Unit+2

toc J || W=F*d*CosPhy d=distance traveled phy= angle between force and direction of motion || N, T or f along the axis of motion || J || KE=1/2mv^2 || object is moving || PE g || Joules J || GPE=mg h GPE=mg y (vertical position) || location of the object is above or below 0 || PE || Joules J || EPE= 1/2kx^2 k= spring force constant(how springy) x=distance of stretching. || stretched or compressed spring. || Law of Conservation of energy (LCE) - energy cannot be created or destroyed but it can be transferred. only work can take out or put in energy in a system.'
 * Term || Definition || symbol || unit || formulas || Cue ||
 * Work || There must be a force, the force must be in the direction of motion and there must be movement. Work is caused be a force acting on the axis of the direction of motion. || W || Joules
 * Kinetic energy || Energy due to an object moving. || KE || Joules
 * gravitational potential energy || something has the ability to fall down. energy possessed by an object because it is above or below zero level.(relative/you define) || GPE
 * elastic potential energy || energy possessed by an object due to an compressed or stretched spring. || EPE

Lab: air bag lab
Objective:How does an air bag protect you during an accident? Hypothesis: the airbag absorbs the force of ones motion so that the distance is made in the air bag rather then in one's head. This is similar to how the flour will absorb the force and work of the falling egg and prevent it from denting itself. Materials: meter stick, bowl, egg, flour, bag. __Procedure:__

//Note: You may want to use the available technology to take "Before" and "After" pics to post in your data table to assist and elaborate on your written descriptions.//


 * 1) Measure the length of your egg #1. Measure the mass of your egg. Record this information.
 * 2) Place an egg in a ziplock bag, squeezing out all of the air in the bag before sealing.
 * 3) Hold a ruler up on the table vertically. Hold the egg vertically at the 2 cm mark. (Keep the excess bag on top.) Drop it.
 * 4) Hold the egg the same exact way at the 4-cm mark and repeat. Continue this process until the egg shell is slightly cracked.
 * 5) Continue until the egg is smashed and the yolk leaks out. Measure the amount of egg still undamaged. How much of the egg is smashed?
 * 6) Fill a bowl with flour and place the bowl inside of the box lid.
 * 7) Measure the length of your egg #2. Measure the mass of your egg. Record this information.
 * 8) Drop the egg from the smash height (Step 5). Measure the amount of egg sticking up out of the flour bed. How much of the egg is buried in the flour? Also, record your qualitative observations.
 * 9) Repeat this, increasing the height in 5-cm increments until the egg is cracked, and then smashed.

__Data and observations__:
 * Egg Mass (kg) || Egg Length(m) || Drop Height (m) || Cracked or Smashed? || How much egg is buried (m) || Initial Gravitational Potential Energy (J) || Work Done (J) || Force Experienced by egg (N) ||
 * 0.0598 || 0.055 || 0.37 || n/a || 0.002 || 0.217 || 0.217 || 108.417 ||
 * 0.0598 || 0.055 || 0.42 ||  || 0.007 || 0.246 || 0.246 || 35.162 ||
 * 0.0598 || 0.055 || 0.47 ||  || 0.011 || 0.275 || 0.275 || 25.040 ||
 * 0.0598 || 0.055 || 0.52 ||  || 0.015 || 0.305 || 0.305 || 20.316 ||
 * 0.0598 || 0.055 || 0.57 ||  || 0.015 || 0.334 || 0.334 || 22.270 ||
 * 0.0598 || 0.055 || 0.62 ||  || 0.018 || 0.363 || 0.363 || 20.186 ||
 * 0.0598 || 0.055 || 0.67 ||  || 0.019 || 0.393 || 0.393 || 20.666 ||
 * 0.0598 || 0.055 || 0.72 ||  || 0.021 || 0.422 || 0.422 || 20.093 ||
 * 0.0598 || 0.055 || 0.77 ||  || 0.023 || 0.451 || 0.451 || 19.620 ||
 * 0.0598 || 0.055 || 0.82 || crack || 0.024 || 0.481 || 0.481 || 20.023 ||
 * 0.0598 || 0.055 || 0.87 || crack || 0.026 || 0.510 || 0.510 || 19.610 ||
 * 0.0598 || 0.055 || 0.92 ||  || 0.027 || 0.539 || 0.539 || 19.969 ||
 * 0.0598 || 0.055 || 1 ||  || 0.034 || 0.586 || 0.586 || 17.236 ||
 * 0.0598 || 0.055 || 1.5 ||  || 0.042 || 0.879 || 0.879 || 20.930 ||
 * 0.0598 || 0.055 || 2 ||  || 0.038 || 1.172 || 1.172 || 30.844 ||
 * 0.0598 || 0.055 || 2.5 ||  || 0.038 || 1.465 || 1.465 || 38.555 ||

__Calculations:__ Show a sample for each, with equation(s), numbers plugged in, and answer with correct units. > || So: .0598*9.8*.37 || > || .216 Joules is the GPE of .37 m || > || || > || W=GPE || > || So mgh=W || > || So .0598*9.8*.37 = w || > || So = .37 Joules || > || mgh=f*d || > || F=mgh/d || > || || > || So (.0598*9.8*.37)/.002 || > || 108.47 N = force || > || || > __Questions:__
 * 1) What is the initial gravitational potential energy?
 * Use: mgh = gpe ||
 * 1) How much work is done in each trial?
 * Use the fact ||
 * 1) How much force was used to stop the egg in each case?
 * Use W=GPE again ||

Conclusion/ analysis: The eggs were able to with stand much higher drops because there flour was there to absorb the impact. the flour increases the time of contact, reducing the force. this is similar to how an airbag protects our heads in a car accident. the air bag produces a drop in force which percents ones skulls from cracking like an egg. Most of the error in the lab was due to measuring issues and problems landing the egg in the flour in the proper direction.
 * 1)  This investigation is an analogy for a person in an automobile collision. What does the egg represent? What does the table represent? What does the flour represent ? The egg represents a persons head, the table is the car or place of impact and the flour is the airbag.
 * 2)  Define the terms: Gravitational Potential Energy, Kinetic Energy and Work. Gravitational potential energy is the energy stored in an object when it is above or below zero when it has the ability to fall. kinetic energy s energy in motion and work is forces acting in the same direction of movement and causing movement.
 * 3)  What factors determine an object's kinetic energy? The height the object is dropped from and the weight of the object and the velocity.
 * 4)  When work is done on an object, what is the effect on the object's kinetic energy? The kinetic energy increases because the object is now in motion.
 * 5)  How does the force needed to stop a moving object depend on the distance the force acts? The more the distance the force acts the more time of collision there is, there fore the force will be smaller.
 * 6)  What difference does a soft landing area make on a passenger during a collision? A soft landing area increases the time of the collision and makes the force smaller.
 * 7)  How does a cushion reduce the force needed to stop a passenger? The Cushion absorbed the impact and makes the time of contact longer, reducing force.
 * 8)  What does the law of conservation of energy have to do with this? Because energy is conserved the energy on both sides of the equation will remain the same and will have to equal the same value. this means that the variables on each side of the equation change proportionally to each other.

KEi+GPEi+EPEi+W in/out=KEf+GPEf+EPEf

HW: Work and Energy
A: What is work? What is its formula? What is necessary to do work? Work is caused be a force acting on the axis of the direction of motion. In the equation for force the F stands for the force, D is displacement and theta is the angle between the force and the displacement vector. If a waiter is supporting a plate while walking across a room the plate is taking in force from his hand. The displacement vector is to the right so the angle of the force is 90 degrees. This means that no work is being done because the cos of 90 is 0, causing the equation to become zero. Work is measured in joules. A lot of practice problems. B:What is potential energy? What types of potential energy are there? What are there formulas? Potential energy is energy stored because of position. There are two types of potential energy, gravitational potential energy and elastic potential energy. Gravitational potential energy is the energy stored in an object as the result of its vertical position or height. The energy is stored as the result of the gravitational attraction of the Earth for the object. The equation for this formula is PEgrav=mass*g*height. More massive objects have greater gravitational potential energy. There is also a direct relation between gravitational potential energy and the height of an object. The higher that an object is elevated, the greater the gravitational potential energy. elastic potential energy is the energy stored in elastic materials as the result of their stretching or compressing. Usually from a spring or band. Fspring= K*x, PEspring= 1/2k*x^2, K= spring constant and x is the amount of compression. C:What is kinetic Energy? Kinetic energy is energy in motion. There are many forms of kinetic energy - vibrational (the energy due to vibrational motion), rotational (the energy due to rotational motion), and translational (the energy due to motion from one location to another). Translational energy has the formula KE=1/2m*v^2. m = mass v=speed of object. Kinetic energy is a scalar quantity and has no direction. D& E: What is mechanical energy and how is it calculated? What is power and its formula? Mechanical energy is described as energy that has the ability to do work. Mechanical energy can come from kinetic energy or stored potential energy. TME= PE+KE, TME = PEgrav + PEspring + KE. Power is the rate at which work is done. Power = work/time. The unit of power is a watt. a Watt is equivalent to a Joule/second Lesson 2 a: What are internal and external forces? external forces include the applied force, normal force, tension force, friction force, and air resistance force. Internal forces include the gravity forces, magnetic force, electrical force, and spring force. if there is positive work, an object will gain energy if there is negative work an object will lose energy. as an object is "forced" from a high elevation to a lower elevation by gravity, some of the potential energy of that object is transformed into kinetic energy. Yet, the sum of the kinetic and potential energies remains constant. B& C: What types of relationships does work form? The quantitative relationship between work and mechanical energy is expressed by the following equation: TMEi+Wext=TMEf,The equation states that the initial amount of total mechanical energy (TMEi) plus the work done by external forces is equal to the final amount of total mechanical energy. The quantitative relationship between work and the two forms of mechanical energy is expressed by the following equation: KE+PE+Wext=KE+PE. Whenever work is done there is an change in mechanical energy. Practice problems... The final equation becomes KEi+GPEi+EPEi+W in/out=KEf+GPEf+EPEf. The following procedure might be useful for constructing work-energy bar charts:
 * analyze the initial and final states of the object in order to make decisions about the presence or absence of the different forms of energy
 * analyze the forces acting upon the object during the motion to determine if __ [|external forces are doing work] __ and whether __ [|the work (if present) is positive or negative] __
 * construct bars on the chart to illustrate the presence and absence of the various forms of energy for the initial and final state of the object; the exact height of the individual bars is not important; what is important is that the sum of the heights on the left of the chart is //balanced by// the sum of the heights on the right of the chart



Lab: elastic collision
__Objective:__ A small sports car hits a heavy truck in a collision. What factors determine the outcome for the passengers of the two vehicles? Which driver will sustain worse injuries? Why? Hypothesis: the sports car will sustain worse injuries because it has less mass than the truck. __Materials:__ List any materials used and draw a labeled diagram of your set-up (alternatively, include a snapshot or video).

__Procedure:__
 * 1) Place a cart on the middle of the track with the spring to the right. On a piece of masking tape, label this the "Target cart." Place a second identical cart on the right end of the track. Mark this as the "Bullet cart".
 * 2) Push the bullet cart very gently towards the target cart so that they collide, with the spring between them.
 * 3) Repeat step 2 several times, giving the bullet cart a bigger push each time. Record your observations.
 * 4) Add 500-g to each cart and repeat the process. Record your observations and compare the results to the first set of collisions.
 * 5) Remove the mass from the target cart and repeat the above steps.
 * 6) Add the mass to the target cart and remove the mass from the bullet cart, and repeat.
 * 7) Get the "Mystery" cart from your teacher. Determine the relative mass of the cart by putting it through a sequence of collisions.

__Data and observations:__
 * Trial || Mass of Bullet Cart (kg) || Mass of Target Cart (kg) || Applied force (qualitative only) || Description and Observations (qualitative) ||
 * 1 || .5 || .5 || Small force || After the contact, the second cart is repelled at a slower speed. ||
 * 2 || .5 || .5 || Medium force || The medium force caused more acceleration an movement ||
 * 3 || .5 || .5 || Large force || Large force caused the most movement. ||
 * 4 || 1 || 1 || Small force || The trial was roughly the same as trial 1 ||
 * 5 || 2 || .5 || Small force || The larger, uneven weight made a huge difference in movement and even a small force/ speed caused a large amount of movement. The target car gains a faster speed than the bullet car ever had. ||
 * 9 || 1 || 1.5 || Small force ||  ||
 * 6 || 1.5 || .5 || Small force || The mystery weight 2 was about 1 kg because the displacement between the measured weight and the mystery weight were very similar. ||
 * 7 || 1.3 || .5 || Small force || Mystery 1 is about .8kg ||
 * 8 ||  ||   ||   ||   ||

Mystery mass 1 is approximately 800g. because when we put .8kg on the bullet cart, and push bullet carts with similar force of trial 1, carts moved same distance as trial 1. Therefore the mass of mystery mass 1 is about .8kg

Mystery mass 2 is approximately 1000g. because when we put 1kg on the bullet cart, and push bullet carts with similar force of trial 1, carts moved same distance as trial 1. Therefore the mass of mystery mass 2 is about 1 kg

% error = (ABS(actual - experimental) / actual) *100 Actual mystery mass 1 was 700g so my percent error is about 14% Actual mass for mystery mass 2 was 1200g so the percent error is 16%

__Questions:__
 * 1)  What is a real-life collision that the collisions in this investigation could represent? this could represent a collision between a big truck and a small car on the highway.
 * 2)  How well did observing collisions enable you to compare the masses of the carts in the last step? Observing the collisions gave me a very qualitative understanding of how the masses relate to each other but it was still very difficult to put an exact value on the weight.
 * 3)  What happened after the collision as the masses changed? as the difference in the m,asses increases so did the energy transferred and the movement of the second, smaller car.
 * 4)  Define the term momentum. Momentum is how fast something is moving relative to its mass.
 * 5)  Which object has greater momentum, a butterfly traveling at 16 km/h or an eagle traveling at the same speed? the eagle because it has more mass.
 * 6)  How does the transfer of momentum occur? As two objects collide the momentum work against each other and form negative momentum. This results in a loss of momentum of the original objects to be transferred to other objects in the collision.
 * 7)  Use momentum to describe what would happen if a skaterboarder was hit by a car .The car has much more mass and therefore more momentum. the skate boarder would be thrown back by the collision and experience negative momentum.

__Conclusion/analysis:__ If a heavy truck and a small sports car moving at the same speed were to collide then the the sports car would be forced backward because of the large difference in momentum. these results were seen in the lab as one added weight to the bullet car. the experimental error was due to different forces pushing the carts without being able to regulate them. Also the collisions were different every time and they are sometimes limited by the space provided. it was very difficult to determine the weight of the mystery mass qualitatively and this accounts for most of the percent error.

Lab: Inelastic collision:
__Objective__: how does the initial momentum compare to the final momentum? Hypothesis: The total momentum of the system will stay the same because it is an isolated system. Even if weight is added to the bullet cart the momentum will stay constant. __Procedure:__ Data:
 * 1) Place 2 photogates on a track, one at 50 cm and the other at 75 cm. Hook these into a Smart Timer and set the Smart Timer to Velocity.
 * 2) Place a cart so that its end is at 75 cm (right in front of the photogate), with the velcro facing 0-cm. Call this the "target cart." Place a second identical cart at the end of the track closest to 0-cm. Call this the "Bullet cart". Make sure that both carts have picket fences.
 * 3) Click "Start" on the Smart Timer, and then push the Bullet cart very gently towards the Target cart so that they collide and stick together. You may need to practice this a few times.
 * 4) Record the velocity right before (the Bullet Cart initially) and right after the collision (the two carts linked). Record these values in your data table.
 * 5) Vary the masses of the carts and repeat the process 5 times.

kgm/s || percent error % ||  || __Calculations:__ Show equation(s), numbers plugged in, and answer with correct units. Add columns in your data table to include these results. __Questions:__ the initial momentum ant the final momentum are about the same for each trial because momentum is conserved.
 * Mass of Bullet Cart (kg) || Mass of Target Cart (kg) || Speed of Bullet Cart (m/s) || Speed of Target cart (m/s) || Combined masses (kg) || Final Velocity of both carts (m/s) || initial momentum kgm/s || final momentum
 * 0.5 || 0.5 || 0.083 || 0 || 1 || 0.04 || 0.0415 || 0.04 || 4 ||  ||
 * 0.5 || 0.5 || 0.066 || 0 || 1 || 0.031 || 0.033 || 0.031 || 3.1 ||  ||
 * 0.5 || 0.5 || 0.083 || 0 || 1 || 0.041 || 0.0415 || 0.041 || 4.1 ||  ||
 * 0.5 || 0.5 || 0.072 || 0 || 1 || 0.034 || 0.036 || 0.034 || 3.4 ||  ||
 * 0.5 || 0.5 || 0.083 || 0 || 1 || 0.04 || 0.0415 || 0.04 || 4 ||  ||
 * 1 || 0.5 || 0.053 || 0 || 1.5 || 0.033 || 0.053 || 0.0495 || 4.95 ||  ||
 * 1 || 0.5 || 0.045 || 0 || 1.5 || 0.029 || 0.045 || 0.0435 || 4.35 ||  ||
 * 1 || 5 || 0.063 || 0 || 1.5 || 0.4 || 0.063 || 0.6 || 60 ||  ||
 * 1 || 0.5 || 0.038 || 0 || 1.5 || 0.022 || 0.038 || 0.033 || 3.3 ||  ||
 * 1 || 0.5 || 0.073 || 0 || 1.5 || 0.049 || 0.073 || 0.0735 || 7.35 ||  ||
 * 0.5 || 1 || 0.12 || 0 || 1.5 || 0.033 || 0.06 || 0.0495 || 4.95 ||  ||
 * 0.5 || 1 || 0.08 || 0 || 1.5 || 0.025 || 0.04 || 0.0375 || 3.75 ||  ||
 * 0.5 || 1 || 0.125 || 0 || 1.5 || 0.041 || 0.0625 || 0.0615 || 6.15 ||  ||
 * 0.5 || 1 || 0.091 || 0 || 1.5 || 0.029 || 0.0455 || 0.0435 || 4.35 ||  ||
 * 0.5 || 1 || 0.125 || 0 || 1.5 || 0.04 || 0.0625 || 0.06 || 6 ||  ||
 * 1.5 || 1 || 0.051 || 0 || 2.5 || 0.029 || 0.0765 || 0.0725 || 7.25 ||  ||
 * 1.5 || 1 || 0.047 || 0 || 2.5 || 0.026 || 0.0705 || 0.065 || 6.5 ||  ||
 * 1.5 || 1 || 0.092 || 0 || 2.5 || 0.055 || 0.138 || 0.1375 || 13.75 ||  ||
 * 1.5 || 1 || 0.052 || 0 || 2.5 || 0.029 || 0.078 || 0.0725 || 7.25 ||  ||
 * 1.5 || 1 || 0.05 || 0 || 2.5 || 0.028 || 0.075 || 0.07 || 7 ||  ||
 * ||  ||   ||   ||   ||   || m1i*v1i || totalmass*Vf ||   ||   ||
 * ||  ||   ||   ||   ||   ||   ||   |||| final-initial/initial*100 ||
 * 1) Find the initial momentum of the bullet cart for each trial.
 * 2) Find the initial momentum of the target cart for each trial.
 * 3) Find the sum of the initial momenta of the two carts for each trial.
 * 4) Find the final momentum of the combined carts for each trial.
 * 5) Find the percent difference between the initial momentum (calc 3) and the final momentum (calc 4).
 * 1) How do the initial momenta compare to the final momenta? Allowing for minor variations due to uncertainties of measurement, are there any patterns?

collisions where the two objects that collide bounce apart are inelastic collisions. Any collisions that do not directly keep their momentum are inelastic.
 * 1) Which types of collisions are definitely inelastic? How do you know?

Elastic collisions are collisions where the original and final momentum are equal.
 * 1) Which types of collisions are definitely elastic? How do you know?

the cue ball is slightly larger than the 15 ball has velocity. this makes its momentum much higher and causes it to pass on its momentum to the 15 ball sending it across the table. __Conclusion/analysis:__ the traffic cops were able to reconstruct he accident because the momentum of the crash was conserved and remained the same both before and after the accident. One of the major causes for experimental error was that release and timing of the cart. Although this was done with photo gate timers the carts would sometimes be mush slower to pass the laser or even start early. This error would account for the slight differences between the initial and final momentum.
 * 1) Use the law of conservation of momentum to describe what happens when a cue ball hits the 15 balls in the middle of the pool table.

Hw: Momentum
A: Momentum is the quantity of motion an object has. If an object is in motion, it has momentum. Momentum=mass*velocity or p=m*v. The equation illustrates that momentum is directly proportional to an object's mass and directly proportional to the object's velocity. The unit for momentum is kg*m/s. Momentum is a vector quantity and involves a direction. B: the more momentum an object has the harder it will be to stop it. an unbalanced force will always accelerate an object, slowing it down or speeding it up. it the velocity of an object changes so does it momentum. The quantity of force*time is known as impulse. m*delta v=change in momentum.The result of the force acting for the given amount of time is that the object's mass either speeds up or slows down (or changes direction). The impulse experienced by the object equals the change in momentum of the object. C: The equations above show that time and force are inversely related. To minimize the effect of the force on an object involved in a collision, the time must be increased. And to maximize the effect of the force on an object involved in a collision, the time must be decreased. This has a lot of practical uses in everyday life. Safety equipment like the airbag and seat belt extend the time of a collision and therefore decrease the force. Often in a collision objects bounce off of each other or rebound. Rebounding involves a change in the direction of an object; the before- and after-collision direction is different.large momentum change is proportional to a large impulse. it is safer for cars to hit, crumble up and stay together in a crash because force will be lower. Lesson 2 a: Law of action reaction. When two objects collide they produce an equal force on each other but their momentum changes. This means if the two objects are of unequal masses than they will experience unequal accelerations.if a person is to push forward they will receive a backwards force. B: The law of momentum conservation states that the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision. The momentum lost by object 1 is equal to the momentum gained by object 2. In a collision the time and forces are equal meaning that the impulses are also equal but opposite in direction. If the impulses are equal than the Change in momentum is also equal in opposite directions. The momentum lost by object 1 is equal to the momentum gained by object 2. C: Isolated systems: An isolated system is a system that is free from the influence of a net external force that could alter momentum. A net external force is a force from objects outside of the system that is not balanced out by other forces.The momentum lost by one object is equal to the momentum gained by another object. D: The equation m1*v1i+m2*v2i=m1*v1f+m2*v2f, for this problem below the momentum of the skater is 0 and the initial momentum of the of the ball is 300. The initial and final momentum do not change, therefore 15vf+60vf=300, Vf =4km/hour. practice problems..

In order for the momentum before the collision to be equal to the momentum after the collision, the after collision velocity must be smaller than the before collision velocity. If the mass is increased by a factor of two than the velocity is decreased by a factor of 2. Mass and velocity are inversely proportional. if the mass increases by 2/3 than divide the original velocity by 2/3. More practice.. E: Momentum conservation in explosions. If the vector sum of all individual parts of the system could be added together to determine the total momentum after the explosion, then it should be the same as the total momentum before the explosion. The impulse of the explosion changes the momentum of the tennis ball as it exits the muzzle at high speed. The cannon experienced the same impulse, changing its momentum from zero to a final value as it recoils backwards. Due to the relatively larger mass of the cannon, its backwards recoil speed is considerably less than the forward speed of the tennis ball. If the ball acquires 50 units of forward momentum, then the cannon acquires 50 units of backwards momentum. The vector sum of the individual momenta of the two objects is 0. Total system momentum is conserved. m1 deltav=-m2delatv []