6.02d Mechanical energy: KE and PE concepts

In this question use \(g = 9.8 \text{ m s}^{-2}\) A ball of mass 0.5 kg is projected vertically upwards with a speed of \(10 \text{ m s}^{-1}\)

1. Calculate the initial kinetic energy of the ball. [1 mark]
2. Assuming that the weight is the only force acting on the ball, use an energy method to show that the maximum height reached by the ball is approximately 5.1 m above the point of projection. [2 marks]
3. 1. A student conducts an experiment to verify the accuracy of the result obtained in part (b). They observe that the ball rises to a height of 4.4 m above the point of projection and concludes that this height difference is due to a resistance force, \(R\) newtons. Find the total work done against \(R\) whilst the ball is moving upwards. [2 marks]
   2. Using a model that assumes \(R\) is constant, find the magnitude of \(R\) [2 marks]
   3. Comment on the validity of the model used in part (c)(ii). [1 mark]

A ball of mass \(m\) kg is held at rest at a height \(h\) metres above a horizontal surface. The ball is released and bounces on the surface. The coefficient of restitution between the ball and the surface is \(e\) Prove that the kinetic energy lost during the first bounce is given by $$mgh(1 - e^2)$$ [4 marks]

A particle \(P\) of mass \(3\) kg is moving on a smooth horizontal surface under the influence of a variable horizontal force \(\mathbf{F}\) N. At time \(t\) seconds, where \(t \geqslant 0\), the velocity of \(P\), \(\mathbf{v}\) m s\(^{-1}\), is given by $$\mathbf{v} = (32\sinh(2t))\mathbf{i} + (32\cosh(2t) - 257)\mathbf{j}.$$

1. 1. By considering kinetic energy, determine the work done by \(\mathbf{F}\) over the interval \(0 \leqslant t \leqslant \ln 2\). [5]
   2. Explain the significance of the sign of the answer to part (a)(i). [1]
2. Determine the rate at which \(\mathbf{F}\) is working at the instant when \(P\) is moving parallel to the \(\mathbf{i}\)-direction. [6]

A particle P of mass 4 kilograms moves in such a way that its position vector at time \(t\) seconds is \(\mathbf{r}\) metres, where $$\mathbf{r} = 3t\mathbf{i} + 2e^{-3t}\mathbf{j}.$$

1. Find the initial kinetic energy of P. [4]
2. Find the time when the acceleration of P is 2 metres per second squared. [3]

\includegraphics{figure_9} Fig. 9 shows the instant of impact of two identical uniform smooth spheres, A and B, each with mass \(m\). Immediately before they collide, the spheres are sliding towards each other on a smooth horizontal table in the directions shown in the diagram, each with speed \(v\). The coefficient of restitution between the spheres is \(\frac{1}{2}\).

1. Show that, immediately after the collision, the speed of A is \(\frac{1}{8}v\). Find its direction of motion. [6]
2. Find the percentage of the original kinetic energy that is lost in the collision. [7]
3. State where in your answer to part (i) you have used the assumption that the contact between the spheres is smooth. [1]

The diagram below shows a woman standing at the end of a diving platform. She is about to dive into the water below. The woman has mass 60 kg and she may be modelled as a particle positioned at the end of the platform which is 10 m above the water. \includegraphics{figure_2} When the woman dives, she projects herself from the platform with a speed of \(7.8\text{ ms}^{-1}\).

1. Find the kinetic energy of the woman when she leaves the platform. [2]
2. Initially, the situation is modelled ignoring air resistance. By using conservation of energy, show that the model predicts that the woman enters the water with an approximate speed of \(16\text{ ms}^{-1}\). [6]
3. Suppose that this model is refined to include air resistance so that the speed with which the woman enters the water is now predicted to be \(13\text{ ms}^{-1}\). Determine the amount of energy lost to air resistance according to the refined model. [3]

By burning a charge, a cannon fires a cannon ball of mass 12 kg horizontally. As the cannon ball leaves the cannon, its speed is 600 ms\(^{-1}\). The recoiling part of the cannon has a mass of 1600 kg.

1. Determine the speed of the recoiling part immediately after the cannon ball leaves the cannon. [3]
2. Find the energy created by the burning of the charge. State any assumption you have made in your solution and briefly explain how the assumption affects your answer. [5]
3. Calculate the constant force needed to bring the recoiling part to rest in 1.2 m. State, with a reason, whether your answer is an overestimate or an underestimate of the actual force required. [4]

Relative to a fixed origin \(O\), the position vector \(\mathbf{r}\) m at time \(t\) s of a particle \(P\), of mass 0.4 kg, is given by $$\mathbf{r} = e^{2t}\mathbf{i} + \sin(2t)\mathbf{j} + \cos(2t)\mathbf{k}.$$

1. Show that the velocity vector \(\mathbf{v}\) and the position vector \(\mathbf{r}\) are never perpendicular to each other. [6]
2. Given that the speed of \(P\) at time \(t\) is \(v\) ms\(^{-1}\), show that $$v^2 = 4e^{4t} + 4.$$ [2]
3. Find the kinetic energy of \(P\) at time \(t\). [1]
4. Calculate the work done by the force acting on \(P\) in the interval \(0 < t < 1\). [2]
5. Determine an expression for the rate at which the force acting on \(P\) is working at time \(t\). [2]

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 point \(O\) is situated a distance \(h\) above a smooth horizontal plane, and a particle \(A\) of mass \(m\) is attached to \(O\) by a light inextensible string of length \(h\). A particle \(B\) of mass \(2m\) is at rest on the plane, directly below \(O\), and is attached to a point \(C\) on the plane, where \(BC = l\), by a light inextensible string of length \(l\). \(A\) is released from rest with the string \(OA\) taut and making an acute angle \(\theta\) with the downward vertical (see diagram). \includegraphics{figure_8} \(A\) moves in a vertical plane perpendicular to \(CB\) and collides directly with \(B\). As a result of this collision, \(A\) is brought to rest and \(B\) moves on the plane in a horizontal circle with centre \(C\). After \(B\) has made one complete revolution the particles collide again.

1. Show that, on the next occasion that \(A\) comes to rest, the string \(OA\) makes an angle \(\phi\) with the downward vertical through \(O\), where \(\cos \phi = \frac{3 + \cos \theta}{4}\). [9]

\(A\) and \(B\) collide again when \(AO\) is next vertical.

1. Find the percentage of the original energy of the system that remains immediately after this collision. [5]
2. Explain why the total momentum of the particles immediately before the first collision is the same as the total momentum of the particles immediately after the second collision. [1]
3. Explain why the total momentum of the particles immediately before the first collision is different from the total momentum of the particles immediately after the third collision. [1]

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 \(21 v ^ { 2 } = 220 x\).