Presentation on theme: "Unit 4 Energy and Momentum"— Presentation transcript:
1Unit 4 Energy and Momentum This unit is a continuation of the study of mechanics that trainees studied in Units 2 and 3. It involves looking at mechanics from a different point of view (using work end energy) that can simplify some otherwise complicated situations.
2Module 4.1 – Momentum and Impulse Momentum and Impulse is another way of looking at the motion of objects, and is a continuation of the last unit. There are some situations where it is difficult to determine all of the individual forces acting on the objects. Looking at these situations from the point of view of momentum and impulse can allow us to more easily analyze the interaction between objects and to predict their behaviour.
3Module 4.1 ObjectivesTo be able to calculate the momentum of an object.To understand the relationship between change in momentum and impulse.To be able to apply conservation of momentum to a variety of one-dimensional collisions and explosions.
4Momentum Momentum depends on mass and velocity Momentum is a vector and is defined as:where momentum p is measured in kg·m/s when mass is measured in kg and velocity is measured in m/s
5ExampleCalculate the momentum of a 1200 kg car travelling at 90.0 km/h north.Solution:
6Check Your LearningWhich has more momentum, a falling raindrop or a transport truck parked along the side of the road?Since the truck is not moving (using the Earth as the frame of reference), it has zero momentum; the raindrop does have momentum, since it is actually moving. Even though the truck has a much larger mass, it does not have any momentum.
7Check Your LearningCalculate the momentum of a kg cricket ball that has a velocity of 110 km/h.The positive sign on the momentum indicates the direction of the momentum. The velocity could also be left in km/h, which would result in a momentum of
8Change in Momentum Change in velocity causes change in momentum This product is referred to as the impulse
9Impulse Change in Momentum and Impulse are equivalent: an impulse of 1 N·s will provide a change in momentum of 1 kg·m/s.
10CollisionsCollisions usually occur in a short time; force changes quicklyThe area of the graph represents the impulse
11Application Smaller time Larger force Larger time Smaller force since impulse is usually fixed
12Example 1An egg when thrown against a brick wall will break; however, an egg thrown at the same speed into a sagging sheet will likely survive. Explain why using the concepts of impulse and momentum.Solution:The change in momentum of the egg is the same in each case, since the mass of the egg does not change and the egg goes from the same initial velocity to rest in each case. Since the change in momentum is the same, the impulse must also be the same; however, the brick wall applies the impulse in a very short time, which means the force must be large. The large force causes the egg to break. The sheet brings the egg to a stop more gradually, thereby lengthening the time that the impulse is applied over. This allows a smaller force to be applied to the egg and does not cause the egg to break.
13Example 2A tennis player hits a ball with a racket, giving it a speed of 43 m/s. If the ball has a mass of kg and is in contact with the racket for 4.5×10-3 s, what is the average force exerted on the ball by the racket?
14Check Your LearningA 0.15 kg cricket ball is approaching a bat with a speed of 95 km/h. After it is hit by the bat, it has a speed in the opposite direction of 115 km/h. The interaction between the bat and the ball lasted s.What was the change in momentum of the ball?It is important to remember that since the ball changes direction, the final velocity must be made negative (if we take the ball’s original direction to be positive)
15Check Your Learning What impulse was applied to the ball? Since impulse is equivalent to change in momentum,What was the average force exerted on the ball?
17Conservation of Momentum Law of Conservation of Momentum - the total momentum of an isolated system remains constant.
18Example 1Someone throws a heavy ball to you when you are standing on a skateboard. You catch it and roll backward with the skateboard. If you were standing on the ground, however, you would be able to avoid moving. Explain both using momentum conservation.Solution:Consider first the case of the person standing on the skateboard when he catches the ball. In this case there is very little friction, so little that it can be ignored compared to the force exerted on the person by the heavy ball. If we define our system to be the ball and the person on the skateboard, conservation of momentum says that the total momentum must stay the same. Before the collision of the ball with the person, the ball has momentum since it is moving. After the person catches the ball, the ball and the person act as one object. The combined momentum of the ball and the person must still be equal to the momentum of the ball before the collision; therefore, the person and the ball must be moving.
19Example 1Now consider the case of the person standing on the ground. If we again define our system to be the ball and the person, there is now a significant outside force acting on the person since friction is now increased. Since there is a significant outside force, the system is no longer isolated and conservation of momentum does not apply. We can, however, define a different system where conservation of momentum will apply. If we now consider our system to be the ball, the person, and the Earth then the friction between the person and the Earth is now an internal force and our system is again isolated. Now the momentum of the ball, person, and the Earth (since they can be considered to be one object after the collision) will be the same as the momentum of the ball before the collision; however, because this combined mass is so much larger than the mass of the ball alone the velocity will be so small that it will be negligible.
20Example 2In the previous example of the person catching the ball while on the skateboard, assume that the ball has a mass of 6.0 kg and is thrown with a speed of 3.4 m/s. The combined mass of the person and the skateboard is 68.0 kg and the person is initially at rest. Find the velocity of the person on the skateboard after catching the ball. Solution: Before the collision, only the ball is moving. After the collision, the ball, person, and the skateboard are moving as one object. It is the total velocity of this combined mass that we are trying to find
21Example 3The person on the skateboard is now standing on the skateboard at rest and holding the heavy ball. The mass of the person and the skateboard is still 68.0 kg and the mass of the ball is still 6.0 kg. The person throws the ball, giving it a velocity of 4.2 m/s to the right, and is amazed to see that he begins rolling to the left. What is the person’s velocity after throwing the ball?Solution:This is an example of an “explosion”. The initial momentum of the system is zero, and must remain at zero since it is an isolated system. We will take right to be the positive direction.
23Check Your LearningA 455 g rubber ball is thrown at a 2.10 kg block with a speed of 5.0 m/s. Assume that the block is at rest on a frictionless surface. The rubber ball rebounds in the opposite direction with a speed of 3.2 m/s. What is the speed of the block after the collision?
24Module Summary In this module, you have learned that Momentum is calculated using the formulaThe change in momentum can be calculated from the change in velocityImpulse is equal to the change in momentum and is given by the formula
25Module SummaryMomentum is conserved in all isolated systems. This is known as the law of conservation of momentum
26Module 4.2 Work, Power and Energy Until now, forces and acceleration have been used to describe and understand motion. We will now look at another model that is used to analyze motion – work and energy. This new model allows us to study situations where the forces are changing throughout the problem, and simplifies some otherwise complex situations. You will also learn in this module what machines can and cannot do to help us to complete tasks.
27Transfer of Mechanical Energy Potential EnergyKinetic EnergyWorkTransfer of Mechanical Energy
28Work Requirements Work requires 3 things: There must be a force There must be a displacementThe displacement must be a result of (or caused by) the force.
30Special CasesForce and Displacement in the same direction (θ = 0°, cos θ = 1)
31Example 1A person uses their hand to lift a 5.3 kg box 1.2 m into the air (at a constant speed). How much work did the person do? Solution:
32Special CasesForce and Displacement are perpendicular to each other (θ = 90°, cos θ = 0)
33Example 2A person walks 3.2 m across the room carrying a 5.3 kg box at a constant speed. How much work was done by the person?Solution:
34Special CasesForce and Displacement are in opposite directions (θ = 180°, cos θ = -1)
35Example 3A person holding a 5.3 kg box in the air lowers the box 1.2 m. How much work is done by the person? Solution:
36Check Your LearningA 12 kg box is being pulled by Bill across the floor. Bill is exerting a force of 45 N, and the coefficient of friction is He pulls the box a distance of 7.2 m.How much work was done by Bill?
37Check Your LearningHow much work was done by friction?
38Check Your LearningCalculate the net work done on the box.orAlthough Bill did 320 J of work, only 140 J of it was actually transferred to the box – the rest was lost to friction as heat energy
39Example 1Consider a 75 kg person who walks up a flight of stairs (shown below) compared to a similar person (same mass) who runs up the stairs. The person walking takes 6.3 s, and the person running takes 2.6 s. Assume both are moving at a constant speed. How much work does each person do?
40Example 1 - Solution Person 1 Since time was not used, person 2 does the same amount of work!!
41PowerSince Person 2 did the work more quickly, we say that they used more power.Power is defined as the rate of doing work (or transferring energy):1 watt (W) = 1 J/s
45Check Your LearningA person is pulling a sled that has a mass of 93 kg. The sled starts from rest and you can assume that there is no friction. The person is pulling with a horizontal force of 225 N over a distance of 12.0 m. What was the average power developed by the person?We must first find the work done by the person:
46Check Your LearningWe must now use the rest of the information to find the time so that we can get the power:
47Kinetic Energy Energy = ability to do work Kinetic Energy is energy of motionConsider the case where a car with an initial velocity is being pushed horizontally along a table with a net force. Since there is a net force in the direction of motion, there is net work being done.
48Kinetic EnergySinceIf we define kinetic energy to be
50Example 1A kg rugby ball is thrown with a horizontal speed of 15.0 m/s.What is its kinetic energy?What was the work done by the person throwing the ball?Solution:
51Example 1The only horizontal force acting on the ball was the one exerted by the person, so the work done by the person is the net work.
52Example 2How much work is required to accelerate a 1250 kg car from 50.0 km/h to 95.0 km/h?Solution:
53Check Your LearningA kg car is travelling at a speed of 50.0 km/h when the driver applies the brakes. The car comes to a stop after travelling 15 m.How much work was done by friction in stopping the car?
54Check Your LearningWhat was the magnitude of the average frictional force applied to the car?
55Check Your LearningIf the same car were travelling at 100. km/h when the driver applied the brakes and the car experienced the same frictional force, how far would the car travel before coming to a stop?Notice that by doubling the car’s speed, the stopping distance is multiplied by four.
56Gravitational Potential Energy Potential energy is stored energy – elastic, chemical, gravitationalAn object with gravitational potential energy has the ability to do work if it is droppedSuppose that we are lifting an object vertically at a constant speed; we must exert an upward force equal to the force of gravity
57Gravitational Potential Energy If we define Gravitational Potential Energy to beThen the work done by the person is nowHeight h used in the calculation of gravitational potential energy must be measured from some reference level, since there is no such thing as an absolute zero position or potential energy.
58Example 1A 225 g book is sitting on a table that is 1.2 m above the floor. It must be lifted to a shelf that is 2.1 m above the floor.What is the book’s final potential energy relative to the floor?What is the book’s final potential energy relative to the table?How much work was required to lift the book to the shelf?
60SolutionWe will choose the floor to be our reference level:
61Work-Energy Theorem System – an object or group of objects Any force exerted by an object in the system on another object in the system is an internal force; any force exerted by an object that is not a part of the system is an external force.Consider a person lifting a box with a force greater than the weight of the box – the potential energy will be increasing (because the box is changing height) and the kinetic energy will be changing because the box will be accelerating.
62Work-Energy Theorem No Friction Define our system to be the box and the Earth and we will ignore friction for now. The force that is being applied by the person is considered an external force.No Friction
63Example 2An 1125 kg elevator is being lifted to a height of 12.5 m from rest. If the cable on the elevator exerts an upward force of 1.2 x104 N what is the final speed of the elevator?Solution:
65Check Your LearningA 34.0 kg child cycles up a hill to a point that is 6.45 m above her starting point. Findthe change in the child’s gravitational potential energy.We will use her starting point as the reference level,
66Check Your Learningthe amount of work done by the child against gravity.
67Mechanical AdvantageSimple machines allow people to use less force in completing tasks, not less workActual Mechanical Advantage (AMA) is defined as:Ideal MachineNo FrictionWork input is equal to work output
69Mechanical AdvantageSmaller force must be applied over larger distance so that the work is the same!MA of 2 means that we apply half the force over twice the distance – work done is the same
70Example 1An ideal machine allows the user to exert a force of 350 N in order to lift a 110 kg mass to a height of 5.2 m.What is the mechanical advantage of this system?Over what distance must the input force be applied?Solution:
71Example 1Since the machine is ideal, the input work must equal the output work
72EfficiencyAnytime we do work to transfer energy, some of our work is wastedUseful work output (or useful energy) is less than work input (total energy) in the real worldEfficiency defined asor
73Example 2An incandescent light bulb transforms 120 J of electric energy into 5.0 J of light energy. A fluorescent bulb requires 65 J of electrical energy to produce the same amount of light.Calculate the efficiency of each type of bulb.Why is the fluorescent bulb more efficient than the incandescent bulb?Solution:In this example, the electrical energy used by the light is the total energy (or input work) and the light energy is the useful energy or work output.
75Example 2The fluorescent bulb is more efficient because it does not generate as much heat as the incandescent bulb. This heat is energy that is wasted.
76Check Your LearningA pulley system is being used to raise a 78 kg mass to a height of 4.2 m. The input force must be exerted over a distance of 9.0 m. The efficiency of the system is 74%.How much work (input) is required to lift this mass?
77Check Your Learning How much force must be applied? What is the actual mechanical advantage?The actual mechanical advantage (AMA) can be found by looking at the ratio of the forces:
78Module Summary In this module, you have learned that: Work can be calculated using the formulaPower is the rate at which work is done and can be calculated usingKinetic Energy is the energy of motion and can calculated using the formula
79Module SummaryThe net work on an object is equal to its change in kinetic energyThe gravitational potential energy of an object can be calculated using the formulaThe work-energy theorem states that the total external work done on a system is equal to the change in energy
80Module SummaryThe actual mechanical advantage of a machine can be found using the formulaThe Efficiency of a machine (or any other energy conversion) is given by the equation
81Module 4.3 HeatWhen dealing with equipment, it is often important to know how it will react to heat and changes in temperature. Heat can cause changes in the size of materials, and also must be considered when looking at the total work and energy involved in a system.
82Module 4.3 Objective 1Upon completion of this module, the participant will be able to:Understand the difference between heat, temperature and thermal energy.
83Heat as Energy Transfer Heat is – amount of energy that is being transferred from one object to another because of a difference in temperature.kilocalorie (kcal) – the amount of heat necessary to raise the temperature of 1 kilogram of water by 1 Celsius degreeWork can also cause an increase in temperatureMechanical Equivalent of Heat:
84Expanded Work-Energy Theorem Work can result in a change in kinetic energy, a change in potential energy, or heatWhere Q is the heat produced by frictionThe external work done on a system or body is equal to the total change in energy of the system
85Example 1Suppose that a 2.0 kg box is pushed along a horizontal surface with a force of 75 N over a distance of 3.8 m. Starting from rest, the box reaches a speed of 10.0 m/s at the end of the distance. How much heat (in kilocalories) was produced and shared between the box and the surface?Solution:If we consider the box and surface to be our system, the force being applied by the person can be used to calculate the external work on the system.
86Example 1Since the box is not changing height, there is no change in gravitational potential energy. There is, however, a change in kinetic energy:
88DefinitionsThermal energy refers to the total energy of all the molecules in an object;Temperature is a physical measurement that is related to the average kinetic energy of individual molecules.Heat refers to a transfer of energy from one object to another because of a difference in temperature.
89Types of Heat TransferConduction – happens through collisions between the molecules in an object. Molecules at the hot end of an object have a lot of kinetic energy and are moving faster. As they collide with nearby molecules, they transfer some of their kinetic energy to these molecules (solids)Convection - heat flows by the movement of molecules from one place to another (liquids and gases)Radiation - heat that is transferred through electromagnetic waves; unlike conduction and convection, no medium is required.
90Check Your LearningWhat happens to the work done when a jar of juice is shaken?The work done in this case would increase the temperature of the juice, since there is no increase in potential energy and no increase in kinetic energy (once the juice stops moving).If two objects of different temperatures are placed in contact, will heat naturally flow from the object with the higher thermal energy to the one with the lower thermal energy?No, heat naturally flows from the object with a higher temperature to the one with a lower temperature. If 50 g of water at 30oC is mixed with 250 g of water at 24oC, heat will flow from the warmer water to the cooler water even though the thermal energy of the cooler water is much greater because there is so much more of it.
91Check Your LearningIs it possible for heat to flow even if the thermal energies of two objects are the same?Yes, heat flow depends on temperature difference not a difference in thermal energies. If a small piece of iron and a large piece of iron have the same thermal energy, the smaller one will be at a higher temperature and heat will flow into the cooler piece of iron.Identify the mechanism for heat transfer in each of the following situations:Heat travelling from the Sun to the Earth.This is an example of radiation – there is no medium (or molecules) to transmit heat using either conduction or convection.
92Check Your LearningWhen boiling a pot of water, the handle of the pot gets hot.This is an example of conduction – the molecules in the pot (a solid) are not free to move around over large distances. Heat is transferred as they collide with neighboring molecules.When boiling a pot of water, the water at the bottom gets hot first and then the water at the top gets hot.This is an example of mostly convection. As the water at the bottom of the pot heats up, it becomes less dense and rises. This carries heat to the top of the water and forces cooler water down.When you hold your hand near a stove element, you feel heat.This is mostly radiation, as hot objects emit electromagnetic radiation. If you are holding your hand above the element, there may also be some convection as the hot air rises.
93Module 4.3 Objective 2Upon completion of this module, the participant will be able to:To understand the mathematical relationship between heat and temperature.
94Specific Heat Capacity Heat flows from an object with a high temperature to one with a low temperature.Change in temperature depends on mass and specific heat capacity (refers to the amount of heat Q needed to raise a particular mass of a certain material by 1 degree Celsius.)If heat is in J and mass is in kg then c is the specific heat capacity measured in
96Example 1A container of water containing 755 g of water is heated, raising its temperature from 5.0°C to 85.0°C. How much heat was absorbed by the water? Solution:
97Heat Transfer in Systems Since heat is a transfer of energy, heat gained by one object can be transferred from another objectSeveral categories of systems:An open system is one in which mass and energy can enter or leave the system.A closed system is one in which energy can enter or leave the system, but mass cannot.A closed system is said to be isolated if no energy can enter or leave the system either.In a closed and isolated system
98Example 2When a 290 g piece of iron is placed at 180.0oC is placed in a 95 g aluminium cup containing 250 g of glycerine at 10.0oC, the final temperature is observed to be 38oC. Calculate the specific heat of the glycerine.Solution:Since the glycerine is already in the aluminium cup, the glycerine and aluminium can be assumed to be at the same temperature.
101Check Your LearningA piece of lead at 120.0oC is placed in 1.3 kg of water that has an initial temperature of 22.0oC. After a period of time, the water and lead reach a common temperature of 23.5oC. Assume a closed and isolated system and ignore the effect of the container holding the water.How much heat did the water gain?
102Check Your Learning How much heat did the lead lose? Since the system is closed and isolated, any heat gained by the water must have been lost by the lead; therefore, the heat lost by the lead was
103Check Your LearningWhat was the mass of the lead?
104Module 4.3 Objective 3Upon completion of this module, the participant will be able to:Understand the mathematical relationship between heat and change of state.
105Latent HeatLatent Heat – the amount of heat needed to change the state of a particular mass of a substanceEnergy goes into separating molecules rather than increasing the temperatureHeat required to change the state of a substance can be found usingWhere ∆H is the latent heat and is measured in J/kg
106Latent HeatIf a substance is melting (going from a solid to a liquid) we must use the latent heat of fusion:If a substance is being vaporized (going from a liquid to a gas) we must use the latent heat of vaporization:For the reverse process (liquid to solid or gas to liquid), energy is released
109ExampleHow much heat does it take to melt 2.0 kg of ice that is at a temperature of -12.0°C? Solution:There are two parts to this problem:Before the ice can melt, the temperature must be raised to the melting point.Once at the melting point, we can calculate how much energy is required for the change of state.
112Check Your LearningWater is being heated with 4.0×106 J of energy, turning it into steam. The mass of the water is 1.5 kg and the water is initially at a temperature of 20.0oC.How much energy did it take for the water to reach its boiling point?For this part, the water must reach a temperature of 100.0oC; there is no change of state involved.
113Check Your LearningHow much energy did it take to convert the water into steam?Once the water has reached its boiling point, there is no change in temperature in converting the water into steam.
114Check Your Learning What was the final temperature of the steam? Since we know the total energy applied to the water, and we know how much energy it took to raise the temperature of the water and to convert it into steam, the remainder of the energy was used to raise the temperature of the steam.
115Module 4.3 Objective 4Upon completion of this module, the participant will be able to:Calculate the expansion or contraction of a solid or liquid when it undergoes a change in temperature.
116Linear ExpansionWhen designing bridges and other types of equipment an allowance must be made for expansion and contractionThe change in length of a solid can be given byWhere α is the coefficient of linear expansion and is measured in °C-1
118Example 1The steel bed of a suspension bridge is m long at 20.0oC. If the extremes of temperature to which it is exposed are -30.0oC and 40.0oC, how much will it contract and expand?Solution:First, we will look at how much it will contract.
119Example 1Now, we will look at how much it will expand.
120Example 1So the steel can decrease in length by as much as 12 cm and increase in length by as much as 4.8 cm.
121Volume ExpansionChange in volume can be applied to liquids as well as solids (gases will be looked at in Chemisty)The change in volume of a substance that undergoes a temperature change is given byWhere β is the coefficient of volume expansion and is measured in °C-1Hollow objects are found to expand similar to solid objects that have the same volume and are made of the same material. A steel tank will expand the same amount as a block of solid steel having the same dimensions.
122Example 2The 70.0 L steel gas tank of a car is filled to the top with gasoline at 20.0oC. The car is then left to sit in the sun and the tank reaches a temperature of 40.0oC. How much gasoline will overflow from the tank?Solution:We must calculate the increase in volume for both the gas tank and the gas.
125Check Your LearningA concrete highway is built of slabs 14 m long, at 20.0oC. How wide should the expansion cracks be between the slabs to prevent buckling if the range of temperatures is -30.0oC to +50.0oC?When the temperature goes down below 20.0oC, the slabs will contract and the gaps will widen. So we must make sure that the gaps are large enough to allow the slabs to expand linearly without touching each other before they reach a temperature of 50.0oC. So the amount of expansion should be equal to the size of the gaps.
127Module Summary In this module you learned: The difference between heat, temperature, and thermal energy.The work-energy theorem can be expanded to include the heat generated by frictionThat the heat applied to or removed from an object is related to the change in temperature using the equation
128Module SummaryThat the latent heat needed for a change of state is given by the equationThat the linear expansion of a solid can be calculated using the equationThat the volume change of a solid or a liquid can be calculated using the equation
129Module 4.4 Conservation of Energy Conservation of energy allows us to analyze many situations that would be very difficult to analyze using kinematics and dynamics because conservation of energy allows us to simply compare two positions in a system. It is not necessary to consider directions and variations at every step of the motion.
130Conservation of Mechanical Energy In the absence of friction, any external work produces a change in mechanical energyIf no work is done, then
131Conservation of Mechanical Energy Conservation of Mechanical Energy - the total mechanical energy at any initial point in an isolated system is equal to the total mechanical energy at any later point (in the absence of friction)Energy can be transformed from one type to another, but the total amount of energy stays the sameExample – a falling book starts with potential energy; as it falls, potential energy gets transformed into kinetic energy
132Example 1Two balls of the same mass are rolled down two different tracks, as shown below. One track is shorter than the other track; the top of each track is at the same height, and the bottom of each track is at the same height. The balls are released from rest. Which ball is going faster when it reaches the bottom, the ball on track A or the ball on track B?
133Example 1Solution:This is an example of a situation that is much easier to analyze using conservation of energy, since a kinematics analysis would involve calculating the accelerations using vectors and using the distances to calculate final velocities.Since both balls are released from rest, they have no initial kinetic energy. Their initial potential energy is the same, since they are released from the same height. When they reach the bottom, neither ball has any potential energy (relative to the bottom of the ramp). Since they had the same initial energy, they must have the same final energy (due to conservation of energy). Since all of the energy is kinetic, having the same kinetic energy (and same mass) means that they are going the same speed. So each ball will reach the bottom with the same speed.
134Example 2A roller coaster cart starts from rest at the top of a hill (point A) that is 15.0 m above the ground, as shown below. How fast will it be going at point C, which is 11.0 m above the ground?
135Example 2Solution:Remember, when dealing with potential energy it is necessary to choose a reference level. We will choose the ground to be the reference level in this solution. We will choose point A as our initial point (since we know information about this point) and point C as our final point (since it is the one we are trying to obtain information about).At point A, the car will have potential energy since it is above the reference level (the ground). It will not have any kinetic energy since it is starting from rest. At point C, the car will have both potential energy (since it is above the reference level) and kinetic energy (since it is moving).
137Example 3 Video Demonstration In a well known lecture demonstration, a bowling ball is hung from a ceiling by a steel wire. The lecturer pulls the ball back and stands against the wall of the room with the ball against his nose.If the lecturer simply releases the ball, he will not be injured. Why not?If the lecturer pushes the ball, when it returns it will likely hit him and cause injury. Why?Video Demonstration
138Example 3Solution:When the ball is released, it is not initially moving so all of its energy is potential. As it swings over and back, its total energy must remain the same (or decrease if it loses any energy) even though the energy gets converted into kinetic energy and back into potential energy. When it returns back to its starting point, all of the energy returns to potential energy. Since it can’t have more energy than it started with, it cannot go any higher than it started. It therefore cannot hit the person.If the ball is given a push, its initial energy is the sum of its initial potential and kinetic energies. After it swings across the room and returns to its starting point, it still has kinetic energy left so it can continue above its starting point. This extra energy that the push gave the ball allows it to swing higher than its original height so it keeps going and hits the person.
139Conservation of Energy Applet The following simulation can be used to generate real-time graphs of kinetic energy, potential energy, and total energy to demonstrate Conservation of Mechanical EnergyConservation of Energy Applet
140Check Your LearningFor each of the following two questions, use the same roller coaster that was used in the example problem.Do the Example 2 again, but this time choose your reference level to be point B. Do you get the same answer?
141Check Your LearningSince our reference level is point B, all height must be measured from this point. We will still use point A as our initial point and point C as our final point.Yes, we do get the exact same answer. It does not matter where we choose our reference level – we will always get the same answer.
142Check Your Learning If the car has a speed of 12.0 m/s at point A, What will its speed be at point C?The car now has kinetic energy at point A since it is moving. We will use the ground as our reference level and we will use point A as our initial position and point C as our final position.
143Check Your LearningWhat is the highest hill (above the ground) that the car could reach on this roller coaster?We will use point A as our initial position again, but for our maximum height we must use some unknown position that occurs later on the roller coaster. At the maximum height, all of the energy will be potential since the kinetic energy will go to zero as the car stops.
144Conservation of Total Energy Consider the situation shown below, where a ball is released from the top of the semi-circular track:Ball would continue to go back and forth forever as the energy is converted back and forth between potential and kinetic energyIn reality the ball will eventually stop, meaning that it has lost all of its mechanical energy. Where did this energy go?
145Conservation of Total Energy In the presence of friction, any external work produces a change in mechanical energyIf no work is done, then
146Conservation of Total Energy Still allows for friction within the system, but no work outside the systemPrinciple of Conservation of Total Energy – even in the presence of friction, the total energy in a closed and isolated system remains constant
147Example 1A 250 g car rolls down a ramp. The car starts from a height of 32 cm, and reaches a speed of 1.7 m/s at the bottom of the ramp. Was mechanical energy conserved?Solution:To determine if mechanical energy was conserved, we must calculate the total initial energy and the total final energy. We will use the bottom of the ramp as the reference level.
148Example 1Since Ei ≠ Ef , mechanical energy was not conserved. There were 0.42 J of energy lost. This energy was generated as heat due to the friction due to the moving parts in the car and between the car and the ramp.
149Example 2Consider the same roller coaster that we looked at in the previous section:If a 215 kg car starts from rest at the top of the first peak (point A) but only reaches a speed of 6.0m/s at point C, how much energy was lost?
150Example 2Solution:The energy lost here was lost due to friction between the car and the track, and will appear as heat. We are therefore trying to find Q. We will use the ground as our reference level, and use point A as our initial point and point C as our final point.
151Problem-Solving Hints In this unit, we have discussed using both the work-energy theorem and the principle of conservation of energy to solve problems. In many situations, either approach might be taken, depending on what you choose as your system.If you choose your system to be an object or objects on which there is an external force doing work, then you must use the work-energy theorem since there will be a change in energy of the system.If you expand your system to include the objects applying the forces (making the forces internal) then there is no external work being done on the system by external forces and conservation of energy can be used.
152Conservation of Energy Hints Draw or visualize a picture of the situation to help indicate where the object is at different times.Determine the system for which energy will be conserved; in other words, determine the system for which there will be no outside forces acting on any of the objects.Choose initial and final positions – these will be the two positions for which you are comparing the total energy.Choose a reference level – this is the point where the gravitational potential energy is zero. Although any level can be chosen, the lowest point in the problem or the ground is often a convenient choice.Ask yourself what kinds of energy (potential and/or kinetic) are present at each of the positions chosen in step 3. These will be included in the conservation of energy equation in the next step.Apply the appropriate conservation of energy equation.
153Check Your LearningIn the roller coaster used in the example, suppose that 3250 J of heat is generated between points A and B. If the 215 kg car starts from rest at point A, how fast will it be going at point B?
154Check Your LearningWe will use point A as the initial point and point B as the final point. We will use the ground as the reference level.
155Energy and CollisionsMomentum is conserved in all collisions when applied to closed and isolated systems. Is energy?Consider two balls going toward one another as shown in the diagram belowonly type of mechanical energy present before the collision is kinetic energy.
156Energy and CollisionsConsider what happens in slow motion and look at the balls on a small scale, we see that they momentarily stop when they come in contact:edges of the ball compress in this process - the kinetic energy of the moving ball is transformed into a type of potential energy known as elastic potential energy.
157Energy and CollisionsWhen the balls bounce away from one another, then the potential energy that was temporarily stored in this collision is transformed back into kinetic energyIf this energy conversion is 100% efficient (meaning that no kinetic energy was lost in this process of transforming kinetic energy into potential energy and back to kinetic energy) than the collision is said to be elastic
158Energy and CollisionsAn elastic collision can be defined as one in which the total amount of kinetic energy is conservedA completely inelastic collision is one in which the objects stick together. Although some kinetic energy must be lost in this type of collision, all of the kinetic energy does not have to be lost since the objects can continue to move together. Even though kinetic energy is lost, it is changed into other forms of energy (often heat) so the total energy is still conserved as always
159Energy and Collisions Momentum is conserved in all collisions In a closed and isolated system,Momentum is conserved in all collisionsMechanical energy is only conserved if the collision is elastic
160Example 1Ball A, mass 3.0 kg, is moving to the left at 2.5 m/s when it collides with ball B, mass 4.0 kg, moving to the right at 4.0 m/s. Ball A rebounds in the opposite direction with a speed of 2.0 m/s.Find the velocity of ball B after the collision.Was the collision elastic? Show your work.
161Example 1Solution:Since we have a collision, the first thing we should do is apply conservation of momentum. Remember that momentum involves vectors, so direction is important. We will use the right as positive.
162Example 1In order to determine if the collision is elastic, we must determine whether or not the total initial kinetic energy is equal to the total final kinetic energy. Since we are calculating the energy we do not have to worry about direction.Since there is less total kinetic energy after the collision than before the collision, the collision was not elastic.
163Example 2Consider a collision between two billiard balls, which we will assume to be elastic. Suppose that the cue ball (ball 1), moving at a speed of 3.0 m/s, collides with another ball (ball 2) of the same mass that is at rest. Find the velocity of each ball after the collision.Solution:Since momentum is conserved in all collisions, we will apply this principle first. For consistency with energy, we will use subscripts instead of primes for all quantities.
164Example 2There are obviously two unknowns in this equation. In order to solve for two unknowns, it is necessary to have two equations. Since the collision can be considered to be elastic, we can apply conservation of kinetic energy:
165Example 2Rearranging the momentum equation we get
166Example 2Substituting this into the energy equation gives
167Example 2We see that mathematically, there are two solutions to the final velocity of ball 2, 0 and 3.0 m/s. It is necessary to look at the situation, however. Ball 2 was initially at rest. It does not make sense that it would remain at rest after the collision; therefore, the velocity of ball 2 must beSubstituting this back into the momentum equation gives
168Check Your LearningA 53.0 g marble moving with a velocity of 2.00 m/s collides with a stationary marble of mass 87.0 g. If the first marble rolls backward with a velocity of m/s, was the collision elastic?Since we have a collision, the first thing we should do is apply conservation of momentum to find the speed of the second marble. Remember that momentum involves vectors, so direction is important.
169Check Your LearningIn order to determine if the collision is elastic, we must determine whether or not the total initial kinetic energy is equal to the total final kinetic energy.
170Check Your LearningSince the initial and final energy are within J (less than 1%) we can assume that the difference is due to rounding (especially since the final energy is slightly larger than the initial energy); since the energies are the same, the collision was elastic.
171Module Summary In this module, you learned The total mechanical energy in a closed and isolated system (in other words, where there is no external work done) remains constant in the absence of friction:If there is friction, then the total energy in the system still remains constant if the heat that is generated is included:
172Module SummaryAn elastic collision is one in which the total mechanical (kinetic) energy remains constant.A completely inelastic collision is one where the objects remain stuck together after the collision. Some kinetic energy is always lost in inelastic collisions.
173Unit Summary Module 4.1 – Momentum and Impulse Momentum is calculated using the formulaThe change in momentum can be calculated from the change in velocityImpulse is equal to the change in momentum and is given by the formulaMomentum is conserved in all isolated systems. This is known as the law of conservation of momentum
174Unit Summary Module 4.2 – Work, Power and Energy Work can be calculated using the formulaPower is the rate at which work is done and can be calculated usingKinetic Energy is the energy of motion and can calculated using the formulaThe net work on an object is equal to its change in kinetic energy
175Unit SummaryThe gravitational potential energy of an object can be calculated using the formulaThe work-energy theorem states that the total external work done on a system is equal to the change in energyThe actual mechanical advantage of a machine can be found using the formulaThe Efficiency of a machine (or any other energy conversion) is given by the equation
176Unit Summary Module 4.3 – Heat The difference between heat, temperature, and thermal energy.The work-energy theorem can be expanded to include the heat generated by frictionThat the heat applied to or removed from an object is related to the change in temperature using the equationThat the latent heat needed for a change of state is given by the equationThat the linear and volume expansion of a substance can be calculated using the equations
177Unit Summary Module 4.4 – Conservation of Energy The total mechanical energy in a closed and isolated system (in other words, where there is no external work done) remains constant in the absence of friction:If there is friction, then the total energy in the system still remains constant if the heat that is generated is included:An elastic collision is one in which the total mechanical (kinetic) energy remains constant.A completely inelastic collision is one where the objects remain stuck together after the collision. Some kinetic energy is always lost in inelastic collisions.