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Physics 42S IBTopic 2 – Mechanics 3Energy and Momentum2.3 – Work, energy and power Nature of science: Theories: Many phenomena can be fundamentally understood through application of the theory of conservation of energy. Over time, scientists have utilized this theory both to explain natural phenomena and, more importantly, to predict the outcome of previously unknown interactions. The concept of energy has evolved as a result of recognition of the relationship between mass and energy. (2.2) Understandings: Kinetic energy Gravitational potential energy Elastic potential energy Work done as energy transfer Power as rate of energy transfer Principle of conservation of energy Efficiency Applications and skills: Discussing the conservation of total energy within energy transformations Sketching and interpreting force–distance graphs Determining work done including cases where a resistive force acts Solving problems involving power Quantitatively describing efficiency in energy transfers Guidance: Cases where the line of action of the force and the displacement are not parallel should be considered Examples should include force–distance graphs for variable forces Theory of knowledge: To what extent is scientific knowledge based on fundamental concepts such as energy? What happens to scientific knowledge when our understanding of such fundamental concepts changes or evolves? Utilization: Energy is also covered in other group 4 subjects (for example, see: Biology topics 2, 4 and 8; Chemistry topics 5, 15, and C; Sports, exercise and health science topics 3, A.2, C.3 and D.3; Environmental systems and societies topics 1, 2, and 3) Energy conversions are essential for electrical energy generation (see Physics topic 5 and sub-topic 8.1) Energy changes occurring in simple harmonic motion (see Physics sub-topics 4.1 and 9.1)Aims: Aim 6: experiments could include (but are not limited to): relationship of kinetic and gravitational potential energy for a falling mass; power and efficiency of mechanical objects; comparison of different situations involving elastic potential energy Aim 8: by linking this sub-topic with topic 8, students should be aware of the importance of efficiency and its impact of conserving the fuel used for energy productionData booklet reference: W=FscosθEK=12mv2Ep=12kx2 ?EP=mg?hpower=FvEfficiency=useful work outtotal work in=useful power outtotal power inWork, Energy, and PowerWorkIf a force is acting on an object as it moves a certain distance, work is being done on the object. We define the work done on an object as:W=FscosθW is the work, in Joules J F is the force, in Newtons Nsis the displacement, in meters mθis the angle between F and sWhen work is done by a force on an object the energy that the object possesses will change. If the force and displacement are in the same direction, energy is added to the object. If the force and displacement are in the opposite directions, energy is removed from the object.When mechanical energy is added to a mass, the speed of a mass typically increases. When mechanical energy is removed from a mass, the speed of the mass typically increases. JoulesThe work being done on an object is the product of force and displacement in the direction of the force. The unit of work is the Joule. 1 J is the work done by a force of 1 N when it moves a body a distance of 1 m in the direction of the force. 1 J=1 N?m1 J=1 kg m s-2? m1 J=1 kg m2 s-2Example 1A force of constant magnitude of 25 N is applied to a body that moves along a straight path. Find the work done after the mass moves a distance of 50. m.Example 2Find the work done by the tension of a string as a mass attached to the end performs one horizontal revolution of uniform circular motion (constant speed).Example 3A mass is being pulled along a level road by a rope attached at an angle of 40.0° with the horizontal. The force in the rope is 20.0 N. What is the work done by this force moving the mass a distance of 8.00 m along the level road?Example 4A 50.0 kg crate is pulled 40.0 m along a horizontal floor by a constant force exerted by a person, FA=100 N, at an angle of 37.0° above the horizontal. The floor exerts a force of friction in the opposite direction of Ff=50.0 N. Determine the work done on the crate by the person.Determine the work done on the crate by friction.Determine the net work done on the crate.Force – Displacement GraphsForce-displacement graph can be used to calculate the work done on a body where the force is not constant. The area under a force-displacement graph of an object gives the work being done to the object.F/ N02505007501000246810d/ mFigure 1 – a force-displacement graphF/ N02505007501000246810d/ mFigure 1 – a force-displacement graphBy analyzing the units, we can see that the area under the graph is equal to the work being done.Area=base×heightArea=m×NArea=JExample 6Find the total change in energy of the body from Figure 1.Homework: Giancoli – Page 161, Questions 1-16Kinetic EnergyWhen work is being done to an object, we can be adding or removing kinetic energy. From our kinematics equations:v2=u2+2asReplacing a=Fm , we get:v2=u2+2Fmsand isolating for Fs:F?s=12mv2-u2F?s is the work done on the mass. This work is the change in kinetic energy. For the kinetic energy of an object:?Ek=12mv2-u2If the initial speed of the object was 0:Ek=12mv2Ekis the kinetic energy, in Joules ( J)mis the mass, in kilograms (kg)vis the speed, in meters per second (m s-1)Example 1How much kinetic energy does a 5.00 kg particle have if it is moving at 2.50 m s-1?Example 2How fast is a 12.5 kg object traveling if it has 100. J of kinetic energy?Work and EnergyTo change the kinetic energy of a mass, and therefore the speed, work must be done to the mass.E1+W=E2Work can also represent the change in kinetic energy of the mass:?EK=E2-E1W=E2-E1Example 3A mass of 5.00 kg is moving with an initial velocity of 12.0 m s-1 is brought to rest by a force over a distance of 12.0 m. What is the magnitude of the force?Example 4A 12.0 kg mass is moving with an initial velocity of +15.0 m s-1. A force of -45.0 N is applied over a displacement of +25.0 m. Using work and energy, calculate the final speed of the mass.Example 5Calculate the work required to accelerate a 1000 kg crate from a speed of 20 m s-1 to a speed of 30 m s-1.Example 6A 50.0 kg mass is moving with an initial speed of 12.0 m s-1. A force of -150. N is applied over a displacement of -13.5 m. Using work and energy, calculate the final speed of the mass.Homework: Giancoli – Page 161, Questions 17-28Gravitational Potential EnergyPotential energy is the amount of work that could be done by a force on an Near the surface of the Earth (or any planet) gravitational potential energy can be approximated by the formula:?EP=mg?hWhere:?EP is the change in gravitational potential energy, in Joules J m is the mass, in kilograms (kg)gis the gravitational field, in Newtons per kilogram N kg-1, or in meters per second squared (m s-2)?his the change in height, in meters (m)The height of the object is always relative to a fixed point, usually the surface of the Earth.This formula for gravitational potential energy can only be used in a uniform gravitational field. Generally, this equation works near the surface of a planet with small distances (0-10 km). If the distance is greater, we must use a more complex equation:EP=-GMmrBut, more on this later… (see Topic 10 – Fields)Work done by Gravity2174392129895mgmgmgdisplacementFigure 2 – work done by gravity on a horizontally moving body00mgmgmgdisplacementFigure 2 – work done by gravity on a horizontally moving bodyIf an object moves horizontally along a surface parallel to the surface of the Earth, then the work done on the object by gravity is zero, since in this case the angle formed between the force and the displacement is 90°.330070776183mgmgmghmgmgmghFigure 3 – work done by gravity on a vertically moving body00mgmgmghmgmgmghFigure 3 – work done by gravity on a vertically moving bodyIf a body is thrown vertically upward to a height of h, then the work done on the object is -mgh. The force of gravity is anti-parallel to the displacement. If a body falls a vertical distance h, then the work done on the object is +mgh. The force of gravity is parallel to the displacement. Example 7A ball of mass m=0.250 kg is thrown from a height of 1.50 m with an initial speed of 15.0 m s-1. Determine the kinetic energy of the ball.Determine the work done by gravity on the ball and the speed of the ball as it reaches a height of 9.00 m directly above where it was thrown.Determine the work done by gravity and the speed of the ball at the point where the ball has displaced 9.00 m 40° above the horizontal from where it was thrown.Conservation of Mechanical EnergyIn a closed, ideal system, the total energy of an object at any one point in time should be the same as the total energy of the same object at any other point in time, i.e. the total energy is constant.If an outside force is applied, then the work done by this force must be considered when calculating the total energy. 152400795020v1v3v2v4v5Figure 4 – a body moving through a uniform gravitational field0v1v3v2v4v5Figure 4 – a body moving through a uniform gravitational fieldIn the following scenario, a ball is thrown into the air. As the ball rises, it gains potential energy. To gain potential energy, it must lose kinetic energy. However, the total energy of the ball remains constant.Example 1A 0.750 kg ball is dropped from rest at a height of 5.5 m. Calculate the total energy and the speed of the ball at a height of 3.2 m and the speed as it hits the ground.h=5.5 mh=3.2 mh=0.0 mFigure 5 – a body falling in a uniform gravitational fieldh=5.5 mh=3.2 mh=0.0 mFigure 5 – a body falling in a uniform gravitational fieldExample 20573405v=4.0 m s-1Figure 6 – A ball falling off a table.h=1.0 m0v=4.0 m s-1Figure 6 – A ball falling off a table.h=1.0 mA 0.50 kg ball rolls off a 1.0 m high flat, level table with a speed of 4.0 m s-1, as in figure 6. Calculate the speed of the ball when it strikes the floor.Example 3A roller coaster with a mass of 350. kg is travelling at a speed of 12.0 m s-1 at point A on the frictionless track in Figure 7. Calculate the height at point B and the speed at point C.vABv=5.00 m s-1h=10.0 mv=?h=?CFigure 7 – A roller coaster moving along a trackvABv=5.00 m s-1h=10.0 mv=?h=?CFigure 7 – A roller coaster moving along a trackExample 4A 500. kg roller coaster is travelling at 5.00 m s-1 along a track where there is friction as in Figure 8. Given the following diagram, what is the work being done by friction?v=5.00 m s-1h=7.50 mv=10.5 m s-1Figure 8 v=5.00 m s-1h=7.50 mv=10.5 m s-1Figure 8 If the work being done on the roller coaster by friction occurs over a distance of 25.0 m, what is the average force of friction?Example 5Determine the minimum speed of the mass in Figure 9 at the initial point such that the mass makes it over the barrier of height h=7.5 m. h=7.5 mv=?Figure 9 h=7.5 mv=?Figure 9 Example 6A pendulum of length 1.000 m is released from rest with the string starting at an angle of 10.0° to the vertical. Find the speed of the mass on the end of the pendulum when it passes through its lowest position.Example 7A body of mass 2.00 kg, initially at rest, slides down a curved path of total length 22.0 m, as in figure 10. The body starts from a vertical height of 5.00 m from the bottom. When it reaches the bottom, its speed is found to equal 6.00 m s-1.?h=5.00 mm=2.00 kgFigure 10?h=5.00 mm=2.00 kgFigure 10Show that there is a force resisting the motion. Assuming the force to have a constant magnitude, determine the magnitude of the force.Example 8A mass of 5.00 kg with an initial velocity of 2.0 m s-1 is acted upon by a force of 55 N in the direction of motion. The motion is opposed by a frictional force. After traveling a distance of 12 m the velocity of the body becomes 15 m s-1. Determine the magnitude of the force of friction. Elastic Potential Energy Work done by a springAccording to Hooke’s Law, a spring applies a force of tension as such:T = k x3136900394970T / Nx / mFigure 11 – a force displacement graph00T / Nx / mFigure 11 – a force displacement graphTo find the work done by extending or compressing a spring we can find the area under a force time graph.W=12TxW=12kx xW=12kx2EP=12kx2We can also derive the equation for elastic potential energy if substitute the average force of tension in the work equation. We can leave out the cosθ as the displacement and the force will always be in the same direction.EP=12(T+T0)sEP=12TxEP=12kxxEP=12kx2Example 1A mass of 8.40 kg rests on top of a vertical spring whose base is attached to the floor. The spring compresses by 5.20 cm. Calculate the spring constant of the spring.Determine the energy stored in the spring.Example 2A mass of 0.250 kg is attached to a spring with a spring constant of k=50.0 N m-1. The mass on the spring is then stretched horizontally a distance of 0.135 m and then released. Assume that the mass and spring rest on a frictionless surface.What is the potential energy of the spring when it is extended 0.135 m?What is the speed of the mass when it returns to the natural length of the spring?What is the speed of the mass when the spring is compressed 0.100 m?Example 3354365694549615.0 kg10.0 mk=200 N m-1Figure 120015.0 kg10.0 mk=200 N m-1Figure 12A mass of 15.0 kg is dropped on to a curved surface at a height of 10.0 m and it slides down the surface. At the end of the curved surface, there is a spring with a spring constant of 200 N m-1 as in Figure 12.What is the potential energy of the mass before it is dropped?What is the speed of the mass at the end of the curved surface?What is the distance that the spring will compress when the mass is brought to rest?Example 4109728568122v=5.60 m s-1?hFigure 130v=5.60 m s-1?hFigure 13A body of mass 4.20 kg with an initial speed 5.60 m s-1 begins to move up an incline as shown in Figure 13:The body will momentarily be brought to rest after colliding with a spring of spring constant 220 N m-1. The body stops at a vertical distance of 0.850 m above its initial position. Determine the amount by which the spring has compressed. There are no frictional forces. Repeat the calculation but now with a constant force of friction opposing the motion along the uphill part of the path. The length of the uphill part of the path is 1.20 m and the frictional force is 15 N.Example 5Determine the spring constant of a spring that makes a maximum extension of 0.178 m when a mass of m=12.5 kg is fixed to the end and is left to oscillate. 1609090174625x=0.178 m0x=0.178 mHomework: Giancoli – Page 162, Questions 29-46PowerWhen a quantity of work ?W is being performed within a time interval ?t the ratio:P=?W?tis the power developed. The units for Power are Watts W which are equivalent to Joules per second. 1 W=1 J s-1If you consider the ratio of work per unit time, we can derive another equation to give us power.P=?W?tP=F ?s?tP=F?s?tP=Fvpower=FvExample 1What is the minimum power required to lift a mass of 50.0 kg up a vertical of 12 m in 5.0 s?Example 2What is the minimum power output of an engine of a car that moves with a speed of 25.0 m s-1 with an overall force of friction of 1.50×103 N?Example 3A 70.0 kg jogger runs up a long flight of stairs in 4.00 s. The vertical height of the stairs is 4.50 m.Estimate the jogger’s power output in watts.How much energy did this require?Example 4A car of a 1400 kg travels on the highway experiences an average retarding force (friction, air resistance, etc.) of 700 N. Calculate the power required under the following circumstances:the car climbs a 10° hill at a steady speed of 80 km h-1.the car accelerates along a level road from 90 km h-1 to 100 km h-1 in a period of 6.0 s to pass another car.Efficiency1993900824865θ=45.0°FfmgRFigure 14 θ=45.0°FfmgRFigure 14 Suppose that a mass is being pulled up along a rough incline with a constant speed, as in figure 14. Let the mass be 15.0 kg and the angle of inclination be 45.0°. The constant frictional force opposing the motion isf=42 N.The forces on the mass are:F=mgsinθ+f=106 N+42 N=148 NIf the force applied raises the mass a distance of 20.0 m along the plane, the work done by the force is:W=148 N?20.0m=2.96×103JThe actual change in energy is ?EP through the change in height of 14.1 m which is:?EP=mgh?EP=15.0 kg?9.81 N kg-1?14.1m=2.07×103JThe efficiency with which the force raised the mass is:Efficiency=useful work outtotal work in=useful power outtotal power in=2.07×103J2.96×103J=0.699ExampleA 0.50 kg battery operated toy train moves with constant velocity 0.30 m s-1along a level class. The power of the motor in the train is 2.0 W and the total force opposing the motion of the train is 5.0 N.What is the efficiency of the train’s motor?Assuming the efficiency and the opposing force stay the same, calculate the speed of the train as it climbs an incline 10° to the horizontal.Homework: Giancoli – Page 164, Questions 57-68Tsokos – Page 96-97, Questions 55-712.4 – Momentum and impulse Nature of science: The concept of momentum and the principle of momentum conservation can be used to analyse and predict the outcome of a wide range of physical interactions, from macroscopic motion to microscopic collisions. (1.9) Understandings: Newton’s second law expressed in terms of rate of change of momentum Impulse and force–time graphs Conservation of linear momentum Elastic collisions, inelastic collisions and explosions Applications and skills: Applying conservation of momentum in simple isolated systems including (but not limited to) collisions, explosions, or water jets Using Newton’s second law quantitatively and qualitatively in cases where mass is not constant Sketching and interpreting force–time graphs Determining impulse in various contexts including (but not limited to) car safety and sports Qualitatively and quantitatively comparing situations involving elastic collisions, inelastic collisions and explosions Guidance: Students should be aware that F = ma is equivalent of F=?pt only when mass is constant Solving simultaneous equations involving conservation of momentum and energy in collisions will not be required Calculations relating to collisions and explosions will be restricted to one-dimensional situations A comparison between energy involved in inelastic collisions (in which kinetic energy is not conserved) and the conservation of (total) energy should be made Data booklet reference: p=mv F=?ptEK=p22m Impulse=Ft=?pInternational-mindedness: Automobile passive safety standards have been adopted across the globe based on research conducted in many countries Theory of knowledge: Do conservation laws restrict or enable further development in physics? Utilization: Jet engines and rockets Martial arts Particle theory and collisions (see Physics sub-topic 3.1)Aims: Aim 3: conservation laws in science disciplines have played a major role in outlining the limits within which scientific theories are developed Aim 6: experiments could include (but are not limited to): analysis of collisions with respect to energy transfer; impulse investigations to determine velocity, force, time, or mass; determination of amount of transformed energy in inelastic collisions Aim 7: technology has allowed for more accurate and precise measurements of force and momentum, including video analysis of real-life collisions and modelling/simulations of molecular collisionsMomentum and ImpulseLinear MomentumLinear Momentum (p) is defined as the product of mass and velocity. Momentum=mass×velocityp=m?vThe S.I. units for momentum are kg m s-1. Alternative units of Ns can also be used. Since velocity is a vector quantity, momentum must also be a vector quantity. The change in momentum, ?p, is called the impulse.In terms of momentum, Newton’s second Law of mechanics can be stated asF=?p?tThe average net force on a body equals the rate of change of a body’s momentum. If a mass of a body is constant, this formula reduces to the familiarF=m a:F=?p?tF=mvf-mvi?tF=m(vf-vi) ?tF=m?v?tF=m aThe advantage of the first equation is that it can also be used in cases where an objects mass is changing (for example, in the motion of a rocket).Example 1A 0.100 kg ball moving at 5.0 m s-1 bounces off a vertical wall without a change in its speed. If the collision with the wall lasted for 0.10 s, what force was exerted on the wall?Example 2A 0.50 kg ball bounces vertically off a hard surface. A graph of the velocity versus time is shown in figure 15. Find the magnitude of the momentum change of the ball during the bounce. The ball stayed in contact with the floor for 0.15 s. What average force did the ball exert on the floor? FNWv/m s-1t/s20-40.30.60.9-2Figure 15 – a velocity time graph and free body diagram for a ball hitting the floorFNWv/m s-1t/s20-40.30.60.9-2Figure 15 – a velocity time graph and free body diagram for a ball hitting the floorExample 3Bullets of mass 30.0 g are being fired from a gun with a speed of 300 m s-1 at a rate of 20 bullets per second. What force is being exerted on the gun?Example 4A cart moves in a horizontal line at a constant speed 3.5 m s-1. Rain starts to fall and fills the cart with water at a rate of σ=0.20 kg s-1. (This means that in one second, 0.20 kg has fallen on the cart.) The cart must keep moving at a constant speed. Determine the magnitude force that must be applied to the cart.Example 5Gravel falls vertically on a conveyor belt at a rate of σ=5.00 kg s-1 as shown in figure 16:gravelbeltFigure 16 – a conveyor belt moving gravelgravelbeltFigure 16 – a conveyor belt moving gravelDetermine:the force that must be applied on the belt to keep it moving at constant speed v=1.75 m s-1.the power that must be supplied by the motor to turn the belt.the rate at which the kinetic energy of the gravel is changing.Explain why the answers to (a) (ii) and (a) (iii) are different.ImpulseThe change in momentum is proportional to the amount of force transferred from one object to another. Mathematically we can write this as:F=?p?tThe impulse is the total change in momentum.Impulse=?p=F ?tThe force generated by impulse is not constant. The average force (F) is often sufficient for a force acting over a period of time.Impulse can also be calculated by finding the area under a Force-time graph.In the graph the area is about 2.5 N s. The maximum force that acted on the body was 1000 N. The average Force is:-158750124460Force/ N02505007501000246810t/ msFigure 17 – a force-time graph00Force/ N02505007501000246810t/ msFigure 17 – a force-time graphF=?p?t=2.5 N s6.0×10-3 s≈470 NExample 1A baseball (m=0.14 kg) has an initial velocity of u=-38 m s-1 as it approaches the bat. The bat is swung and hits the ball. The baseball leaves the bat with a velocity of v=+58 m s-1. Determine the impulse applied to the ball by the bat. Determine the average force in the following example if the time of contact was ?t=1.6×10-3 s.Example 2Consider the force-time graph in Figure 18.F/N151050-5-10t/s123456Figure 18F/N151050-5-10t/s123456Figure 18The force acts on a body of mass 3.0 kg initially at rest. Calculate:the initial acceleration of the body.the speed at t=4.0 s.the speed at t=6.0 s.Example 3A ball of mass 0.250 kg moves on a frictionless horizontal surface and hits a vertical wall at a speed of 5.0 m s-1. The ball rebounds with a speed of 4.0 m s-1. If the ball was in contact with the wall for 0.150 s, find the average force that acted on the ball.The force in the previous example is assumed to vary with time. Deduce the maximum force that acted on the ball if the force acted on the ball as in Figure 19.136478136307F/Nt/s0.150 sFigure 190F/Nt/s0.150 sFigure 19Example 4A force of 1000.0 N acts on a body of 40.0 kg, that is initially at rest, for a time interval of 0.0500 s. What is the velocity of the mass?Example 5A body of mass 12.5 kg moves with a velocity of +5.70 m s-1. A force of -250 N acts on the body for a period of time t=0.450 s. Determine the final velocity of the body.Example 6The following graph shows the applied force to a cart that has a mass of 15.0 kg. 3259455104775F/ N02505007501000246810t/ msFigure 2000F/ N02505007501000246810t/ msFigure 20Determine the impulse of the force on the cart. The cart starts from rest. Determine the final momentum and speed of the cart. Example 73328416384073F/ N02505007501000246810t/ msFigure 21F/ N02505007501000246810t/ msFigure 21A body with a mass of 2.50 kg is in motion with a velocity of +1.50 m s-1. It experiences a non- uniform force shown in the following graph. Calculate the initial momentum of the body.Calculate the impulse acting on the body.Determine the final velocity of the body.2995074-183338Force/ N0-100-200-300-40012345t/s00Force/ N0-100-200-300-40012345t/sExample 8A body with a mass of 45.0 kg is in motion with a velocity of +18.5 m s-1. It experiences a non- uniform force shown in the following graph.Calculate the initial momentum of the body.Calculate the impulse acting on the body.Determine the final velocity of the body.3767455314960F/Nt/sFigure 220F/Nt/sFigure 22Consider 2 bodies with the same momentum. A body that is brought to rest with a greater average force requires less time stop. A body that is brought to rest with a smaller average force requires more time.The graph in Figure 21 represents 2 bodies undergoing the same impulse. Example 9A 70.0 kg person lands on firm ground after jumping from a height of 5.00 m.Calculate the impulse experienced by the person from the ground.Calculate the average force exerted on the legs if the person lands stiff legged, change in distance of 1.0 cm. Calculate the average force exerted on the legs if the legs bend so the distance moved is 0.50 m. Conservation of MomentumThe law of conservation of linear momentum states that the total linear momentum of a system of interacting bodies or particles remains constant. This is true for a closed system provided that no resultant external force interacts with the system.In effect, the total momentum before and after a collisions or impulse must be equal.Example 1Two masses of 2.0 kg and 3.0 kg move to the right with speeds of 4.0 m s-1 and 5.0 m s-1 respectively. What is the total momentum of the system?Example 2A system consists of 2 masses; one mass of 2.0 kg moves to the right with a speed of 10.0 m s-1 and another mass of 4.0 kg moves to the left with a speed of 8.0 m s-1. What is the total momentum of the system?Example 3A mass of 3.0 kg is at rest is hit by a second mass of 5.0 kg and a velocity of +4.0 m s-1. The smaller mass moves away with a velocity of+3.0 m s-1. What happens to the larger mass?Example 4Two masses of 2.0 kg and 4.0 kg are held together with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a speed of 6.0 m s-1, what is the speed of the other mass?Homework: Giancoli – Page 187, Questions 1-20Conservation of Momentum in 2 DimensionsWhen a collision happens in 2 dimensions, the momentum must be conserved in all directions, i.e. the horizontal momentum must be conserved and the vertical momentum must be conserved. Example 1Two identical curling stones of mass 19.5 kg collide. The first stone hits the stationary second stone with a velocity of 5.00 m s-1 [N]. If the velocity of the first stone is 3.20 m s-1 [N30°W] after the collision, find the velocity of the second stone after the collision.Example 2A mass of?22.5?kg traveling at a velocity of?14.0?m?s-1?[N] collides with a mass of?17.0?kg traveling at a velocity of?10.5?m?s-1?[S]. If the?17.0?kg mass has a velocity of?8.50?m?s-1??S40°W?after the collision, what is the velocity of the?22.5?kg mass?Example 3Consider a stationary body of mass 12.0 kg that is hit by a 4.0 kg mass moving at 12 m s-1. The collision is not head on and the bodies move away from each other at an angle to the original direction of motion as shown in Figure 21. How can we find the speeds of the two bodies after the collision?60°30°Figure 2160°30°Figure 21Elastic CollisionsConsider the following scenario where 2 masses collide:Before the collision:193675977908.0 kg10 m s-112 kgFigure 22 – two bodies before a collision08.0 kg10 m s-112 kgFigure 22 – two bodies before a collision p=m?v=80 kg m s-1Ek=12m v2=400 JScenario 1: Totally inelastic collision1143001409708.0 kg4 m s-112 kgFigure 22 (a) – a totally inelastic collision008.0 kg4 m s-112 kgFigure 22 (a) – a totally inelastic collisionp=m?v=80 kg m s-1Ek=12m v2=160 JThe momentum is conserved however kinetic energy is lost. In collisions that are completely inelastic the speed of separation is 0 m s-1. The masses ‘stick’ together. In this scenario the maximum amount of kinetic energy is lost.Scenario 2: Inelastic collision114300755658.0 kg6 m s-112 kg1 m s-1Figure 22 (b) – an inelastic collision008.0 kg6 m s-112 kg1 m s-1Figure 22 (b) – an inelastic collisionp=m1?v1+m2?v2=80 kg m s-1Ek=12m1 v12+12m2 v22=220 JThe momentum is conserved however kinetic energy is still lost. In collisions that are inelastic the speed of separation does not equal the speed of approach.vapp=10 m s-1vsep=5 m s-1Scenario 3: Elastic collision1155701739908.0 kg8 m s-112 kg2 m s-1Figure 22 (c) – an elastic collision008.0 kg8 m s-112 kg2 m s-1Figure 22 (c) – an elastic collisionp=m1?v1+m2?v2=80 kg m s-1Ek=12m1 v12+12m2 v22=400 JThe momentum and kinetic energy are conserved. In collisions that are elastic the speed of separation equals the speed of approach.vapp=10 m s-1vsep=10 m s-1Calculating velocities for elastic collisions-6350396875v=3 m s-1v=2 m s-1ABFigure 23 – calculations for elastic collisions0v=3 m s-1v=2 m s-1ABFigure 23 – calculations for elastic collisionsTo determine the velocities of particles after an elastic collision we need to use relative velocities. Particle A has a mass of 3.0 kg and particle B has a mass of 4.0 kg.The total momentum of the system is +1.0 kg m s-1 and the total energy of the system is 21.5 J.-6350415290v=5 m s-1v=0 m s-1Figure 23 (a) – relative velocities in B’s frame of referenceAB0v=5 m s-1v=0 m s-1Figure 23 (a) – relative velocities in B’s frame of referenceABWe will count particle B as stationary and consider the relative velocity of particle A by adding +2.00 m s-1 to each velocity.The total momentum of the system before and after the collision can be calculated as:pA1+pB1=pA2+pB2We can isolate for the final momentum of particle B:pA1=pA2+pB2pA1-pA2=pB2mAvA1-mAvA2=mBvB2mAvA1-vA2=mBvB2The total kinetic energy would be calculated in a similar fashion:EKA1+EKB1=EKA2+EKB2We can isolate the final kinetic energy of particle BEKA1=EKA2+EKB212mAvA12=12mAvA22+12mBvB22mAvA12=mAvA22+mBvB22mAvA12-mAvA22=mBvB22mAvA12-vA22=mBvB22We can then divide the energy equation by the momentum equation:mAvA12-vA22=mBvB22 ; mAvA1-vA2=mBvB2mAvA12-vA22mAvA1-vA2=mBvB22mBvB2And simplify to:vA12-vA22vA1-vA2=vB2vA1+vA2vA1-vA2vA1-vA2=vB2vA1+vA2=vB2This equation shows that the velocity of separation must equal the velocity of approach:vA1=vB2-vA2Replacing the velocity for particle B into the original momentum equation would give us:vA1+vA2=vB2pA1=pA2+pB2mAvA1=mAvA2+mBvB2mAvA1-mAvA2=mBvA1+vA2And solve for the final velocity of particle AmAvA1-mAvA2=mBvA1+mBvA2mAvA1-mBvA1=mAvA2+mBvA2vA1mA-mB=vA2(mA+mB)vA1mA-mBmA+mB=vA2For this example:5 m s-13.0 kg-4.0 kg3.0 kg+4.0 kg=vA2-0.71 m s-1=vA2And the final velocity of particle B:vA1+vA2=vB25.0 m s-1+-0.71 m s-1=vB2+4.3 m s-1=vB2The relative velocities of the particles are as follows:v2=0.71 m s-1v2=4.3 m s-1Figure 23 (b) – relative velocities after the collision from B’s initial frame of referenceABv2=0.71 m s-1v2=4.3 m s-1Figure 23 (b) – relative velocities after the collision from B’s initial frame of referenceABWe can revert the velocities back from their relative velocity to their actual velocity by subtracting +2.00 m s-1. The actual velocities are:v2=2.71 m s-1v2=2.3 m s-1ABFigure 23 (c) –velocities after the collision v2=2.71 m s-1v2=2.3 m s-1ABFigure 23 (c) –velocities after the collision The total momentum of the system is +1.0 kg m s-1 and the total energy of the system is 21.5 J. As both momentum and kinetic energy are conserved the collision is perfectly elastic. The speed of separation is equal to the speed of approach.vapp=5.0 m s-1vsep=5.0 m s-1Example 1A 0.300 kg toy train and 0.600 kg toy train are involved in an elastic collision on a straight section of a model rail. The 0.300 kg train, traveling at +3.50 m s-1 strikes the 0.600 kg train that has a velocity of -2.00 m s-1. Determine the velocities of the trains after the collision.Example 2Two balls are involved in an elastic head-on collision. If the first ball has a mass m1=2.5 kg was traveling at +4.0 m s-1 and the second ball has a mass m2=4.0 kg was traveling at -2.0 m s-1, determine the velocity of each ball after collision.Example 3The velocity-time graph in Figure 23 shows a collision between 2 carts on a rail track. The first, a mass of 400. kg moving at 30.0 m s-1 collides with a 600. kg cart that is at rest. 262255093980v / ms-112162024280480.20.40.60.81.0t/ s-4Mass 1Mass 2Figure 2400v / ms-112162024280480.20.40.60.81.0t/ s-4Mass 1Mass 2Figure 24What evidence is there that the collision is elastic?Calculate the impulse applied to each carriage.Calculate the average force applied to each carriage.Homework: Giancoli – Page 188, Questions 21-25Tsokos – Page 109-110, Questions 72-83 ................
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