6.02d Mechanical energy: KE and PE concepts

5 A uniform rod of mass 4 kg and length 2.4 m can rotate in a vertical plane about a fixed horizontal axis through one end of the rod. The rod is released from rest in a horizontal position and a frictional couple of constant moment 20 Nm opposes the motion.

1. Find the angular acceleration of the rod immediately after it is released.
2. Find the angle that the rod makes with the horizontal when its angular acceleration is zero.
3. Find the maximum angular speed of the rod.
4. The rod first comes to instantaneous rest after rotating through an angle \(\theta\) radians from its initial position. Find an equation for \(\theta\), and verify that \(2.0 < \theta < 2.1\).

6 \includegraphics[max width=\textwidth, alt={}, center]{ab760a4b-e0ec-4256-838f-ed6c762ff18b-3_716_483_890_790} Two small smooth pegs \(P\) and \(Q\) are fixed at a distance \(2 a\) apart on the same horizontal level, and \(A\) is the mid-point of \(P Q\). A light rod \(A B\) of length \(4 a\) is freely pivoted at \(A\) and can rotate in the vertical plane containing \(P Q\), with \(B\) below the level of \(P Q\). A particle of mass \(m\) is attached to the rod at \(B\). A light elastic string, of natural length \(2 a\) and modulus of elasticity \(\lambda\), passes round the pegs \(P\) and \(Q\) and its two ends are attached to the rod at the point \(X\), where \(A X = a\). The angle between the rod and the downward vertical is \(\theta\), where \(- \frac { 1 } { 2 } \pi < \theta < \frac { 1 } { 2 } \pi\) (see diagram). You are given that the elastic energy stored in the string is \(\lambda a ( 1 + \cos \theta )\).

1. Show that \(\theta = 0\) is a position of equilibrium, and show that the equilibrium is stable if \(\lambda < 4 m g\).
2. Given that \(\lambda = 3 m g\), show that \(\ddot { \theta } = - k \frac { g } { a } \sin \theta\), stating the value of the constant \(k\). Hence find the approximate period of small oscillations of the system about the equilibrium position \(\theta = 0\).

7 \includegraphics[max width=\textwidth, alt={}, center]{ab760a4b-e0ec-4256-838f-ed6c762ff18b-4_783_783_255_641} A uniform circular disc with centre \(C\) has mass \(m\) and radius \(a\). The disc is free to rotate in a vertical plane about a fixed horizontal axis passing through a point \(A\) on the disc, where \(A C = \frac { 1 } { 2 } a\). The disc is slightly disturbed from rest in the position with \(C\) vertically above \(A\). When \(A C\) makes an angle \(\theta\) with the upward vertical the force exerted by the axis on the disc has components \(R\) parallel to \(A C\) and \(S\) perpendicular to \(A C\) (see diagram).

1. Show that the angular speed of the disc is \(\sqrt { \frac { 4 g ( 1 - \cos \theta ) } { 3 a } }\).
2. Find the angular acceleration of the disc, in terms of \(a , g\) and \(\theta\).
3. Find \(R\) and \(S\), in terms of \(m , g\) and \(\theta\).
4. Find the magnitude of the force exerted by the axis on the disc at an instant when \(R = 0\).

6 \includegraphics[max width=\textwidth, alt={}, center]{6e3d5f5e-7ffa-4111-903d-468fb4d20192-4_640_608_267_715} A smooth wire forms a circle with centre \(O\) and radius \(a\), and is fixed in a vertical plane. The highest point on the wire is \(A\). A small ring \(R\) of mass \(m\) moves along the wire. A light elastic string, with natural length \(\frac { 1 } { 2 } a\) and modulus of elasticity \(2 m g\), has one end attached to \(A\) and the other end attached to \(R\). The string \(A R\) makes an angle \(\theta\) (measured anticlockwise) with the downward vertical (see diagram), and you may assume that the string does not become slack.

1. Taking \(A\) as the reference level for gravitational potential energy, show that the total potential energy of the system is \(m g a \left( 6 \cos ^ { 2 } \theta - 4 \cos \theta + \frac { 1 } { 2 } \right)\).
2. Show that there are two positions of equilibrium for which \(0 \leqslant \theta < \frac { 1 } { 2 } \pi\).
3. For each of these positions of equilibrium, determine whether it is stable or unstable.

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}

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 \(t = 1\).

7 Fig. 7.1 shows one end of a light inextensible string attached to a block A of mass 4.4 kg . The other end of the string is attached to a block B of mass 5.2 kg . Block A is in contact with a smooth horizontal plane. The string is taut and passes over a small smooth pulley at the end of the plane. Block B is inside a hollow vertical tube and the vertical sides of B are in contact with the tube. Initially B is 1.6 m above the horizontal base of the tube. \begin{figure}[h] \end{figure} The blocks are released from rest. It may be assumed that in the subsequent motion A does not reach the pulley and the string remains taut. Block B reaches the base of the tube with speed \(3.5 \mathrm {~ms} ^ { - 1 }\).

1. Given that the frictional force exerted by the tube on B is constant, use an energy method to show that the magnitude of this force is 14.21 N . Blocks A and B remain attached to the opposite ends of a light inextensible string, but A is now in contact with a rough plane inclined at \(\theta ^ { \circ }\) to the horizontal, as shown in Fig. 7.2. The string connecting A and B is taut and passes over a small smooth pulley at the top of the plane. Block B is inside the same hollow vertical tube as before with the vertical sides of B in contact with the tube. It may be assumed that the frictional force exerted by the tube on B remains unchanged. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{b20e2254-955e-466c-8161-9614d8ccdba0-8_623_723_552_260} \captionsetup{labelformat=empty} \caption{Fig. 7.2}
   \end{figure} The coefficient of friction between block A and the plane is \(\frac { 3 } { 11 }\).
   The blocks are released from rest, with block B 1.6 m above the base of the tube. It may be assumed that in the subsequent motion A does not reach the pulley and the string remains taut.
2. Given that block B reaches the base of the tube with speed \(0.7 \mathrm {~ms} ^ { - 1 }\), show that \(\theta\) satisfies the equation \(3 \cos \theta + 11 \sin \theta = k\),
   where \(k\) is a constant to be determined. \section*{END OF QUESTION PAPER} \section*{}

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 sack of beans of mass 40 kg is pulled from rest at point A up a non-uniform slope onto and along a horizontal platform. Fig. 6 shows this slope AB and the platform BC , which is a vertical distance of 12 m above A. \begin{figure}[h] \end{figure}

1. Calculate the gain in the gravitational potential energy of the sack when it is moved from A to the platform. The sack has a speed of \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) by the time it reaches C at the far end of the platform. The total work done against friction in moving the sack from A to C is 484 J . There are no other resistances to the sack's motion.
2. Calculate the total work done in moving the sack between the points A and C . At point C , travelling at \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\), the sack starts to slide down a straight chute inclined at \(\alpha\) to the horizontal. Point D at the bottom of the chute is at the same vertical height as A , as shown in Fig. 6. The chute is rough and the coefficient of friction between the chute and the sack is 0.6 . During this part of the motion, again the only resistance to the motion of the sack is friction.
3. Use an energy method to calculate the value of \(\alpha\) given that the sack is travelling at \(3 \mathrm {~ms} ^ { - 1 }\) when it reaches D . For safety reasons the sack needs to arrive at D with a speed of less than \(3 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The value of \(\alpha\) can be adjusted to try to achieve this.
4. (A) Find the range of values of \(\alpha\) which achieve a safe speed at D .
   (B) Comment on whether adjusting \(\alpha\) is a practical way of achieving a safe speed at D .

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.

4 A child throws a ball of mass \(m \mathrm {~kg}\) vertically upwards with a speed of \(7.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The ball leaves the child's hand at a height of 1.6 m above horizontal ground.

1. Ignoring any possible air resistance, use an energy method to determine the maximum height reached by the ball above the ground. In fact, the ball only reaches a height of 4.1 m above the ground. For the rest of this question you should assume that the air resistance may be modelled as a constant force acting in the opposite direction to the ball's motion.
2. Show that the ball does 0.568 mJ of work against air resistance per metre travelled.
3. Calculate the speed of the ball just before it hits the ground. The ball bounces off the ground and first comes instantaneously to rest 2.8 m above the ground.
4. Determine the coefficient of restitution between the ball and the ground. In the first impact between the ball and the ground, the magnitude of the impulse exerted on the ball by the ground is 12 Ns .
5. Determine the value of \(m\).

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 .

2. A particle of mass 0.5 kg is moving under the action of a single force \(\mathbf { F N }\) so that its velocity \(\mathrm { v } \mathrm { ms } ^ { - 1 }\) at time \(t\) seconds is given by $$\mathbf { v } = 3 t ^ { 2 } \mathbf { i } - 8 t \mathbf { j } + 2 \mathrm { e } ^ { - t } \mathbf { k }$$

1. Find an expression for the acceleration of the particle at time \(t \mathrm {~s}\).
2. Determine an expression for F.v at time \(t \mathrm {~s}\).
3. Find the kinetic energy of the particle at time \(t \mathrm {~s}\).
4. Describe the relationship between the kinetic energy of a particle and the rate of working of the force acting on the particle. Verify this relationship using your answers to part (b) and part (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 \(D\).

5. Two smooth spheres \(A\) and \(B\), of equal radii, are moving on a smooth horizontal plane when they collide. Immediately after the collision sphere \(A\) has velocity ( \(- 2 \mathbf { i } - 5 \mathbf { j }\) ) \(\mathrm { ms } ^ { - 1 }\) and sphere \(B\) has velocity \(( \mathbf { i } + 3 \mathbf { j } ) \mathrm { ms } ^ { - 1 }\). When the spheres collide, their line of centres is parallel to the vector \(\mathbf { i }\) and the coefficient of restitution between the spheres is \(\frac { 2 } { 5 }\). Sphere \(A\) has mass 4 kg and sphere \(B\) has mass 2 kg .

1. Find the velocity of \(A\) and the velocity of \(B\) immediately before the collision. After the collision, sphere \(A\) continues to move with velocity ( \(- 2 \mathbf { i } - 5 \mathbf { j }\) ) \(\mathrm { ms } ^ { - 1 }\) until it collides with a smooth vertical wall. The impulse exerted by the wall on \(A\) is \(32 \mathbf { j }\) Ns.
2. State whether the wall is parallel to the vector \(\mathbf { i }\) or to the vector \(\mathbf { j }\). Give a reason for your answer.
3. Find the speed of \(A\) after the collision with the wall.
4. Calculate the loss of kinetic energy caused by the collision of sphere \(A\) with the wall.

5. A particle \(A\), of mass \(m \mathrm {~kg}\), has position vector \(11 \mathbf { i } + 6 \mathbf { j }\) and a velocity \(2 \mathbf { i } + 7 \mathbf { j }\). At the same moment, second particle \(B\), of mass \(2 m \mathrm {~kg}\), has position vector \(7 \mathbf { i } + 10 \mathbf { j }\) and a velocity \(5 \mathbf { i } + 4 \mathbf { j }\).

1. If the particles continue to move with these velocities, prove that the particles will collide. Given that the particles coalesce after collision, find the common velocity of the particles after collision.
2. Determine the impulse exerted by \(A\) on \(B\).
3. Calculate the loss of kinetic energy caused by the collision.

1. A small ball of mass 0.3 kg is released from rest from a point 3.6 m above horizontal ground. The ball falls freely under gravity, hits the ground and rebounds vertically upwards.

In the first impact with the ground, the ball receives an impulse of magnitude 4.2 Ns . The ball is modelled as a particle.

1. Find the speed of the ball immediately after it first hits the ground.
2. Find the kinetic energy lost by the ball as a result of the impact with the ground.

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 \(h\).