6.02i Conservation of energy: mechanical energy principle

The diagram shows a slide, \(ABC\), at a water park. The shape of the slide may be modelled by two circular arcs, \(AB\) and \(BC\), in the same vertical plane. Arc \(AB\) has radius \(7\) m and subtends an angle \(\alpha\) at its centre \(D\), where \(\cos \alpha = \frac{9}{10}\). Arc \(BC\) has radius \(5\) m and subtends an angle of \(45°\) at its centre, \(O\). The straight line \(DBO\) is vertical. \includegraphics{figure_6} Users of the slide are required to sit in a rubber ring and are released from rest at point \(A\). A girl decides to use the slide. The combined mass of the girl and the rubber ring is \(50\) kg.

1. When the rubber ring is at a point \(P\) on the circular arc \(BC\), its speed is \(v\) ms\(^{-1}\) and \(OP\) makes an angle \(\theta\) with the upward vertical.
   1. Show that \(v^2 = 111.72 - 98\cos\theta\). [4]
   2. Find, in terms of \(\theta\), the reaction between the rubber ring and the slide at \(P\). [4]
   3. Show that, according to this model, the rubber ring loses contact with the slide before reaching \(C\). [3]
   4. In reality, there will be resistive forces opposing the motion of the rubber ring. Explain how this fact will affect your answer to (iii). [1]
2. Show that the rubber ring will remain in contact with the slide along the arc \(AB\). [3]

A particle \(P\), of mass 3 kg, is attached to a fixed point \(O\) by a light inextensible string of length 4 m. Initially, particle \(P\) is held at rest at a point which is \(2\sqrt{3}\) m horizontally from \(O\). It is then released and allowed to fall under gravity.

1. Show that the speed of \(P\) when it first begins to move in a circle is \(\sqrt{3g}\). [4]
2. In the subsequent motion, when the string first makes an angle of 45° with the downwards vertical,
   1. calculate the speed \(v\) of \(P\),
   2. determine the tension in the string. [8]

\includegraphics{figure_7} A smooth wire is shaped into a circle of radius 2.5 m which is fixed in a vertical plane with its centre at a point \(O\). A small bead \(B\) is threaded onto the wire. \(B\) is held with \(OB\) vertical and is then projected horizontally with an initial speed of \(8.4 \mathrm{ms}^{-1}\) (see diagram).

1. Find the speed of \(B\) at the instant when \(OB\) makes an angle of 0.8 radians with the downward vertical through \(O\). [3]
2. Determine whether \(B\) has sufficient energy to reach the point on the wire vertically above \(O\). [3]

\includegraphics{figure_8} As shown in the diagram, \(AB\) is a long thin rod which is fixed vertically with \(A\) above \(B\). One end of a light inextensible string of length 1 m is attached to \(A\) and the other end is attached to a particle \(P\) of mass \(m_1\) kg. One end of another light inextensible string of length 1 m is also attached to \(P\). Its other end is attached to a small smooth ring \(R\), of mass \(m_2\) kg, which is free to move on \(AB\). Initially, \(P\) moves in a horizontal circle of radius 0.6 m with constant angular velocity \(\omega \mathrm{rad s}^{-1}\). The magnitude of the tension in string \(AP\) is denoted by \(T_1\) N while that in string \(PR\) is denoted by \(T_2\) N.

1. By considering forces on \(R\), express \(T_2\) in terms of \(m_2\). [2]
2. Show that
   1. \(T_1 = \frac{49}{4}(m_1 + m_2)\), [2]
   2. \(\omega^2 = \frac{49(m_1 + 2m_2)}{4m_1}\). [3]
3. Deduce that, in the case where \(m_1\) is much bigger than \(m_2\), \(\omega \approx 3.5\). [2] In a different case, where \(m_1 = 2.5\) and \(m_2 = 2.8\), \(P\) slows down. Eventually the system comes to rest with \(P\) and \(R\) hanging in equilibrium.
4. Find the total energy lost by \(P\) and \(R\) as the angular velocity of \(P\) changes from the initial value of \(\omega \mathrm{rad s}^{-1}\) to zero. [5]

\includegraphics{figure_4} As shown in the diagram, \(AB\) is a long thin rod which is fixed vertically with \(A\) above \(B\). One end of a light inextensible string of length \(1\) m is attached to \(A\) and the other end is attached to a particle \(P\) of mass \(m_1\) kg. One end of another light inextensible string of length \(1\) m is also attached to \(P\). Its other end is attached to a small smooth ring \(R\), of mass \(m_2\) kg, which is free to move on \(AB\). Initially, \(P\) moves in a horizontal circle of radius \(0.6\) m with constant angular velocity \(\omega\) rad s\(^{-1}\). The magnitude of the tension in string \(AP\) is denoted by \(T_1\) N while that in string \(PR\) is denoted by \(T_2\) N.

1. By considering forces on \(R\), express \(T_2\) in terms of \(m_2\). [2]
2. Show that
   1. \(T_1 = \frac{4g}{5}(m_1 + m_2)\). [2]
   2. \(\omega^2 = \frac{4g(m_1 + 2m_2)}{4m_1}\). [3]
3. Deduce that, in the case where \(m_1\) is much bigger than \(m_2\), \(\omega \approx 3.5\). [2]

In a different case, where \(m_1 = 2.5\) and \(m_2 = 2.8\), \(P\) slows down. Eventually the system comes to rest with \(P\) and \(R\) hanging in equilibrium.

1. Find the total energy lost by \(P\) and \(R\) as the angular velocity of \(P\) changes from the initial value of \(\omega\) rad s\(^{-1}\) to zero. [5]

At a demolition site, bricks slide down a straight chute into a container. The chute is rough and is inclined at an angle of \(30°\) to the horizontal. The distance travelled down the chute by each brick is \(8\) m. A brick of mass \(3\) kg is released from rest at the top of the chute. When it reaches the bottom of the chute, its speed is \(5\) m s\(^{-1}\).

1. Find the potential energy lost by the brick in moving down the chute.

(2)

1. By using the work-energy principle, or otherwise, find the constant frictional force acting on the brick as it moves down the chute.

(5)

1. Hence find the coefficient of friction between the brick and the chute.

(3) Another brick of mass \(3\) kg slides down the chute. This brick is given an initial speed of \(2\) m s\(^{-1}\) at the top of the chute.

1. Find the speed of this brick when it reaches the bottom of the chute.

(5)

A bungee jumper of mass 80 kg steps off a high bridge with an elastic rope attached to her ankles. She is assumed to fall vertically from rest and the air resistance she experiences is modelled as a constant force of 32N. The rope has natural length 4 m and modulus of elasticity of 470 N. By considering energy, determine the total distance she falls before first coming to instantaneous rest. [6]

\includegraphics{figure_3} A light elastic string has natural length \(8a\) and modulus of elasticity \(5mg\). A particle \(P\) of mass \(m\) is attached to the midpoint of the string. The ends of the string are attached to points \(A\) and \(B\) which are a distance \(12a\) apart on a smooth horizontal table. The particle \(P\) is held on the table so that \(AP = BP = L\) (see diagram). The particle \(P\) is released from rest. When \(P\) is at the midpoint of \(AB\) it has speed \(\sqrt{80ag}\).

1. Find \(L\) in terms of \(a\). [5]
2. Find the initial acceleration of \(P\) in terms of \(g\). [3]

A particle \(P\) of mass 5.6 kg is attached to one end of a light rod of length 2.1 m. The other end of the rod is freely hinged to a fixed point \(O\). The particle is initially at rest directly below \(O\). It is then projected horizontally with speed \(5 \text{ ms}^{-1}\). In the subsequent motion, the angle between the rod and the downward vertical at \(O\) is denoted by \(\theta\) radians, as shown in the diagram. \includegraphics{figure_2}

1. Find the speed of \(P\) when \(\theta = \frac{1}{4}\pi\). [4]
2. Find the value of \(\theta\) when \(P\) first comes to instantaneous rest. [2]

\(A\) and \(B\) are two points a distance of 5 m apart on a horizontal ceiling. A particle \(P\) of mass \(m\) kg is attached to \(A\) and \(B\) by light elastic strings. The particle hangs in equilibrium at a distance of 4 m from \(A\) and 3 m from \(B\) so that angle \(APB = 90°\) (see diagram). \includegraphics{figure_4} The string joining \(P\) to \(A\) has natural length 2 m and modulus of elasticity \(\lambda_A\) N. The string joining \(P\) to \(B\) also has natural length 2 m but has modulus of elasticity \(\lambda_B\) N.

1. 1. Show that \(\lambda_B = \frac{3}{4}\lambda_A\). [4]
   2. Find an expression for \(\lambda_A\) in terms of \(m\) and \(g\). [3]
2. Find, in terms of \(m\) and \(g\), the total elastic potential energy stored in the strings. [2]

The string joining \(P\) to \(A\) is detached from \(A\) and a second particle, \(Q\), of mass \(0.3m\) kg is attached to the free end of the string. \(Q\) is then gently lowered into a position where the system hangs vertically in equilibrium.

1. Find the distance of \(Q\) below \(B\) in this equilibrium position. [4]

\includegraphics{figure_10} \(A\) and \(B\) are points at the upper and lower ends, respectively, of a line of greatest slope on a plane inclined at \(30°\) to the horizontal. The distance \(AB\) is \(20\text{m}\). \(M\) is a point on the plane between \(A\) and \(B\). The surface of the plane is smooth between \(A\) and \(M\), and rough between \(M\) and \(B\). A particle \(P\) is projected with speed \(4.2\text{m s}^{-1}\) from \(A\) down the line of greatest slope (see diagram). \(P\) moves down the plane and reaches \(B\) with speed \(12.6\text{m s}^{-1}\). The coefficient of friction between \(P\) and the rough part of the plane is \(\frac{\sqrt{3}}{6}\).

1. Find the distance \(AM\). [8]
2. Find the angle between the contact force and the downward direction of the line of greatest slope when \(P\) is in motion between \(M\) and \(B\). [3]

One end of a light spring of length 0.5 m is attached to a fixed point \(F\). A particle \(P\) of mass 2.5 kg is attached to the other end of the spring and hangs in equilibrium 0.55 m below \(F\). Another particle \(Q\), of mass 1.5 kg, is attached to \(P\), without moving it, and both particles are then released.

1. Show that the modulus of elasticity of the spring is 250 N. [2]
2. Prove that the motion is simple harmonic. [4]
3. Find
   1. the amplitude of the motion, [1]
   2. the greatest speed of the particles, [1]
   3. the period of the motion, [1]
   4. the time taken for the spring to be extended by 0.1 m for the first time. [3]

One end of a light inextensible string of length \(l\) is attached to a fixed point \(O\). The other end is attached to a particle \(P\) of mass \(m\). The particle hangs at rest vertically below \(O\). The particle is then given a horizontal speed \(u\).

1. 1. Show that when \(OP\) has turned through an angle \(\theta\) the tension in the string is given by $$T = mg(3\cos \theta - 2) + \frac{mu^2}{l}$$ as long as the string remains taut. [5]
   2. Deduce that \(u^2 \geq 5gl\) in order for the particle to perform complete circles. [1]
2. 1. In the case \(u^2 = 3gl\), find the angle that \(OP\) makes with the downward vertical at \(O\) at the instant when the string becomes slack. [2]
   2. Describe the nature of the motion while the string is slack. [1]

A light elastic string of natural length \(2a\) and modulus of elasticity \(\lambda\) is stretched between two points \(A\) and \(B\), which are \(3a\) apart on a smooth horizontal table. A particle of mass \(m\) is attached to the mid-point of the string, pulled aside to \(A\) and released.

1. Prove that, while one part of the string is taut and the other part is slack, the particle is describing simple harmonic motion. [2]
2. Find the speed of the particle when the slack part of the string becomes taut. [2]
3. Prove that the total time for the particle to reach the mid-point of the string for the first time is $$\sqrt{\frac{ma}{\lambda}} \left( \frac{\pi}{3} + \frac{1}{\sqrt{2}} \sin^{-1} \frac{1}{\sqrt{7}} \right).$$ [8]

3 A light elastic string has natural length 0.8 m and modulus of elasticity 16 N . One end of the string is attached to a fixed point \(O\), and a particle \(P\) of mass 0.4 kg is attached to the other end of the string. The particle \(P\) hangs in equilibrium vertically below \(O\).

1. Show that the extension of the string is 0.2 m . \(P\) is projected vertically downwards from the equilibrium position. \(P\) first comes to instantaneous rest at the point where \(O P = 1.4 \mathrm {~m}\).
2. Calculate the speed at which \(P\) is projected.
3. Find the speed of \(P\) at the first instant when the string subsequently becomes slack.

2 A particle \(P\) of mass 0.3 kg is attached to one end of a light elastic string of natural length 0.6 m and modulus of elasticity 45 N . The other end of the string is attached to a fixed point \(O\). The particle \(P\) is released from rest at \(O\) and falls vertically. Find the extension of the string when \(P\) is at its lowest position.

7 A small ball \(B\) of mass 0.2 kg moves in a narrow fixed smooth cylindrical tube \(O A\) of length 1 m , closed at the end \(A\). When the ball has displacement \(x \mathrm {~m}\) from \(O\), it has velocity \(v \mathrm {~ms} ^ { - 1 }\) in the direction \(O A\) and experiences a resisting force of magnitude \(\frac { k } { 1 - x } \mathrm {~N}\).

1. \includegraphics[max width=\textwidth, alt={}, center]{10abedc3-c814-47c0-8ed4-849ef325feca-4_186_805_488_715} The tube is fixed in a horizontal position and \(B\) is projected from \(O\) towards \(A\) with velocity \(1.2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) (see diagram). Given that \(B\) comes to instantaneous rest after travelling 0.55 m , show that \(k = 0.1803\), correct to 4 significant figures.
2. The tube is now fixed in a vertical position with \(O\) above \(A\). The ball \(B\) is released from rest at \(O\). Calculate the speed of \(B\) after it has descended 0.1 m . \end{document}

5 \includegraphics[max width=\textwidth, alt={}, center]{77976dad-c055-45fd-93fe-e37fa8e9ae22-3_343_691_254_725} A light inextensible rope has a block \(A\) of mass 5 kg attached at one end, and a block \(B\) of mass 16 kg attached at the other end. The rope passes over a smooth pulley which is fixed at the top of a rough plane inclined at an angle of \(30 ^ { \circ }\) to the horizontal. Block \(A\) is held at rest at the bottom of the plane and block \(B\) hangs below the pulley (see diagram). The coefficient of friction between \(A\) and the plane is \(\frac { 1 } { \sqrt { 3 } }\). Block \(A\) is released from rest and the system starts to move. When each of the blocks has moved a distance of \(x \mathrm {~m}\) each has speed \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\).

1. Write down the gain in kinetic energy of the system in terms of \(v\).
2. Find, in terms of \(x\),
   (a) the loss of gravitational potential energy of the system,
   (b) the work done against the frictional force.
3. Show that \(21 v ^ { 2 } = 220 x\).