# Good Example Of Report On Elastic And Inelastic Collisions

Type of paper: Report

Topic: Glider, Traffic, Crash, Collision, Experiment, Conservation, Velocity, Wind

Pages: 9

Words: 2475

Published: 2021/01/06

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## Introduction

The conservation of momentum is one of the greatest concepts in physics. An analysis of momentum can be done by performing an experiment involving collision of bodies. Collisions can be elastic or inelastic; but in all cases, they obey the law of the conservation of momentum. The law of the conservation of momentum states tha for a system of moving bodies that are not acted upon by an external force, the total momentum remains constant (Cameron and Craig, 2011). Momentum (P) is a product of mass (m) and velocity (v); P = mv (Avison, 1989). Given that momentum is conserved, it can be deduced that the final momentum of a system of bodies is equal to the initial momentum of a system;

m1v1i+m2v2i=m1v1f+m2v2f, kg*ms [1]

This experiment involves the investigation of the conservation of momentum and kinetic energy for elastic collision, and the conservation of momentum for perfectly inelastic collisions. The hypothesis is that the momentum of a system of bodies is conserved. Collision is described as an act where a body collides with another or other bodies with each body exerting a momentary force on each other. Examples of collisions in real life include two vehicles on head-on collision during an accident, a meteor striking the earth, a footballer kicking a ball (foot coming in contact with a ball), and a baseball bat coming in contact with a baseball. Basically, collisions can occur in any of the three forms. While some are elastic, others are inelastic. Collisions can be also perfectly inelastic. While momentum is only conserved in an inelastic collision, both the momentum and kinetic energy are conserved in an elastic collision. A perfectly inelastic collision, such as that of a meteorite striking the earth, occur when the bodies colliding stick to each other after collision while. Collisions are best analyzed in terms of impulse and momentum. The momentum of an object is a vector quantity expressed as a product of mass and velocity. Since the direction of the momentum vector occurs in the same direction as that of the velocity, differentiating momentum with respect to time yields force (product of mass and acceleration) as per the Newton’s Second Law of motion. In a collision involves several objects colliding with each own velocity and momentum, the total linear momentum of the system is achieved by summation of individual momenta of the particles. Whether a collision is elastic or inelastic, the vector sum of momentum prior to and after collision is equal (See Eq. 1 and Eq.2). However, for elastic collisions, the kinetic energy is also conserved (See Eq.3 and Eq.4)

pi=pf [2]

m1v1i+ m2mv2i=m1v1f+ m2v2f, kg*ms [3]

Ki=Kf [4]

12m1v1i2+12m2v2i2=12m1v1f2+12m2v2f2, kg*ms [5]

All measures were taken to ensure that the experiment was free of errors such as reading errors. The experiment was also carried out while observing all the procedures and safety concerns. The experiment involved the testing of both elastic and inelastic collisions. Before testing began, all testing equipment were confirmed available and in good working condition. The experiment shall be therefore divided into two parts: Part A and Part B. Part A was about the investigation of inelastic collision while Part B was about investigation of elastic collision.

## Equipment Used

The following equipment was used in the experiment.

Air track

photogates

## Gliders

Accessories

Wax

Wax receptacle

Procedure

There are two parts to this experiment: Part A and Part B.

Part A

The purpose of Part A of this experiment was to investigate whether there is a conservation of momentum and kinetic energy when a moving body collides with a stationary body. In this experiment, one glider was placed stationary at a distance while another glider was to be set in motion. The glider moving glider stroke the stationary glider resulting in collision. All records were tabulated and analyzed accordingly.

## The apparatus were set up accordingly

The air track was first checked if it was level. Then three photogates (G1, G2, and G3) were position in such a way that G1 was approximately 60cm away from the left end of the air track, G2 near the middle of the air track, and G3 approximately 60 cm from the right end of the air track.

A flag was placed on top of each of the two gliders (1 and 2). While glider 1 starts out from the left end of the track, glider 2 was always on the right of glider 1. An elastic bumper was placed was placed to the right of glider 1, while to the left of glider 1, an attachment that held the glider to the electromagnet was placed. However, on both sides of glider 2, electromagnet attachments were placed.

An investigation of collision between two bodies of different masses was then carried out. Glider 1 was chosen to be heavier than glider 2. Masses were added to glider, symmetrically to each wing of the glider. In this experiment, one mass per each wing was added.

The width of the flags were measured and the values recorded together with the uncertainties into the software. Each of the gliders were weighed with their attachments and additional masses. The value of glider masses and their uncertainties were recorded.

The air supply pump and the electro-magnet were turned on using the electro-magnet switch in the measurement window and glider 1 attached to the launcher. Glider 2 was positioned in such a way that its flag was located a few centimeters to the right of photogate G2. This was done while ensuring that the flag on glider 1 was to the left of photogate G2 at the time of collision. Photogate G3 was used to measure velocity of glider G2 after collision.

Glider 1 was launched by clicking on the GO button. Records of blocking times of glider 1 and glider 2 flags were tabulated. The measurements were repeated until a total of 10 trials were completed.

## Part B

Inelastic Collision with a Stationary Object

The purpose of the experiment was to investigate the conservation of momentum and kinetic energy collision is perfectly inelastic. In this experiment, two gliders were used. Unlike in Part A where the gliders collided and moved separately, this experiment involved one glider colliding with another one and moving together in the same direction. The two got stuck to each other and moved together as a single object with the same velocity.

Since photogates G1 and G2 were used in the experiment, photogate G3 was moved away from the air track. Each of the photogates G1 and G2were moved approximate 60cm away from the left and right end of the air track respectively.

Wax receptacle was placed on the left side of glider 2 while ensuring that the needle was inserted in the wax receptacle.

## Some additional masses were added to glider 1 while ensuring that they were symmetrically placed to each wing of the gliders.

Each of the gliders were then weighed together with their additional masses. The masses and their uncertainties were recorded into the measurement window.

## Glider 1 was attached to the launcher and position glider 2 between the photogates.

Then glider 1 was launched. Just after collision, it was observed that the gliders stuck together and continued through photogate G2.

## The time intervals was recorded on the screen

The measurements were repeated until a total of 10 trials were completed.

Initial Data and Preliminary Analysis

Initial Data

Calculated Data:

Analysis

The velocity for Glider 1 was found by taking the width of the glider’s flag and dividing it by the average time it took out of ten trials for glider 1 to pass photogate 1.

vG1i=wG1tavg=0.025m0.0547s=0.4570ms

The Uncertainty in Velocity of glider 1 initially is found by first finding the uncertainty in width and time of velocity of glider 1. The uncertainty in width and time are both taken by calculation partial derivatives with respect to either distance or time.

∆vwG1i=∂w∂t=1tavg*∆w=0.0005m0.0547s=0.009140ms

∆vtG1i=∂t∂w=wG1tavg2*∆t=0.025m0.0547s2*0.001s=0.008355ms

## Thus, total uncertainty can now be found by squaring the values above, adding them, and then taking the square root.

∆vG1i=∆vwG1i2+∆vtG1i2=0.009140ms2+0.008355ms2=0.0124ms

These formulas could be repeated to complete the steps necessary to finding the velocities at photogate 2 and photogate 3. The velocity recorded at photogate 2 is the velocity of the first glider after the collision. And the velocity at the second photogate is the second glider after the collision.

vG1f=wG1tavg=0.025m0.2421s=0.1033ms

∆vwG1f=∂w∂t=1tavg*∆w=0.0005m0.2421s=0.002065ms

∆vtG1f=∂t∂w=wG1tavg2*∆t=0.025m0.2421s2*0.001s=0.0004265ms

∆vG1f=∆vwG1f2+∆vtG1f2=0.002065ms2+0.0004265ms2=0.002109ms

## The final velocity and its uncertainty for glider 2 is shown below:

vG2f=wG2tavg=0.025m0.0449s=0.5568ms

∆vwG2f=∂w∂t=1tavg*∆w=0.0005m0.0449s=0.01114ms

∆vtG2f=∂t∂w=wG2tavg2*∆t=0.025m0.0449s2*0.001s=0.0124ms

∆vG2f=∆vwG2f2+∆vtG2f2=0.01114ms2+0.0124ms2=0.0167ms

Momentum is conserved and the initial momentum will be equal to the momentum after the elastic collision. Momentum is equal to mass times velocity.

pi=m1vG1i=0.3041kg0.4570ms=0.1390kg*ms

## The uncertainty in the measurement for momentum is shown by taking partial derivatives in respect to either mass or velocity.

∆pi,m=∂p∂m∆m=vG1i∆m=0.4570ms0.0001kg=0.0000457kg*ms

∆pi,v=∂p∂v∆v=m1∆vG1i=0.3041kg0.0124ms=0.003771kg*ms

## Thus, the uncertainty for momentum:

∆pi=∆pi,m2+∆pi,v2=0.0000457kg*ms2+0.003771kg*ms2=0.003771kg*ms

Similarly, the final momentum can be found with the same calculations using mass and velocity of the gliders after the collision took place.

pf=m1vG1f+m2vG2f=0.3041kg0.1033ms+0.2049kg0.5568ms=0.1455kg*ms

∆pf,m1=∂p∂m1∆m=vG1f∆m=0.1033ms0.0001kg=0.00001033kg*ms

∆pf,m2=∂p∂m2∆m=vG2f∆m=0.5568ms0.0001kg=0.00005568kg*ms

∆pf,v1=∂p∂vG1∆vG1=m1∆vG1=0.3041kg0.002109ms=0.0006413kg*ms

∆pf,v2=∂p∂vG2∆vG2=m2∆vG2=0.2049kg0.0167ms=0.003422kg*ms

∆pf=∆pf,m12+∆pf,v12+∆pf,m22+∆pf,vG22=0.00001033kg*ms2+0.00005568kg*ms2+0.0006413kgms2+0.003422kgms2=0.003482kg*ms

In order to find the total change in energy one must first calculate the change in total kinetic energy. Here, the velocity is now squared and multiplied by the mass first to find the initial kinetic energy and then to find the final kinetic energy.

KEi=12m1vG1i2=120.3041kg0.4570ms2=0.03176J

∆KEi,m=∂KE∂m∆m=vG1i22∆m=0.4570ms220.0001kg=0.00001044J

∆KEi,v=∂KE∂v∆v=m1vG1i∆vG1i=0.3041kg0.4570ms0.0124ms=0.001723J

∆KEi=∆KEi,m2+∆KEi,v2=0.00001044J2+(0.001723J)2=0.001723J

Now that the initial kinetic energy and its uncertainty has been calculated, the final kinetic energy must be calculated to find the total kinetic energy.

KEf=12m1vG1f2+12m2vG2f2=120.3041kg0.1033ms2+120.2049kg0.5568ms2=0.03338J

∆KEf,m=∂KE∂m∆m=vG1f22∆m=0.1033ms220.0001kg=0.0000005335J

∆KEf,v=∂KE∂v∆v=m1vG1f∆vG1f=0.3041kg0.1033ms0.002109ms=0.00006625J

∆KEf,m=∂KE∂m∆m=vG2f22∆m=0.5568ms220.0001kg=0.00001550J

∆KEf,v=∂KE∂v∆v=m2vG2f∆vG2f=0.2049kg0.5568ms0.0167ms=0.001905J

∆KEf=∆KEf,m12+∆KEf,v12+∆KEf,m22+∆KEf,v22=0.0000005335J2+0.00006625J2+0.00001550J2+0.001905J2=0.001906J

In part B, a perfectly inelastic collision took place where the two sliders collided and stuck together. The same calculations are done for part B just as were done in part A.

vi,G1=wG1tavg=0.025m0.0566s=0.4417ms

∆vi,wG1=∂w∂t=1tavg*∆w=0.0005m0.0566s=0.008834ms

∆vi,tG1=∂t∂w=wG1tavg2*∆t=0.025m0.0566s2*0.001s=0.007804ms

∆vi,G1=∆vi,wG12+∆vi,tG12=0.008834ms2+0.007804ms2=0.0118ms

## Similarly, the final velocity must be found for the two gliders together.

vf,G1∧2=wG2tavg=0.025m0.0925s=0.2432ms

∆vf,wG1∧2=∂w∂t=1tavg*∆w=0.0005m0.0925s=0.005405ms

∆vf,tG1∧2=∂t∂w=wG2tavg2*∆t=0.025m0.0925s2*0.001s=0.002922ms

∆vf,G1∧2=∆vf,wG1∧22+∆vf,tG1∧22=0.0054052+0.002922ms2=0.006144ms

## The momentum before should equal the momentum after the inelastic collision in order to find the total conservation of energy.

pi=m1vG1i=0.3037kg0.4417ms=0.1341kg*ms

∆pi,m=∂p∂m∆m=vG1i∆m=0.4417ms0.0001kg=0.00004417kg*ms

∆pi,v=∂p∂v∆v=m1∆vG1i=0.3037kg0.0118ms=0.003584kg*ms

∆pi=∆pi,m2+∆pi,v2=0.00004417kg*ms2+0.003584kg*ms2=0.003584kg*ms

## The final momentum is found using the same calculations and when the mass of glider 1 and glider 2 are added.

pf=mG1+G2vf,G1∧2=0.5088kg0.2432ms=0.1238kg*ms

∆pf,m1+m2=∂p∂m∆m=vf,G1∧2∆m=0.2432ms0.0001kg=0.00002432kg*ms

∆pf,v=∂p∂v∆v=mG1+G2∆vf,G1∧2=0.5088kg0.006144ms=0.003126kg*ms

∆pf=∆pf,m2+∆pf,v2=0.00002432kg*ms2+0.003126kg*ms2=0.003126kg*ms

The change is momentum will give a basic estimate for the conservation of energy but a more precise value can be determined by calculating the kinetic energy of the gliders during this perfectly inelastic collision.

KEi=12m1vG1i2=120.3037kg0.4417ms2=0.0296J

∆KEi,m=∂KE∂m∆m=vG1i22∆m=0.4417ms220.0001kg=0.000009755J

∆KEi,v=∂KE∂v∆v=m1vG1i∆vG1i=0.3037kg0.4417ms0.0118ms=0.001583J

∆KEi=∆KEi,m2+∆KEi,v2=0.000009755J2+0.001583J2=0.001583J

## The final kinetic energy can be found using the same derivations.

KEf=12mG1+G2vf,G1∧22=120.5088kg0.2432ms2=0.01505J

∆KEf,m1+m2=∂KE∂m∆m=vf,G1∧222∆m=0.2432ms220.0001kg=0.000002957J

∆KEf,vG1∧2=∂KE∂v∆v=mG1+G2vf,G1∧2∆vf,G1∧2=0.5088kg0.2432ms0.006144ms=0.0007603J

∆KEf=∆KEf,m2+∆KEf,v2=0.000002957J2+0.0007603J2=0.0007603J

Finally, the total momentum and kinetic energies are provide for part A and part B thus allowing to find the total conservation of energy within the system. With the uncertainties of each value, the error can properly be calculated as well. This is best shown by the error bars.

## Conclusion

Generally, the accuracy and precision of measuring instruments was high. Precision can be increased by use of high accuracy measuring instruments for time interval, weight and also making the gliding surface more frictionless. There was an observed differences in the weight of the measured masses. This show there might have been a discrepancy in the measuring equipment. Reading errors that require the aid of an eye such as a ruler can be improved by using other digital measuring devices that are free of alignment errors and that can also measure very accurately. The measurement of the width of the flags should be improved. The analysis of values in this experiment demonstrate that momentum was conserved in all the collisions; whether they were elastic or inelastic. This experiment also demonstrate that KE is conserved in only elastic collisions. The small differences observed in the calculated initial and final momentum can be attributed to errors in the experiment. This experiment offered an opportunity to learn calculations, collection of data, and analysis of errors and uncertainties in measuring equipment. It further demonstrated practically the important concept of the conservation of momentum. This helped reinforce the theoretical knowledge that had been taught in class.

## Reference

Avison, J. (1989).The World of Physics. Cheltenham: Thomas Nelson and Sons.

Cameron, S.C. & Craig, C. (2011). Scientific Theories, Laws, and Principles. Mark Twain Publishers.

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