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Motion II Momentum and Energy
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Momentum Obviously there is a big difference between a truck moving 60 mi/hr and a baseball moving 60 mi/hr. Obviously there is a big difference between a truck moving 60 mi/hr and a baseball moving 60 mi/hr. We want a way to quantify this. We want a way to quantify this. Newton called it Quantity of Motion. We call it Momentum. Newton called it Quantity of Motion. We call it Momentum.
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Definition: Momentum = (mass)x(velocity) p = mv Momentum is a vector. Momentum is a vector. Combination of mass and velocity Combination of mass and velocity Large momentum for massive objects moving slowly or small objects moving quickly Large momentum for massive objects moving slowly or small objects moving quickly
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Newton’s 2 nd Law Note: Sometimes we write t instead of just t. They both represent the time interval for whatever change occurs in the numerator. Note: Sometimes we write t instead of just t. They both represent the time interval for whatever change occurs in the numerator.
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Newton’s 2 nd Law Since p = mv, we have p = (mv) Since p = mv, we have p = (mv) If m is constant (not true for rockets and ballons), p = m v and v/ t = a If m is constant (not true for rockets and ballons), p = m v and v/ t = a
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Rearrange Newton’s 2 nd Law This is the impulse This is the impulse Impulse is the total change in momentum Impulse is the total change in momentum
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Example Baseball Baseball p= p f – p i p= p f – p i = mv f – mv i F = p/ t F = p/ t t is the time the bat is in contact with the ball t is the time the bat is in contact with the ball v: bring the ball to rest, accelerate in other direction v: bring the ball to rest, accelerate in other direction
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Ion Propulsion Engine: very weak force (0.1 N, about the weight of a paper clip) but for very long times (months) Ion Propulsion Engine: very weak force (0.1 N, about the weight of a paper clip) but for very long times (months) Large momentum change Large momentum change Very efficient Very efficient
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A 70 kg person traveling at +15 m/s stops. What is the impulse? A. 525 kgm/s B. 1050 kgm/s C. -525 kgm/s D. -1050 kgm/s E. 2100 kgm/s F. -2100 kgm/s
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Example 70 kg person traveling at 15 m/s stops. 70 kg person traveling at 15 m/s stops. p = p f – p i = 0 kgm/s – 1050 kg·m/s = -1050 kg·m/s p = p f – p i = 0 kgm/s – 1050 kg·m/s = -1050 kg·m/s If the person is stopped by the windshield t = 0.01 s If the person is stopped by the windshield t = 0.01 s What is the force involved? What is the force involved?
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Force to stop 70 kg in 0.01s A. -1050 N B. -10500 N C. -105000 N D. -1050000 N
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Example 70 kg person traveling at 15 m/s stops. 70 kg person traveling at 15 m/s stops. p = p f – p i = 0 kgm/s – 1050 kg·m/s = -1050 kg·m/s p = p f – p i = 0 kgm/s – 1050 kg·m/s = -1050 kg·m/s If the person is stopped by an airbag t = 0.40 s If the person is stopped by an airbag t = 0.40 s What is the force involved? What is the force involved?
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Force to stop 70 kg in 0.40s A. -1050 N B. -2625 N C. -5250 N D. -105000 N
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Is the impulse the same in both cases? A. Yes B. No
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Crumple zones on cars increase the time for a car to stop. This greatly reduces the average force during accident. Crumple zones on cars increase the time for a car to stop. This greatly reduces the average force during accident. Air bags do the same thing Air bags do the same thing
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Conservation of Momentum Momentum is always conserved within a system Momentum is always conserved within a system “Momentum is conserved in all collisions.” “Momentum is conserved in all collisions.” Momentum of a single object can be changed Momentum of a single object can be changed
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Conservation of Momentum Newton’s 3 rd Law requires F 1 = -F 2 Newton’s 3 rd Law requires F 1 = -F 2 p 1 / t = - p 2 / t p 1 / t = - p 2 / t Or Or p 1 / t + p 2 / t = 0 p 1 / t + p 2 / t = 0 Or Or p 1 + p 2 = 0 p 1 + p 2 = 0 NO NET CHANGE IN MOMENTUM NO NET CHANGE IN MOMENTUM
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Example: Cannon 1000 kg cannon 1000 kg cannon 5 kg cannon ball 5 kg cannon ball Before it fires P c + P b = 0 P c + P b = 0 P total must be zero after it fires. If V b = 400 m/s M c V c + M b V b = 0 M c V c + M b V b = 0 V c = -M b V b /M c V c = -M b V b /M c = -(5kg)(400m/s)/(1000kg) = -(5kg)(400m/s)/(1000kg) = -2m/s = -2m/s
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ENERGY & WORK In the cannon example, both the cannon and the cannon ball have the same momentum (equal but opposite). In the cannon example, both the cannon and the cannon ball have the same momentum (equal but opposite). Question: Why does a cannon ball do so much more damage than the cannon? Question: Why does a cannon ball do so much more damage than the cannon?
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ENERGY & WORK In the cannon example, both the cannon and the cannon ball have the same momentum (equal but opposite). In the cannon example, both the cannon and the cannon ball have the same momentum (equal but opposite). Question: Why does a cannon ball do so much more damage than the cannon? Question: Why does a cannon ball do so much more damage than the cannon? Answer: ENERGY Answer: ENERGY
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ENERGY & WORK E ssentially, energy is the ability to make something move. E ssentially, energy is the ability to make something move. What transfers or puts energy into objects? What transfers or puts energy into objects? A push A push
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WORK We will begin by considering the “Physics” definition of work. We will begin by considering the “Physics” definition of work. Work = Force x displacement indirection of force Work = Force x displacement indirection of force W=F d W=F d 1) if object does not move, NO WORK is done. 1) if object does not move, NO WORK is done. 2) if direction of motion is perpendicular to force, NO WORK is done 2) if direction of motion is perpendicular to force, NO WORK is done
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Example Push a block for 5 m using a 10 N force. Push a block for 5 m using a 10 N force. Work = (10 N) x (5 m) Work = (10 N) x (5 m) = 50 Nm = 50 Nm = 50 kgm 2 /s 2 = 50 kgm 2 /s 2 = 50 J (Joules) = 50 J (Joules)
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I have not done any work if I push against a wall all day long without moving it. A. True B. False
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Doing WORK on a body changes its ENERGY Doing WORK on a body changes its ENERGY ENERGY is the capacity to do work. ENERGY is the capacity to do work. The release of energy does work, and doing work on something increases its energy The release of energy does work, and doing work on something increases its energy Two basics type of energy: Kinetic and Potential. Two basics type of energy: Kinetic and Potential.
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Kinetic energy: Energy of motion Kinetic energy: Energy of motion Potential energy: Energy of situation Potential energy: Energy of situation Position Position Chemical structure Chemical structure Electrical charge Electrical charge
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Kinetic Energy: Energy of Motion Push a ball of mass m with a constant force. Push a ball of mass m with a constant force. KE = W = F d = (ma)d KE = (ma)(½ at 2 ) = ½ m(at) 2 =½ mv 2 KE =½ mv 2
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If the speed of the truck doubles what happens to its kinetic energy? A. Increases by ½ B. Doubles C. Quadruples
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A 6000 kg truck travels at 10 m/s. What is its momentum? A. 3000 kg∙m/s B. 6000 kg∙m/s C. 30000 kg∙m/s D. 60000 kg∙m/s E. 300000 kg∙m/s F. 600000 kg∙m/s
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Potential Energy Lift an object through a height, h. Lift an object through a height, h. PE = W = F d= (mg)h PE = mgh Note: We need to decide where to set h=0.
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Other types of energy Electrical Electrical Chemical Chemical Thermal Thermal Nuclear Nuclear
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Conservation of Energy We may convert energy from one type to another, but we can never destroy it. We may convert energy from one type to another, but we can never destroy it. Examples using only PE and KE (Mechanical Energy). Examples using only PE and KE (Mechanical Energy). KE + PE = const. OR KE i + PE i = KE f + PE f
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Conservation of Energy Amount of mechanical energy (combined KE & PE) stays the same only if there is no friction Amount of mechanical energy (combined KE & PE) stays the same only if there is no friction
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Drop a rock off a 100m cliff Call bottom of cliff h = 0 Call bottom of cliff h = 0 Initial KE = 0; Initial PE = mgh Initial KE = 0; Initial PE = mgh Final PE = 0 Final PE = 0 Use conservation of energy to find final velocity Use conservation of energy to find final velocity mgh = ½mv 2 v 2 = 2gh
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5 kg rock at top of 100 m cliff. What is its potential energy? A. 4900 J B. 9800 J C. 500 J D. 980 J
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For our 100 m high cliff For our 100 m high cliff Note that this is the same as we got from using kinematics with constant acceleration Note that this is the same as we got from using kinematics with constant acceleration
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5 kg rock falls from a 100 m cliff. What is its kinetic energy at bottom? A. 500 J B. 2450 J C. 4900 J D. 9800 J
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A 5 kg ball is dropped from rest from the top of a 300 m building in Chicago. From conservation of energy, what is its speed at the bottom? A. 5880 m/s B. 2940 m/s C. 76.7 m/s D. 54.2 m/s
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Pendulum At bottom, all energy is kinetic. At bottom, all energy is kinetic. At top, all energy is potential. At top, all energy is potential. In between, there is a mix of both potential and kinetic energy. In between, there is a mix of both potential and kinetic energy. Bowling Ball Video Bowling Ball Video Bowling Ball Video Bowling Ball Video
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Energy Content of a Big Mac 540 Cal = 540000 cal 540 Cal = 540000 cal 1 Cal = 4186 Joules 1 Cal = 4186 Joules
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How high can we lift a 1kg object with the energy content of 1 Big Mac? PE = mgh PE = mgh h = PE/(mg) h = PE/(mg) PE = 2,260,440 J PE = 2,260,440 J m = 1 kg m = 1 kg g = 9.8 m/s 2 g = 9.8 m/s 2
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2 Big Macs have enough energy content to lift a 1 kg object up to the orbital height of the International Space Station. 2 Big Macs have enough energy content to lift a 1 kg object up to the orbital height of the International Space Station. 150 Big Macs could lift a person to the same height. 150 Big Macs could lift a person to the same height. It would take about 4,000,000 Big Macs to get the space shuttle to the ISS height. It would take about 4,000,000 Big Macs to get the space shuttle to the ISS height. Note: McDonald’s sells approximately 550,000,000 Big Macs per year. Note: McDonald’s sells approximately 550,000,000 Big Macs per year.
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Cannon: Energy KE b = ½ mv 2 = 0.5 (5kg)(400m/s) 2 = 400,000 kgm 2 /s 2 = 400,000 kgm 2 /s 2 = 400,000 J KE c = ½ mv 2 = 0.5 (1000kg)(2m/s) 2 = 2,000 kgm 2 /s 2 = 2,000 kgm 2 /s 2 = 2,000 J Note: Cannon Ball has 200 times more ENERGY
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When I drop a rock, it start off with PE which gets converted in to KE as it falls. At the end, it is not moving, so it has neither PE or KE. What happened to the energy? 1. Energy is not conserved in this case. 2. The energy is converted into Thermal energy 3. The energy is transferred to whatever it is dropped on.
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Important Equations
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