Questions M4 (303 questions)

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OCR MEI M4 2008 June Q1
1 A rocket in deep space starts from rest and moves in a straight line. The initial mass of the rocket is \(m _ { 0 }\) and the propulsion system ejects matter at a constant mass rate \(k\) with constant speed \(u\) relative to the rocket. At time \(t\) the speed of the rocket is \(v\).
  1. Show that while mass is being ejected from the rocket, \(\left( m _ { 0 } - k t \right) \frac { \mathrm { d } v } { \mathrm {~d} t } = u k\).
  2. Hence find an expression for \(v\) in terms of \(t\).
  3. Find the speed of the rocket when its mass is \(\frac { 1 } { 3 } m _ { 0 }\).
OCR MEI M4 2008 June Q2
2 A car of mass \(m \mathrm {~kg}\) starts from rest at a point O and moves along a straight horizontal road. The resultant force in the direction of motion has power \(P\) watts, given by \(P = m \left( k ^ { 2 } - v ^ { 2 } \right)\), where \(v \mathrm {~ms} ^ { - 1 }\) is the velocity of the car and \(k\) is a positive constant. The displacement from O in the direction of motion is \(x \mathrm {~m}\).
  1. Show that \(\left( \frac { k ^ { 2 } } { k ^ { 2 } - v ^ { 2 } } - 1 \right) \frac { \mathrm { d } v } { \mathrm {~d} x } = 1\), and hence find \(x\) in terms of \(v\) and \(k\).
  2. How far does the car travel before reaching \(90 \%\) of its terminal velocity?
OCR MEI M4 2008 June Q3
3 A circular disc of radius \(a \mathrm {~m}\) has mass per unit area \(\rho \mathrm { kg } \mathrm { m } ^ { - 2 }\) given by \(\rho = k ( a + r )\), where \(r \mathrm {~m}\) is the distance from the centre and \(k\) is a positive constant. The disc can rotate freely about an axis perpendicular to it and through its centre.
  1. Show that the mass, \(M \mathrm {~kg}\), of the disc is given by \(M = \frac { 5 } { 3 } k \pi a ^ { 3 }\), and show that the moment of inertia, \(I \mathrm {~kg} \mathrm {~m} ^ { 2 }\), about this axis is given by \(I = \frac { 27 } { 50 } M a ^ { 2 }\). For the rest of this question, take \(M = 64\) and \(a = 0.625\).
    The disc is at rest when it is given a tangential impulsive blow of 50 N s at a point on its circumference.
  2. Find the angular speed of the disc. The disc is then accelerated by a constant couple reaching an angular speed of \(30 \mathrm { rad } \mathrm { s } ^ { - 1 }\) in 20 seconds.
  3. Calculate the magnitude of this couple. When the angular speed is \(30 \mathrm { rads } ^ { - 1 }\), the couple is removed and brakes are applied to bring the disc to rest. The effect of the brakes is modelled by a resistive couple of \(3 \dot { \theta } \mathrm { Nm }\), where \(\dot { \theta }\) is the angular speed of the disc in \(\mathrm { rad } \mathrm { s } ^ { - 1 }\).
  4. Formulate a differential equation for \(\dot { \theta }\) and hence find \(\dot { \theta }\) in terms of \(t\), the time in seconds from when the brakes are first applied.
  5. By reference to your expression for \(\dot { \theta }\), give a brief criticism of this model for the effect of the brakes.
OCR MEI M4 2008 June Q4
4 A uniform smooth pulley can rotate freely about its axis, which is fixed and horizontal. A light elastic string AB is attached to the pulley at the end B . The end A is attached to a fixed point such that the string is vertical and is initially at its natural length with B at the same horizontal level as the axis. In this position a particle P is attached to the highest point of the pulley. This initial position is shown in Fig. 4.1. The radius of the pulley is \(a\), the mass of P is \(m\) and the stiffness of the string AB is \(\frac { m g } { 10 a }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{636e1d23-3bbb-469f-8fc9-1f64da865126-3_451_517_607_466} \captionsetup{labelformat=empty} \caption{Fig. 4.1}
\end{figure} \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{636e1d23-3bbb-469f-8fc9-1f64da865126-3_456_451_607_1226} \captionsetup{labelformat=empty} \caption{Fig. 4.2}
\end{figure}
  1. Fig. 4.2 shows the system with the pulley rotated through an angle \(\theta\) and the string stretched. Write down the extension of the string and hence find the potential energy, \(V\), of the system in this position. Show that \(\frac { \mathrm { d } V } { \mathrm {~d} \theta } = m g a \left( \frac { 1 } { 10 } \theta - \sin \theta \right)\).
  2. Hence deduce that the system has a position of unstable equilibrium at \(\theta = 0\).
  3. Explain how your expression for \(V\) relies on smooth contact between the string and the pulley. Fig. 4.3 shows the graph of the function \(\mathrm { f } ( \theta ) = \frac { 1 } { 10 } \theta - \sin \theta\). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{636e1d23-3bbb-469f-8fc9-1f64da865126-3_538_1342_1706_404} \captionsetup{labelformat=empty} \caption{Fig. 4.3}
    \end{figure}
  4. Use the graph to give rough estimates of three further values of \(\theta\) (other than \(\theta = 0\) ) which give positions of equilibrium. In each case, state with reasons whether the equilibrium is stable or unstable.
  5. Show on a sketch the physical situation corresponding to the least value of \(\theta\) you identified in part (iv). On your sketch, mark clearly the positions of P and B .
  6. The equation \(\mathrm { f } ( \theta ) = 0\) has another root at \(\theta \approx - 2.9\). Explain, with justification, whether this necessarily gives a position of equilibrium.
OCR MEI M4 2009 June Q1
1 A raindrop increases in mass as it falls vertically from rest through a stationary cloud. At time \(t \mathrm {~s}\) the velocity of the raindrop is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and its mass is \(m \mathrm {~kg}\). The rate at which the mass increases is modelled as \(\frac { m g } { 2 ( v + 1 ) } \mathrm { kg } \mathrm { s } ^ { - 1 }\). Resistances to motion are neglected.
  1. Write down the equation of motion of the raindrop. Hence show that $$\left( 1 - \frac { 1 } { v + 2 } \right) \frac { \mathrm { d } v } { \mathrm {~d} t } = \frac { 1 } { 2 } g .$$
  2. Solve this differential equation to find an expression for \(t\) in terms of \(v\). Calculate the time it takes for the velocity of the raindrop to reach \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
  3. Describe, with reasons, what happens to the acceleration of the raindrop for large values of \(t\).
OCR MEI M4 2009 June Q2
2 A uniform rigid rod AB of mass \(m\) and length \(4 a\) is freely hinged at the end A to a horizontal rail. The end B is attached to a light elastic string BC of modulus \(\frac { 1 } { 2 } m g\) and natural length \(a\). The end C of the string is attached to a ring which is small, light and smooth. The ring can slide along the rail and is always vertically above B . The angle that AB makes below the rail is \(\theta\). The system is shown in Fig. 2. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{9763e6c4-e372-46ef-a666-3ccb185aa5d2-2_277_707_1398_717} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure}
  1. Find the potential energy, \(V\), of the system when the string is stretched and show that $$\frac { \mathrm { d } V } { \mathrm {~d} \theta } = 4 m g a \cos \theta ( 2 \sin \theta - 1 )$$
  2. Hence find any positions of equilibrium of the system and investigate their stability.
OCR MEI M4 2009 June Q3
3 A uniform circular disc has mass \(M\) and radius \(a\). The centre of the disc is at point C .
  1. Show by integration that the moment of inertia of the disc about an axis through C and perpendicular to the disc is \(\frac { 1 } { 2 } M a ^ { 2 }\). The point A on the disc is at a distance \(\frac { 1 } { 10 } a\) from its centre.
  2. Show that the moment of inertia of the disc about an axis through A and perpendicular to the disc is \(0.51 M a ^ { 2 }\). The disc can rotate freely in a vertical plane about an axis through A that is horizontal and perpendicular to the disc. The disc is held slightly displaced from its stable equilibrium position and is released from rest. In the motion that follows, the angle that AC makes with the downward vertical is \(\theta\).
  3. Write down the equation of motion for the disc. Assuming \(\theta\) remains sufficiently small throughout the motion, show that the disc performs approximate simple harmonic motion and determine the period of the motion. A particle of mass \(m\) is attached at a point P on the circumference of the disc, so that the centre of mass of the system is now at A .
  4. Sketch the position of P in relation to A and C . Find \(m\) in terms of \(M\) and show that the moment of inertia of the system about the axis through A and perpendicular to the disc is \(0.6 M a ^ { 2 }\). The system now rotates at a constant angular speed \(\omega\) about the axis through A .
  5. Find the kinetic energy of the system. Hence find the magnitude of the constant resistive couple needed to bring the system to rest in \(n\) revolutions.
OCR MEI M4 2009 June Q4
4 A parachutist of mass 90 kg falls vertically from rest. The forces acting on her are her weight and resistance to motion \(R \mathrm {~N}\). At time \(t \mathrm {~s}\) the velocity of the parachutist is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) and the distance she has fallen is \(x \mathrm {~m}\). While the parachutist is in free-fall (i.e. before the parachute is opened), the resistance is modelled as \(R = k v ^ { 2 }\), where \(k\) is a constant. The terminal velocity of the parachutist in free-fall is \(60 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
  1. Show that \(k = \frac { g } { 40 }\).
  2. Show that \(v ^ { 2 } = 3600 \left( 1 - \mathrm { e } ^ { - \frac { g x } { 1800 } } \right)\). When she has fallen 1800 m , she opens her parachute.
  3. Calculate, by integration, the work done against the resistance before she opens her parachute. Verify that this is equal to the loss in mechanical energy of the parachutist. As the parachute opens, the resistance instantly changes and is now modelled as \(R = 90 v\).
  4. Calculate her velocity just before opening the parachute, correct to four decimal places.
  5. Formulate and solve a differential equation to calculate the time it takes after opening the parachute to reduce her velocity to \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
OCR MEI M4 2010 June Q1
1 At time \(t\) a rocket has mass \(m\) and is moving vertically upwards with velocity \(v\). The propulsion system ejects matter at a constant speed \(u\) relative to the rocket. The only additional force acting on the rocket is its weight.
  1. Derive the differential equation \(m \frac { \mathrm {~d} v } { \mathrm {~d} t } + u \frac { \mathrm {~d} m } { \mathrm {~d} t } = - m g\). The rocket has initial mass \(m _ { 0 }\) of which \(75 \%\) is fuel. It is launched from rest. Matter is ejected at a constant mass rate \(k\).
  2. Assuming that the acceleration due to gravity is constant, find the speed of the rocket at the instant when all the fuel is burnt.
OCR MEI M4 2010 June Q2
2 A particle of mass \(m \mathrm {~kg}\) moves horizontally in a straight line with speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at time \(t \mathrm {~s}\). The total resistance force on the particle is of magnitude \(m k v ^ { \frac { 3 } { 2 } } \mathrm {~N}\) where \(k\) is a positive constant. There are no other horizontal forces present. Initially \(v = 25\) and the particle is at a point O .
  1. Show that \(v = 4 \left( k t + \frac { 2 } { 5 } \right) ^ { - 2 }\).
  2. Find the displacement from O of the particle at time \(t\).
  3. Describe the motion of the particle as \(t\) increases. Section B (48 marks)
OCR MEI M4 2010 June Q3
3 A uniform rod AB of mass \(m\) and length \(4 a\) is hinged at a fixed point C , where \(\mathrm { AC } = a\), and can rotate freely in a vertical plane. A light elastic string of natural length \(2 a\) and modulus \(\lambda\) is attached at one end to B and at the other end 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 \(2 a\) above C . The string is always vertical. This system is shown in Fig. 3 with the rod inclined at \(\theta\) to the horizontal. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{cb86219c-e0b1-4f75-b8b2-50b5a233aa54-2_387_613_1763_767} \captionsetup{labelformat=empty} \caption{Fig. 3}
\end{figure}
  1. Find an expression for \(V\), the potential energy of the system relative to C , and show that \(\frac { \mathrm { d } V } { \mathrm {~d} \theta } = a \cos \theta \left( \frac { 9 } { 2 } \lambda \sin \theta - m g \right)\).
  2. Determine the positions of equilibrium and the nature of their stability in the cases
    (A) \(\lambda > \frac { 2 } { 9 } m g\),
    (B) \(\lambda < \frac { 2 } { 9 } m g\),
    (C) \(\lambda = \frac { 2 } { 9 } m g\). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{cb86219c-e0b1-4f75-b8b2-50b5a233aa54-3_522_755_342_696} \captionsetup{labelformat=empty} \caption{Fig. 4.1}
    \end{figure}
  3. Show, by integration, that the moment of inertia of the cone about its axis of symmetry is \(\frac { 3 } { 10 } M a ^ { 2 }\). [You may assume the standard formula for the moment of inertia of a uniform circular disc about its axis of symmetry and the formula \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\) for the volume of a cone.] A frustum is made by taking a uniform cone of base radius 0.1 m and height 0.2 m and removing a cone of height 0.1 m and base radius 0.05 m as shown in Fig. 4.2. The mass of the frustum is 2.8 kg . \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{cb86219c-e0b1-4f75-b8b2-50b5a233aa54-3_391_517_1352_813} \captionsetup{labelformat=empty} \caption{Fig. 4.2}
    \end{figure} The frustum can rotate freely about its axis of symmetry which is fixed and vertical.
  4. Show that the moment of inertia of the frustum about its axis of symmetry is \(0.0093 \mathrm {~kg} \mathrm {~m} ^ { 2 }\). The frustum is accelerated from rest for \(t\) seconds by a couple of magnitude 0.05 N m about its axis of symmetry, until it is rotating at \(10 \mathrm { rad } \mathrm { s } ^ { - 1 }\).
  5. Calculate \(t\).
  6. Find the position of G , the centre of mass of the frustum. The frustum (rotating at \(10 \mathrm { rad } \mathrm { s } ^ { - 1 }\) ) then receives an impulse tangential to the circumference of the larger circular face. This reduces its angular speed to \(5 \mathrm { rad } \mathrm { s } ^ { - 1 }\).
  7. To reduce its angular speed further, a parallel impulse of the same magnitude is now applied tangentially in the horizontal plane through G at the curved surface of the frustum. Calculate the resulting angular speed.
OCR MEI M4 2011 June Q1
1 A raindrop of mass \(m\) falls vertically from rest under gravity. Initially the mass of the raindrop is \(m _ { 0 }\). As it falls it loses mass by evaporation at a rate \(\lambda m\), where \(\lambda\) is a constant. Its motion is modelled by assuming that the evaporation produces no resultant force on the raindrop. The velocity of the raindrop is \(v\) at time \(t\). The forces on the raindrop are its weight and a resistance force of magnitude \(k m v\), where \(k\) is a constant.
  1. Find \(m\) in terms of \(m _ { 0 } , \lambda\) and \(t\).
  2. Write down the equation of motion of the raindrop. Solve this differential equation and hence show that \(v = \frac { g } { \lambda - k } \left( \mathrm { e } ^ { ( \lambda - k ) t } - 1 \right)\).
  3. Find the velocity of the raindrop when it has lost half of its initial mass.
OCR MEI M4 2011 June Q2
2 A small ring of mass \(m\) can slide freely along a fixed smooth horizontal rod. A light elastic string of natural length \(a\) and stiffness \(k\) has one end attached to a point A on the rod and the other end attached to the ring. An identical elastic string has one end attached to the ring and the other end attached to a point B which is a distance \(a\) vertically above the rod and a horizontal distance \(2 a\) from the point A . The displacement of the ring from the vertical line through B is \(x\), as shown in Fig. 2. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{0166dd50-5069-47f4-a015-d01a9c54faf4-2_405_1063_1270_539} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure}
  1. Find an expression for \(V\), the potential energy of the system when \(0 < x < a\), and show that $$\frac { \mathrm { d } V } { \mathrm {~d} x } = 2 k x - k a - \frac { k a x } { \sqrt { a ^ { 2 } + x ^ { 2 } } }$$
  2. Show that \(\frac { \mathrm { d } ^ { 2 } V } { \mathrm {~d} x ^ { 2 } }\) is always positive.
  3. Show that there is a position of equilibrium with \(\frac { 1 } { 2 } a < x < a\). State, with a reason, whether it is stable or unstable.
OCR MEI M4 2011 June Q3
3 A car of mass 800 kg moves horizontally in a straight line with speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at time \(t\) seconds. While \(v \leqslant 20\), the power developed by the engine is \(8 v ^ { 4 } \mathrm {~W}\). The total resistance force on the car is of magnitude \(8 v ^ { 2 } \mathrm {~N}\). Initially \(v = 2\) and the car is at a point O . At time \(t\) s the displacement from O is \(x \mathrm {~m}\).
  1. Find \(v\) in terms of \(x\) and show that when \(v = 20 , x = 100 \ln 1.9\).
  2. Find the relationship between \(t\) and \(x\), and show that when \(v = 20 , t \approx 19.2\). The driving force is removed at the instant when \(v\) reaches 20 .
  3. For the subsequent motion, find \(v\) in terms of \(t\). Calculate \(t\) when \(v = 2\).
OCR MEI M4 2011 June Q4
4 In this question you may assume without proof the standard results in Examination Formulae and Tables (MF2) for
  • the moment of inertia of a disc about an axis through its centre perpendicular to the disc,
  • the position of the centre of mass of a solid uniform cone.
Fig. 4 shows a uniform cone of radius \(a\) and height \(2 a\), with its axis of symmetry on the \(x\)-axis and its vertex at the origin. A thin slice through the cone parallel to the base is at a distance \(x\) from the vertex. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{0166dd50-5069-47f4-a015-d01a9c54faf4-3_497_748_1283_699} \captionsetup{labelformat=empty} \caption{Fig. 4}
\end{figure} The slice is taken to be a thin uniform disc of mass \(m\).
  1. Write down the moment of inertia of the disc about the \(x\)-axis. Hence show that the moment of inertia of the disc about the \(y\)-axis is \(\frac { 17 } { 16 } m x ^ { 2 }\).
  2. Hence show by integration that the moment of inertia of the cone about the \(y\)-axis is \(\frac { 51 } { 20 } M a ^ { 2 }\), where \(M\) is the mass of the cone. [You may assume without proof the formula for the volume of a cone.] The cone is now suspended so that it can rotate freely about a fixed, horizontal axis through its vertex. The axis of symmetry of the cone moves in a vertical plane perpendicular to the axis of rotation. The cone is released from rest when its axis of symmetry is at an acute angle \(\alpha\) to the downward vertical. At time \(t\), the angle the axis of symmetry makes with the downward vertical is \(\theta\).
  3. Use an energy method to show that \(\dot { \theta } ^ { 2 } = \frac { 20 g } { 17 a } ( \cos \theta - \cos \alpha )\).
  4. Hence, or otherwise, show that if \(\alpha\) is small the cone performs approximate simple harmonic motion and find the period. RECOGNISING ACHIEVEMENT
OCR MEI M4 2012 June Q1
1 A rocket in deep space has initial mass \(m _ { 0 }\) and is moving in a straight line at speed \(v _ { 0 }\). It fires its engine in the direction opposite to the motion in order to increase its speed. The propulsion system ejects matter at a constant mass rate \(k\) with constant speed \(u\) relative to the rocket. At time \(t\) after the engines are fired, the speed of the rocket is \(v\).
  1. Show that while mass is being ejected from the rocket, \(\left( m _ { 0 } - k t \right) \frac { \mathrm { d } v } { \mathrm {~d} t } = u k\).
  2. Hence find an expression for \(v\) at time \(t\).
OCR MEI M4 2012 June Q2
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.
OCR MEI M4 2012 June Q3
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]
\includegraphics[alt={},max width=\textwidth]{c4d3b482-5d09-4128-891d-4499fa49670c-3_339_563_534_737} \captionsetup{labelformat=empty} \caption{Fig. 3}
\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.
OCR MEI M4 2012 June Q4
4
  1. Show by integration that the moment of inertia of a uniform circular lamina of radius \(a\) and mass \(m\) about an axis perpendicular to the plane of the lamina and through its centre is \(\frac { 1 } { 2 } m a ^ { 2 }\). A closed hollow cylinder has its curved surface and both ends made from the same uniform material. It has mass \(M\), radius \(a\) and height \(h\).
  2. Show that the moment of inertia of the cylinder about its axis of symmetry is \(\frac { 1 } { 2 } M a ^ { 2 } \left( \frac { a + 2 h } { a + h } \right)\). For the rest of this question take the cylinder to have mass 8 kg , radius 0.5 m and height 0.3 m .
    The cylinder is at rest and can rotate freely about its axis of symmetry. It is given a tangential impulse of magnitude 55 Ns at a point on its curved surface. The impulse is perpendicular to the axis.
  3. Find the angular speed of the cylinder after the impulse. A resistive couple is now applied to the cylinder for 5 seconds. The magnitude of the couple is \(2 \dot { \theta } ^ { 2 } \mathrm { Nm }\), where \(\dot { \theta }\) is the angular speed of the cylinder in rad s \({ } ^ { - 1 }\).
  4. Formulate a differential equation for \(\dot { \theta }\) and hence find the angular speed of the cylinder at the end of the 5 seconds. The cylinder is now brought to rest by a constant couple of magnitude 0.03 Nm .
  5. Calculate the time it takes from when this couple is applied for the cylinder to come to rest.
OCR MEI M4 2013 June Q1
1 An empty railway truck of mass \(m _ { 0 }\) is moving along a straight horizontal track at speed \(v _ { 0 }\). The point P is at the front of the truck. The horizontal forces on the truck are negligible. As P passes a fixed point O , sand starts to fall vertically into the truck at a constant mass rate \(k\). At time \(t\) after P passes O the speed of the truck is \(v\) and \(\mathrm { OP } = x\).
  1. Find an expression for \(v\) in terms of \(m _ { 0 } , v _ { 0 } , k\) and \(t\), and show that \(x = \frac { m _ { 0 } v _ { 0 } } { k } \ln \left( 1 + \frac { k t } { m _ { 0 } } \right)\).
  2. Find the speed of the truck and the distance OP when the mass of sand in the truck is \(2 m _ { 0 }\).
OCR MEI M4 2013 June Q2
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]
\includegraphics[alt={},max width=\textwidth]{bc637a95-b469-493b-8fd4-d3b12049878b-2_328_732_1032_667} \captionsetup{labelformat=empty} \caption{Fig. 2.1}
\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.
OCR MEI M4 2013 June Q3
3 A model car of mass 2 kg moves from rest along a horizontal straight path. After time \(t \mathrm {~s}\), the velocity of the car is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The power, \(P \mathrm {~W}\), developed by the engine is initially modelled by \(P = 2 v ^ { 3 } + 4 v\). The car is subject to a resistance force of magnitude \(6 v \mathrm {~N}\).
  1. Show that \(\frac { \mathrm { d } v } { \mathrm {~d} t } = ( 1 - v ) ( 2 - v )\) and hence show that \(t = \ln \frac { 2 - v } { 2 ( 1 - v ) }\).
  2. Hence express \(v\) in terms of \(t\). Once the power reaches 4.224 W it remains at this constant value with the resistance force still acting.
  3. Verify that the power of 4.224 W is reached when \(v = 0.8\) and calculate the value of \(t\) at this instant.
  4. Find \(v\) in terms of \(t\) for the motion at constant power. Deduce the limiting value of \(v\) as \(t \rightarrow \infty\).
OCR MEI M4 2013 June Q4
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.
OCR MEI M4 2014 June Q1
1 A sports car of mass 1.2 tonnes is being tested on a horizontal, straight section of road. After \(t \mathrm {~s}\), the car has travelled \(x \mathrm {~m}\) from the starting line and its velocity is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The engine produces a driving force of 4000 N and the total resistance to the motion of the car is given by \(\frac { 40 } { 49 } v ^ { 2 } \mathrm {~N}\). The car crosses the starting line with speed \(10 \mathrm {~m} \mathrm {~s} ^ { - 1 }\).
  1. Write down an equation of motion for the car and solve it to show that \(v ^ { 2 } = 4900 - 4800 \mathrm { e } ^ { - \frac { 1 } { 735 } x }\).
  2. Hence find the work done against the resistance to motion over the first 100 m beyond the starting line.
OCR MEI M4 2014 June Q2
2 On a building site, a pulley system is used for moving small amounts of material up to roof level. A light pulley, which can rotate freely, is attached with its axis horizontal to the top of some scaffolding. A light inextensible rope hangs over the pulley with a counterweight of mass \(m _ { 1 } \mathrm {~kg}\) attached to one end. Attached to the other end of the rope is a bag of negligible mass into which \(m _ { 2 } \mathrm {~kg}\) of roof tiles are placed, where \(m _ { 2 } < m _ { 1 }\). This situation is shown in Fig. 2. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{c3ac9277-d34d-4d0e-9f9b-d0bce8c741af-2_554_711_1098_678} \captionsetup{labelformat=empty} \caption{Fig. 2}
\end{figure} Initially the system is held at rest with the rope taut, the counterweight at the top of the scaffolding and the bag of tiles on the ground. When the counterweight is released, the bag ascends towards the top of the scaffolding. At time \(t \mathrm {~s}\) the velocity of the counterweight is \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) downwards. The counterweight is made from a bag of negligible mass filled with sand. At the moment the counterweight is released, this bag is accidentally ripped and after this time the sand drops out at a constant rate of \(\lambda \mathrm { kg } \mathrm { s } ^ { - 1 }\).
  1. Find the equation of motion for the counterweight while it still contains sand, and hence show that $$v = g t + \frac { 2 g m _ { 2 } } { \lambda } \ln \left( 1 - \frac { \lambda t } { m _ { 1 } + m _ { 2 } } \right) .$$
  2. Given that the sand would run out after 10 seconds and that \(m _ { 2 } = \frac { 4 } { 5 } m _ { 1 }\), find the maximum velocity attained by the counterweight towards the ground. You may assume that the scaffolding is sufficiently high that the counterweight does not hit the ground before this velocity is reached.