...linear motion relations, the kinematic equations; 2) using the principles of conservation of energy and momentum. In this paper, we aim to validate the law of conservation of momentum. We do so by comparing results from two experiments conducted with a single ballistic launcher/pendulum apparatus. Hypothesis: The initial velocity of a ballistic pendulum can be determined using the law of conservation of momentum. Momentum should be conserved, based on the law of conservation of energy. If momentum is conserved, the velocity found using the law of conservation of momentum equation should equal the velocity found using projectile motion. Due to the law of conservation of momentum, the total momentum before the pendulum is swung equals the net momentum after the pendulum is swung. Introduction/Purpose The ballistic pendulum is a device where a ball is shot into and captured by a pendulum. The pendulum is initially at rest but acquires energy from the collision with the ball. Using conservation of energy it is possible to find the initial velocity of the ball. In this ball-pendulum system we cannot use the conservation of mechanical energy to relate the quantities because energy is transferred from mechanical to nonconservative forces. In the absence of external forces, the momentum of the system does not change no matter how complicated the collision. The initial momentum is always equal to the final momentum. Figure 1: Experimental Setup for Part 1 In the first part of the...
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...------------------------------------------------- Experiment 9 – Conservation of Linear Momnetum [Document subtitle] March 29, 2016 Engineering Physics 150 March 29, 2016 Engineering Physics 150 Objectives: The objectives of this laboratory experiment is to investigate the velocities and momentm of two carts before and after various types of collisions. Theory: * When objects collide and assuming there are no external forces are acting on the colliding objects, the principle of the conservation of momentum always holds. * For a two-object collision, momentum conservation is stated mathematically by the equation: * PTotali =PTotalf * m1v1i+m2v2i=m1v1f+m2v2f * When working with a complete inelastic collision, the two objects stick together after the collision, and the momentum conservation equation becomes: * PTotali =PTotalf * m1v1i+m2v2i=(m1+m2)vf * During this experiment, photogates will measure the motion of two carts before and after elastic collision. The cart masses can be measured by using a simple mass scale. * Then, total momentum of the two carts before collision will be compared to the total momentum of the two carts after collision. Equipment: 850 Universal | Dynamic Track | Two dynamic carts | Two picket fences | Mounting brackets | Mass Balance | Mass Bar | Two Photogates | Data: Part A – Elastic Collision with approx. equal masses: Trial | V1 i (m/s) | V2 i (m/s) | V1 f (m/s) | V2 f (m/s)...
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...experiments. Many questions were raised about the nature of photons after Einstein’s discoveries in 1905. Compton, curious to whether or not the law of conservation of momentum applied to photons, decided to study collisions between photons and electrons. He sought free electrons as these would be ideal to conduct his studies with since there would be no outside forces acting upon the electrons. However, free electrons rarely exist in nature. Alternatively, Compton chose a metal with a very low work function and a high frequency photon source. This way the energy required to eject the photoelectron would be considered insignificant compared to the total energy of the system. Compton used X – rays for his experiment given that they produce very high frequency photons. Read more in Physics « The Photoelectric EffectHow an Amplifier Works » Through his experiments Compton discovered that when a high energy photon collides with an electron there is a transfer of energy. Also, he discovered that the electron and the lower energy photon scatter after the collision. The equation that describes the conservation of energy in the collision is hf= hf’+ ½ mev2 (Figure 2) where hf is the energy of the scattered photon and ½ mev2 is the kinetic energy of the electron The photon loses energy in the collision because it is transferred to the electron in the form of kinetic energy. Since the energy of a photon is dependent upon its frequency, this transfer of energy...
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... Law of Conservation of Mechanical Energy Mechanical energy is stored in an object due to virtue of its motion. Potential energy and kinetic energy are the two types of mechanical energy, and the law of conservation of mechanical energy is associated with the conservation of these two energies in a system. Mechanical energy is basically a combination of potential and kinetic energy. Principle of Conservation of Mechanical Energy According to the law of conservation of mechanical energy, in an isolated system, that is, in the absence of non-conservative forces like friction, the initial total energy of the system equals to the total energy of the system. Simply stated, the total mechanical energy of a system is always constant (in case of absence of non-conservative forces). For instance, if a ball is rolled down a frictionless roller coaster, the initial and final energies remain constant. Conservative forces are those that don't depend on the path taken by an object. For example, gravity, spring and electrical forces are examples of mechanical energy. Conservation of Mechanical Energy Equation The quantitative relationship between work and energy is stated by the mechanical energy equation. UT = Ki + Pi + Wext = Kf + Pf, where, UT = Total mechanical energy Ki = Initial kinetic energy Kf = Final kinetic energy Pi = Initial potential energy Pf = Final potential energy Wext = External work done This is a general equation for mechanical energy conservation. In case...
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...Franke Horstmann Lab Report—Momentum Lab Survey Physical Science 09/09/2014 Abstract This experiment helped validate the law of conservation of momentum. The test was done by conducting two experiments with car A and car B, car A’s mass being known and car B’s mass unknown. Experiment A was done by sticking the cars in the center of the track and setting off the puncher then recording the car velocity. Experiment B was done by placing car B in front of the second photogate and car A at the start of the track, again the trigger of car A was set off and it collided with car B and the car velocity was recorded. Our results show momentum and kinematics methods show highly accurate results. Introduction Newton’s 3rd Law states that for every action there is an equal and opposite reaction. There were two experiments conducted in order to validate the law of momentum conservation. This law states that when there is a collision of two objects the total momentum of the two before the collision should be equal to the total momentum of the two objects after the impact. Materials * Car track with slots * 2 rolling cars * 2 Photogates * Weights * Data Acquisition Board * Computer with measurement software * 2 picket fences * Ruler * Balance scale Procedure In order to perform the experiment the car track and balance scale were checked in order to make sure they were well-adjusted so that the results were accurate. Next...
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...Momentum and Energy in 2D Name__________________________________________ Date ________ Explosions: Conservation of Momentum: Note: If particles are not emitted at 900 Then must use Law of Sines: P1 / sin(1800-((1 + (2)) = P2 / sin(3 = P3 / sin(2 Or Law of Cosines P32 = P22 + P12 - 2P1P2cos(2 P22 = P32 + P12 - 2P3P1cos(3 P12 = P32 + P22 - 2P3P2cos(1800-((1 + (2)) Collisions: Conservation of Momentum: Examples: An unstable nucleus of mass 17 x 10-27 kg, initially at rest, disintegrates into three particles. One of the particles, of mass 5.0 x 10-27 kg, moves along the positive y-axis with a speed of 6.0 x 106 m/s. Another particle of mass 8.4 x 10-27 kg, moves along the positive x-axis with a speed of 4.0 x 106 m/s. Determine the third particle’s speed and direction of motion. (Assume that mass is conserved) Write down what you know mnucl = 17 x 10-27 kg m1 = 5.0 x 10-27 kg m2 = 8.4 x 10-27 kg m3 = mnucl – (m1 + m2) = 17 x 10-27 kg – (5.0 x 10-27 kg + 8.4 x 10-27 kg) = 3.6 x 10-27 kg vnuci = 0 m/s v1 = 6.0 x 106 m/s @ y-axis v2 = 4.0 x 106 m/s @ x-axis v3 = ? Draw a vector diagram of the momentum Solve for the momentum of the third particle and then find its velocity Right angled triangle so use Pythagorean Theorem P12 + P22 = P32 therefore:...
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...Abstract: The purpose of this lab was to understand momentum and energy and how they are related in the event of a collision. This experiment played around with that and different mass on carts was projected to elastic and inelastic collisions. One cart would be stationary while another was plunged towards it or they were both moving. As a result the elastics final kinetic energy was pretty much equal to the initial kinetic energy. In the inelastic case the final kinetic energy is normally less than the initial kinetic energy. Introduction: There were two main goals when performing this experiment. This first goal being to accurately demonstrate linear momentum conservation for collisions and classify whether they are elastic or inelastic. The second goal was to measure the energy loss in inelastic collisions. During this experiment we measured initial and final velocities, momentum, change in momentum, initial and final kinetic energies, and change in kinetic energies. The change in kinetic energies was what distinguished whether it was an inelastic or elastic collision. Theory and Derivations: The velocities of two carts were measured as a function of time. Two carts were attached to rotary motion sensors...
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...Collisions and Conservation of Momentum Visit the website http://phet.colorado.edu/en/simulation/collision-lab & complete the following: 1. In the green box on the right side of the screen, select the following settings: 1 dimension, velocity vectors ON, momentum vectors ON, reflecting borders ON, momenta diagram ON, elasticity 0%. Look at the red and green balls on the screen and the vectors that represent their motion. a. Which ball has the greater velocity? b. Which has the greater momentum? 2. Explain why the green ball has more momentum but less velocity than the red ball (HINT: what is the definition of momentum?). 3. Push “play” and let the balls collide. After they collide and you see the vectors change, click “pause”. Click “rewind” and watch the momenta box during the collision. Watch it more than once if needed by using “play”, “rewind”, and “pause”. Zoom in on the vectors in the momenta box with the control on the right of the box to make it easier to see if necessary. a. What happens to the momentum of the red ball after the collision? b. What about the green ball? c. What about the total momentum of both the red and green ball? 4. Change the mass of the red ball to match that of the green ball. a. Which ball has greater momentum now? b. How has the total momentum changed? c. Predict what will happen to the motion of the balls after they collide. 5...
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...Collisions and Conservation of Momentum Visit the website http://phet.colorado.edu/en/simulation/collision-lab & complete the following: 1. In the green box on the right side of the screen, select the following settings: 1 dimension, velocity vectors ON, momentum vectors ON, reflecting borders ON, momenta diagram ON, elasticity 0%. Look at the red and green balls on the screen and the vectors that represent their motion. a. Which ball has the greater velocity? b. Which has the greater momentum? 2. Explain why the green ball has more momentum but less velocity than the red ball (HINT: what is the definition of momentum?). 3. Push “play” and let the balls collide. After they collide and you see the vectors change, click “pause”. Click “rewind” and watch the momenta box during the collision. Watch it more than once if needed by using “play”, “rewind”, and “pause”. Zoom in on the vectors in the momenta box with the control on the right of the box to make it easier to see if necessary. a. What happens to the momentum of the red ball after the collision? b. What about the green ball? c. What about the total momentum of both the red and green ball? 4. Change the mass of the red ball to match that of the green ball. a. Which ball has greater momentum now? b. How has the total momentum changed? c. Predict what will happen to the motion of the balls after they collide. 5...
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...Section 2 Conservation of Momentum Lab performed: 2/18/13 Report submitted: 2/20/13 Sample Calculations Results The magnitudes of the masses for the gliders before the collision were: mass XA1 is equal to 0.107m, and mass XB1 is equal to 0.101m. The magnitudes after the collision of the masses were: mass XA2 equal to 0.0890m, and mass XB2 equal to 0.0820m. The momentum of the masses before collision were: mass PA1 equal to 0.3737 (kg m/s), and mass PB1 equal to 0.3551(kg m/s). The momentum after collision of the masses were: PA2 equal to 0.3109 (kg m/s), and mass PB2 equal to 0.2886 (kg m/s). When considering the direction of the vectors the area of uncertainty is small when considering the area of both parallelograms of uncertainty, because the overlapping of the parallelograms is only a small portion of each. The momentum was partially conserved within the error range of the parallelograms, or the portion where they overlap. Conclusions 1. It is necessary that they glide on a cushion of air, so that they can avoid any friction which would slow down their movement and could possibly keep them from colliding. The friction would differ for each mass, and would change all the predicted values. If the masses were not on an air cushion, it is impossible to predict that the two masses would ever collide because of the differing frictions for each mass. 2.If the masses were doubled, the before and after collision velocities...
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...Conservation of Momentum Partial Lab Report Results Summary Elastic Collision Initial Momentum = .414 N | Initial Momentum = 0 N | Initial Kinetic Energy = .084 J | Initial KE = 0 J | V1’ = .0596 m/s | V2’ = .462 m/s | Final Momentum = .061 N | Final Momentum = .354 N | Final KE = .00182 J | Final KE = 0.082 J | Inelastic Collision Initial Momentum = n/a | Initial Momentum = n/a | Initial Kinetic Energy = n/a | Initial KE = n/a | V1’ = n/a | V2’ = n/a | Final Momentum = n/a | Final Momentum = n/a | Final KE = n/a | Final KE = n/a | Discussion For the first part of our lab we were able to successfully show that both kinetic energy and momentum were conserved during the collision. As stated in the lab procedure, we kept one mass heavier than the other and made the heavier object collide with the lighter object at rest. Looking back, there wasn’t too much difference between the sizes of the mass. There was only a 300 gram difference in the system and that may have helped in keeping the collision elastic. The speed was recorded as .404 m/s and that also may have helped in causing the object at rest to bounce off the moving object. We were not able to complete the inelastic collision part of the lab because we were not able to make the objects stick together. We were not sure if the plane the objects were set on was level and this may have caused the objects to keep bouncing away from each other. We also kept the weight and velocity (roughly) the same...
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...in three unique scenarios of both elastic and inelastic collisions, I was able to conclude that speed and velocity are both conserved. This conclusion was made through the change in velocity of a cart relative to its "crasher". For example: throughout the different-mass elastic collisions, the speed that which the cart started and ended were equivalent - indicating that there was conservation of speed, in addition to the complete transfer of velocity. However, in the different-mass elastic collisions, the transfer of the speed of the cart was not complete, but instead, the lighter cart moved quicker than the heavier cart. This shows us that although force may be the same, the transfer of momentum shows us why the lighter cart moves more quickly than the slower. Throughout our previous unit, we described the constant velocity of objects in motion. That laid the basis for this next unit, where we will be studying why and how the object moves the way it does, specifically the "push" or "pull" of force. The heavier cart in a same-direction elastic collision seems to push the lighter cart, which causes an increase in speed for the lighter cart. Although we may have brushed on the surface of movement, this unit will pave the path for further investigation on velocity as well as momentum. According to today's lab, it is possible to measure the mass of the carts and then multiple the mass by the velocity to determine momentum. These two things will be related to almost everything...
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... properties of matter and heat. • resolve and interpret quantitative and qualitative problems in an analytical manner. • acquire an overall perspective of the inter-relationship between the various topics covered and their applications to the real world. • acquire laboratory skills including the proper handling and use of laboratory apparatus and materials. 11. Learning Outcome of Unit: At the end of the course, students will be able to: 1. Identify and practice the use of units and dimensional analysis, uncertainty significant figures and vectors analysis. 2. Apply and solve problems related to translational and rotational kinematics and dynamics in one and two dimensions. 3. Apply and solve problems related to the conservation of energy. 4. Identify and compare the state and properties of matter, and fluid mechanics with various related principles, and kinetic theory of gases. 5. Identify the difference in temperature scales and solve problems related to the concept of heat and heat transfer. 6. Carry out some procedures and techniques for practical investigations and interpret the results by the...
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... properties of matter and heat. • resolve and interpret quantitative and qualitative problems in an analytical manner. • acquire an overall perspective of the inter-relationship between the various topics covered and their applications to the real world. • acquire laboratory skills including the proper handling and use of laboratory apparatus and materials. 11. Learning Outcome of Unit: At the end of the course, students will be able to: 1. Identify and practice the use of units and dimensional analysis, uncertainty significant figures and vectors analysis. 2. Apply and solve problems related to translational and rotational kinematics and dynamics in one and two dimensions. 3. Apply and solve problems related to the conservation of energy. 4. Identify and compare the state and properties of matter, and fluid mechanics with various related principles, and kinetic theory of gases. 5. Identify the difference in temperature scales and solve problems related to the concept of heat and heat transfer. 6. Carry out some procedures and techniques for practical investigations and interpret the results by the...
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...KINEMATICS 1. INTRODUCTION TO KINEMATICS Dynamics is the study of moving objects. The subject is divided into kinematics and kinetics. Kinematics is the study of a body’s motion independent of the forces on the body. It is the study of the geometry of motion without consideration of the causes of motion. Kinematics deals only with relationships among the position, velocity, acceleration, and time. Kinetics deals with both forces and motion. 2. PARTICLES AND RIGID BODIES Bodies in motion can be considered particles if rotation is absent or insignificant. Particles do not possess rotational kinetic energy. All parts of a particle have the same instantaneous displacement, velocity, and acceleration. A rigid body does not deform when loaded and can be considered as a combination of two or more particles that remain at a fixed, finite distance from each other. 3. COORDINATE SYSTEMS The position of a particle is specified with reference to a coordinate system. A coordinate can represent a position along an axis, as in the rectangular coordinate system or it can represent an angle, as in the polar, cylindrical, and spherical coordinate systems. In general, the number of degrees of freedom is equal to the number of coordinates required to completely specify the state of an object. If each of the coordinates is independent of the others, the coordinates are shown as holonomic coordinates. 4. CONVENTIONS OF REPRESENTATION ...
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