm {~ms} ^ { - 1 }\), the rider switches off the engine so that the snowmobile coasts freely. When the snowmobile passes B, it has a speed of \(18 \mathrm {~ms} ^ { - 1 }\). The resistances to motion can be modelled as a single, constant force of magnitude 120 N .

1. Calculate the distance the snowmobile travels from A to B. The rider now turns the snowmobile around and brings it back to B, so that it faces up the slope. Starting from rest, the snowmobile ascends the slope so that it passes A with a speed of \(7 \mathrm {~ms} ^ { - 1 }\). It takes 30 seconds for the snowmobile to travel from B to A. The resistances to motion can still be modelled as a single, constant force of magnitude 120 N .
2. Show that the snowmobile develops an average power of 2856 W during this time. The snowmobile can develop a maximum power of 6000 W . At a later point in the journey, the rider and snowmobile reach a different slope inclined at \(12 ^ { \circ }\) to the horizontal. The resistances to motion can still be modelled as a single, constant force of magnitude 120 N .
3. Determine the maximum speed with which the rider and snowmobile can ascend. The power developed by a vehicle is sometimes given in the non-SI unit mechanical horsepower \(( \mathrm { hp } ) .1 \mathrm { hp }\) is the power required to lift 550 pounds against gravity, starting and ending at rest, by 1 foot in 1 second.
4. Given that 1 metre \(\approx 3.28\) feet and \(1 \mathrm {~kg} \approx 2.2\) pounds, determine the number of watts that are equivalent to 1 hp .

2 A ball P of mass \(m \mathrm {~kg}\) is held at a height of 12.8 m above a horizontal floor. P is released from rest and rebounds from the floor. After the first bounce, P reaches a maximum height of 5 m above the floor. Two models, A and B , are suggested for the motion of P .
Model A assumes that air resistance may be neglected.

1. Determine, according to model A , the coefficient of restitution between P and the floor. Model B assumes that the collision between P and the floor is perfectly elastic, but that work is done against air resistance at a constant rate of \(E\) joules per metre.
2. Show that, according to model \(\mathrm { B } , \mathrm { E } = \frac { 39 } { 89 } \mathrm { mg }\).
3. Show that both models predict that P will attain the same maximum height after the second bounce.

4 The diagram shows three beads, A, B and C, of masses \(0.3 \mathrm {~kg} , 0.5 \mathrm {~kg}\) and 0.7 kg respectively, threaded onto a smooth wire circuit consisting of two straight and two semi-circular sections. The circuit occupies a vertical plane, with the two straight sections horizontal and the upper section 0.45 m directly above the lower section. \includegraphics[max width=\textwidth, alt={}, center]{a87d62b8-406d-44cd-9ffa-384005329566-5_361_961_450_248} Initially, the beads are at rest. A and B are each given an impulse so that they move towards each other, A with a speed of \(8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and B with a speed of \(1.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). In the subsequent collision between A and \(\mathrm { B } , \mathrm { A }\) is brought to rest.

1. Show that the coefficient of restitution between A and B is \(\frac { 1 } { 3 }\). Bead B next collides with C.
2. Show that the speed of B before this collision is \(4.37 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), correct to 3 significant figures. In this collision between B and C , B is brought to rest.
3. Determine whether C next collides with A or with B .
4. Explain why, if B has a greater mass than C , B could not be brought to rest in their collision.

5 In the diagram below, points \(\mathrm { A } , \mathrm { B }\) and C lie in the same vertical plane. The slope AB is inclined at an angle of \(30 ^ { \circ }\) to the horizontal and \(\mathrm { AB } = 5 \mathrm {~m}\). The point B is a vertical distance of 6.5 m above horizontal ground. The point C lies on the horizontal ground. \includegraphics[max width=\textwidth, alt={}, center]{a96a0ebe-8f4f-4d79-9d11-9d348ef72314-6_601_1285_395_244} Starting at A , a particle P , of mass \(m \mathrm {~kg}\), moves along the slope towards B , under the action of a constant force \(\mathbf { F }\). The force \(\mathbf { F }\) has a magnitude of 50 N and acts at an angle of \(\theta ^ { \circ }\) to AB in the same vertical plane as A and B . When P reaches \(\mathrm { B } , \mathbf { F }\) is removed, and P moves under gravity landing at C . It is given that

• the speed of P at A is \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\),
• the speed of P at B is \(6 \mathrm {~ms} ^ { - 1 }\),
• the speed of P at C is \(12 \mathrm {~m} \mathrm {~s} ^ { - 1 }\),
• 58 J of work is done against non-gravitational resistances as P moves from A to B ,
• 42 J of work is done against non-gravitational resistances as P moves from B to C .
  1. By considering the motion from B to C, show that \(m = 4.33\) correct to 3 significant figures.
  2. By considering the motion from A to B , determine the value of \(\theta\).
  3. Calculate the power of \(\mathbf { F }\) at the instant that P reaches B .

4 The diagram shows the path of a particle P of mass 2 kg as it moves from the origin O to C via A and B . The lengths of the sections \(\mathrm { OA } , \mathrm { AB }\) and BC are given in the diagram. The units of the axes are metres. \includegraphics[max width=\textwidth, alt={}, center]{5c1cfe41-d7a2-4f69-ae79-67d9f023c246-4_670_1322_404_246} P , starting from O , moves along the path indicated in the diagram to C under the action of a constant force of magnitude \(T \mathrm {~N}\) acting in the positive \(x\)-direction. As P moves, it does \(R \mathrm {~J}\) of work for every metre travelled against resistances to motion. It is given that

• the speed of P at O is \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\),
• the speed of P at A is \(11 \mathrm {~m} \mathrm {~s} ^ { - 1 }\),
• the speed of P at C is \(15 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

You should assume that both \(x\) - and \(y\)-axes lie in a horizontal plane.

1. By considering the entire path of P from O to C , show that $$20 \mathrm {~T} - 30 \mathrm { R } = 108 .$$
2. By formulating a second equation, determine the values of \(T\) and \(R\).
3. It is now given that the \(x\)-axis is horizontal, and the \(y\)-axis is directed vertically upwards. By considering the kinetic energy of P at B , show that the motion as described above is impossible.

6 A smooth solid hemisphere of radius \(a\) is fixed with its plane face in contact with a horizontal surface.
The highest point on the hemisphere is H , and the centre of its base is O . A particle of mass \(m\) is held at a point S on the surface of the hemisphere such that angle HOS is \(30 ^ { \circ }\), as shown in Fig. 6. The particle is projected from S with speed \(0.8 \sqrt { a g }\) along the surface of the hemisphere towards H . \begin{figure}[h] \end{figure}

1. Show that the particle passes through H without leaving the surface of the hemisphere. After passing through H , the particle passes through a point Q on the surface of the hemisphere, where angle \(\mathrm { HOQ } = \theta ^ { \circ }\).
2. State, in terms of \(g\) and \(\theta\), the tangential component of the acceleration of the particle when it is at Q . The particle loses contact with the hemisphere at Q and subsequently lands on the horizontal surface at a point L .
3. Find the value of \(\cos \theta\) correct to 3 significant figures.
4. Show that \(\mathrm { OL } = k a\), where \(k\) is to be found correct to 3 significant figures.

5 A young man of mass 60 kg swings on a trapeze. A simple model of this situation is as follows. The trapeze is a light seat suspended from a fixed point by a light inextensible rope. The man's centre of mass, G , moves on an arc of a circle of radius 9 m with centre O , as shown in Fig. 5. The point C is 9 m vertically below O . B is a point on the arc where angle COB is \(45 ^ { \circ }\). \begin{figure}[h] \end{figure}

1. Calculate the gravitational potential energy lost by the man if he swings from B to C . In this model it is also assumed that there is no resistance to the man's motion and he starts at rest from B.
2. Using an energy method, find the man's speed at C . A new model is proposed which also takes into account resistance to the man's motion.
3. State whether you would expect any such model to give a larger, smaller or the same value for the man's speed at C . Give a reason for your answer. A particular model takes account of the resistance by assuming that there is a force of constant magnitude 15 N always acting in the direction opposing the man's motion. This new model also takes account of the man 'pushing off' along the arc from B to C with a speed of \(1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
4. Using an energy method, find the man's speed at C .

6. The diagram shows a rollercoaster at an amusement park where a car is projected from a launch point \(O\) so that it performs a loop before instantaneously coming to rest at point \(C\). The car then performs the same journey in reverse. \includegraphics[max width=\textwidth, alt={}, center]{b430aa50-27e3-46f7-afef-7b8e75d46e1f-5_677_1733_552_166} The loop section is modelled by considering the track to be a vertical circle of radius 10 m and the car as a particle of mass \(m\) kg moving on the inside surface of the circular loop. You may assume that the track is smooth. At point \(A\), which is the lowest point of the circle, the car has velocity \(u \mathrm {~ms} ^ { - 1 }\) such that \(u ^ { 2 } = 60 g\). When the car is at point \(B\) the radius makes an angle \(\theta\) with the downward vertical.

1. Find, in terms of \(\theta\) and \(g\), an expression for \(v ^ { 2 }\), where \(v \mathrm {~ms} ^ { - 1 }\) is the speed of the car at \(B\).
2. Show that \(R \mathrm {~N}\), the reaction of the track on the car at \(B\), is given by $$R = m g ( 4 + 3 \cos \theta ) .$$
3. Explain why the expression for \(R\) in part (b) shows that the car will perform a complete loop.
4. This model predicts that the car will stop at \(C\) at a vertical height of 30 m above \(A\). However, after the car has completed the loop, the track becomes rough and the car only reaches a point \(D\) at a vertical height of 28 m above \(A\). The resistance to motion of the car beyond the loop is of constant magnitude \(\frac { m g } { 32 } \mathrm {~N}\). Calculate the length of the rough track between \(A\) and \(D\).

2. \begin{figure}[h] \end{figure} A particle of mass em is at rest on a smooth horizontal plane between two smooth fixed parallel vertical walls, as shown in the plan view in Figure 2. The particle is projected along the plane with speed \(u\) towards one of the walls and strikes the wall at right angles. The coefficient of restitution between the particle and each wall is \(e\) and air resistance is modelled as being negligible. Using the model,

1. find, in terms of \(m , u\) and \(e\), an expression for the total loss in the kinetic energy of the particle as a result of the first two impacts. Given that \(e\) can vary such that \(0 < e < 1\) and using the model,
2. find the value of \(e\) for which the total loss in the kinetic energy of the particle as a result of the first two impacts is a maximum,
3. describe the subsequent motion of the particle.

1. A plane is inclined to the horizontal at an angle \(\alpha\), where \(\tan \alpha = \frac { 3 } { 4 }\)

A particle \(P\) is held at rest at a point \(A\) on the plane.
The particle \(P\) is then projected with speed \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from \(A\), up a line of greatest slope of the plane. In an initial model, the plane is modelled as being smooth and air resistance is modelled as being negligible. Using this model and the principle of conservation of mechanical energy,

1. find the speed of \(P\) at the instant when it has travelled a distance \(\frac { 25 } { 6 } \mathrm {~m}\) up the plane from \(A\). In a refined model, the plane is now modelled as being rough, with the coefficient of friction between \(P\) and the plane being \(\frac { 3 } { 5 }\) Air resistance is still modelled as being negligible.
   Using this refined model and the work-energy principle,
2. find the speed of \(P\) at the instant when it has travelled a distance \(\frac { 25 } { 6 } \mathrm {~m}\) up the plane from \(A\).

1. A stone of mass 0.5 kg is projected vertically upwards with a speed \(U \mathrm {~ms} ^ { - 1 }\) from a point \(A\). The point \(A\) is 2.5 m above horizontal ground.

The speed of the stone as it hits the ground is \(25 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) The motion of the stone from the instant it is projected from \(A\) until the instant it hits the ground is modelled as that of a particle moving freely under gravity.

1. Use the model and the principle of conservation of mechanical energy to find the value of \(U\). In reality, the stone will be subject to air resistance as it moves from \(A\) to the ground.
2. State how this would affect your answer to part (a). The ground is soft and the stone sinks a vertical distance \(d \mathrm {~cm}\) into the ground. The resistive force exerted on the stone by the ground is modelled as a constant force of magnitude 2000 N and the stone is modelled as a particle.
3. Use the model and the work-energy principle to find the value of \(d\), giving your answer to 3 significant figures.

3. \begin{figure}[h] \end{figure} Figure 1 shows part of the end elevation of a building which sits on horizontal ground. The side of the building is vertical and has height \(h\). A small stone of mass \(m\) is at rest on the roof of the building at the point \(A\). The stone slides from rest down a line of greatest slope of the roof and reaches the edge \(B\) of the roof with speed \(\sqrt { 2 g h }\) The stone then moves under gravity before hitting the ground with speed \(W\).
In a model of the motion of the stone from \(\boldsymbol { B }\) to the ground

• the stone is modelled as a particle
• air resistance is ignored

Using the principle of conservation of mechanical energy and the model,

1. find \(W\) in terms of \(g\) and \(h\). In a model of the motion of the stone from \(\boldsymbol { A }\) to \(\boldsymbol { B }\)
   Using this model,
2. find, in terms of \(m\) and \(g\), the magnitude of the frictional force acting on the stone as it slides down the roof,
3. use the work-energy principle to find \(d\) in terms of \(h\).