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[일반 물리 실험 영어 레포트] Conservation of momentum and impulse

Conservation of Momentum and ImpulseAbstract:During collision, a factor known as ‘momentum’ of all objects in the process of collision is conserved. in other words, the sum of momentum before the collision equals to the sum of momentum after collision occurred. There are two types of collision which we consider – elastic and inelastic collision other than explosion. In this experiment, magnetic bumpers are used to demonstrate elastic collision, velcro bumpers were used to demonstrate inelastic collision, and spring bumpers for explosion. Momentum was conserved in all collisions, while kinetic energy was only conserved for elastic collision.Even though momentum is conserved, it changes when impulse is done on an object. Impulse is present due to the force applied to the object over time. Hence, the experiment also examined different amount of force applied to each bumper over time, changing the impulse that the fixed end stop exerted on the cart.Introduction:Momentum of an object dependormula:Moreover, the major disparity between elastic and inelastic collision is that the total kinetic energy is conserved in elastic, while the initial sum of kinetic energy differs from final sum of kinetic energy. In elastic collision, the relative velocity of the two objects is constant if the mass of two objects involved in collision is the same. Additionally, kinetic energy in inelastic collision is lost as energy is converted to other form such as vibrational energy.Impulse is defined as change in momentum.Since and , impulse can also be shown as:As a result, impulse also equals to the area of the F•t curve.Methods:Set up the experiment by connecting two carts to PASCO software.Elastic collision:Put on magnetic bumper to both cartsPut a cart in the middle of the track and start recording.Push another cart from the end of the track towards the middle so that the two carts start colliding.Repeat steps 1-3 after putting masses to one of the cartsInelastic collision:Put on velcro bul.Place a cart connected to PASCO software in the middle of the track.Rubber:Attach rubber bumper to the end-stop and the cartStart recording on PASCO softwarePush the cart in the direction of the end-stop and observe change in velocityRepeat after adding masses.Magnet:Attach magnetic bumper to both the end-stop and the cartStart recording.Push the cart and observe change in velocityRepeat after adding masses.Spring:Attach spring bumper to the end-stop.Start recording.Push the cart and observe change in velocityRepeat after adding masses.Results:Conservation of momentum:Elastic collision:Equal mass:Inequal mass:Inelastic collision:Equal mass:Inequal mass:Explosion:Equal mass:Inequal mass:Impulse:Magnetic bumper:Spring bumper:Rubber bumper:Discussion:Elastic collision:Equal mass:MomentumRed cartBlue cartsumBefore(kg•m/s)0.0000.1450.145After(kg•m/s)0.1420.0000.142%difference2.07%Kinetic EnergyRed cartBlue cartsumBefore(kg•m/s)0.0000.03820.0382After(kg•m/s)0.03650.0000.0365%difference4.45kg•m/s)0.0000.09800.0980After(kg•m/s)0.02180.0420.0638%difference34.8%Explosion:Equal mass:MomentumRed cartBlue cartsumBefore(kg•m/s)0.332-0.356-0.0240After(kg•m/s)-0.2710.248-0.0222%difference7.18%Kinetic EnergyRed cartBlue cartsumBefore(kg•m/s)0.1990.2310.430After(kg•m/s)0.1320.1120.0148%difference96.56%Inequal mass:MomentumRed cartBlue cartsumBefore(kg•m/s)0.238-0.271-0.0325After(kg•m/s)-0.2580.228-0.0304%difference6.45%Kinetic EnergyRed cartBlue cartsumBefore(kg•m/s)0.1020.06880.171After(kg•m/s)0.06250.04860.1111%difference35.0%As shown from the tables presented above, the momentum and kinetic energy of each condition is calculated and its percentage difference between the initial and final velocity is determined. For most conditions, the momentum has shown to be conserved. However, the presence of percentage difference is still present. Conservation of momentum only applies to isolated system, but due to the condition of the experiment, the system may not have been fully isolated nal velocity(m/s)-0.552-0.455-0.473Change in momentum(kg•m/s)-0.347-0.316-0.345Impulse(Ns)-0.32-0.29-0.31% diff5.47%5.80%7.25%As shown from the table above, the percentage difference of the impulse determined from the area under Force-time curve and the calculated value was found to be small. Hence, this shows that the data obtained from the experiment is highly reflective of the actual value. As shown from the findings, the initial and final velocity of three different conditions are relatively similar. However, when comparing the time span taken for three different conditions, the magnetic bumper took the longest time of 0.12 seconds. This is almost double the time of spring bumper (0.06 s) and rubber bumper (0.07 s). The greatest impulse was shown from rubber bumper, then spring bumper, and the least force was shown from magnetic.From the findings, we can determine that in elastic collision through magnetic bumper, the greatest time span and the smallest maximum force (impulse) is sf

자연과학| 2021.01.05| 11페이지| 2,500원| 조회(145)

물리보고서, 일반물리, 일반물리실험, 운동량, 영어레포트, 실험레포트, 충격량, 물리..

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[일반 물리 실험 영어 레포트] Projectile Motion

Projectile MotionAbstract:In the experiment, we recorded a trajectory of a launched ball in a projectile motion by marking the point of the ball every frame of the recording of the ball’s projectile motion. We launched the ball at 30º, 45º, 60º. The parabolic graph that we ended up with, and the gravitational acceleration that we derived from the parabolic graph agreed with the theoretical result, with the value of -9.8 m/s2. In addition, the angle that allowed the object to travel the longest distance was 45º, even though the ball stayed in the air for the longest time when the angle was 60º.Introduction:A lot of objects in our daily lives are in projectile motion as gravity being the only force working on the object. As a result, the vertical component of the launched ball’s velocity is constant, while only the horizontal component of the velocity of the object is experiencing constant acceleration of -9.8m/s2. Since the ball was launched at different angle, the maximum and total dis of Earth, and the radius of Earth respectively, we arrive at the value of -9.8m/s2, the value of acceleration at the surface of the Earth.In projectile motion, only vertical acceleration of the ball exists at -9.8m/s2. Hence, the horizontal component of the velocity remains constant, and the horizontal acceleration remains at 0. The constant vertical acceleration allows us to arrive at the conclusion.Alternative approach can be explained by the angle in which the object is launched at along with its initial velocity. The initial velocity of x-position is and the y-position is . By integrating the two equations with respect to time, the horizontal and vertical direction will be shown by the function of time, which are:Through two equations shown above, the equation of projectile motion is then obtained. The equation for x-component can be rearranged with respect to time, . By substituting the equation into y-component function, we can obtain:During projectile motion, energy is always cher.Turn on the camera and launch the PASCO program.Record the video as the ball gets launched.In PASCO software, stop recording when the ball reaches the ground.Replay the recording of the flight frame by frame and mark each point of the ball to collect the points which forms a parabola.Repeat the steps 2-7 after setting the angle to 45º, 60ºExport the tables and create the graph using the data.Results:30º:Time [s]Position x [m]5.7905.8230.045.8560.15.890.165.9230.225.9560.295.990.366.0230.426.0560.496.090.56Time [s]Position y [m]5.7905.8230.025.8560.045.890.055.9230.055.9560.045.990.026.023-0.026.056-0.066.09-0.12Position x [m]Position y [m]000.040.020.10.040.160.050.220.050.290.040.360.020.42-0.020.49-0.060.56-0.1245º:Time [s]Position x [m]3.18803.2210.043.2540.093.2880.153.3210.23.3540.253.3880.33.4210.363.4540.413.4880.473.5210.533.5540.59Time [s]Position y [m]3.1882.64E-043.2210.033.2540.073.2880.093.3210.113.3540.113.3880.13.4210.093.4540.053.4880.013.521-0.053.554-0.12Position ant velocity can then be calculated.Angle(m/s)(m/s)30º1.902.1945º1.622.2960º1.102.20Moreover, the initial velocity of y-component can be calculated using the equation.Angle(m/s)30º1.1045º1.6260º1.92Furthermore, by using the equation , we are able to denote the acceleration of gravity by differentiating the equation twice, which results in . Hence, we can check the percentage error by comparing the calculated value of acceleration from the quadratic equation from the graph with 9.81 m/s2.Angleg (m/s2)Percentage error30º9.483.39 %45º10.163.55 %60º9.760.53%The flight time, horizontal distance and the maximum height of the projectile motion in different angles can be calculated from the equations shown in “Theoretical Background”. By following the equations, the calculated data are summarized in the table below.AngleFlight timeHorizontal distanceMaximum height30º0.305 s0.491 m0.0613 m45º0.393 s0.268 m0.134 m60º0.442 s0.493 m0.185 mComparison between theoretical and experimental value:The e similar to theoretical finding. However, all the graphs commonly display that the theoretical graph shows a slightly higher projectile. This is possibly due to the coordinate settings using the software along with minor errors present in experimental environment which could have contributed to the difference between experimental and theoretical findings.Conclusions:To conclude, projectile motion of the launched ball in three different angles are successfully displayed from our findings. As shown from three conditions, the graph of x-component vs. time display linear relationship between the two factors whereas the graph of y-component vs. time display quadratic relationship.In this experiment, the flight time increased as the angle increased, as well as the maximum height of the projectile increased. However, the relationship between other factors were not clearly shown. This is possibly due to the additional factors such as air resistance which we have not considered in the explorati.pdf

자연과학| 2021.01.05| 8페이지| 2,500원| 조회(140)

물리보고서, 일반물리, 일반물리실험, 영어레포트, 실험레포트, 포물선운동, 물리실험..

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[일반 물리 실험 영어 레포트] Newton's 2nd Law

Newton’s 2nd LawAbstract:Force exerted onto an object is responsible for causing acceleration in an object. The objective of this lab was to apply the Newton’s second law for a one-dimensional system.Since acceleration can be calculated using the formula F=ma, knowing the force gives insight into the acceleration, velocity, and the position of an object. In this experiment, masses of known value hanging off in free fall and the mass of the cart were used as the way of applying force to the cart. As the object fell down, we observed the change in force, acceleration, velocity and displacement. From the experiment, we were able to derive a-t, v-t. d-t graph and use the information to understand Newton’s 2nd law.Introduction:Force is responsible for all the moments we see in daily lives. To be more precise, force cause acceleration in an object, and the acceleration results change in velocity and position of the object. As the force is bigger, the acceleration on the object is larger, and is the mass of an object and is acceleration. From the acceleration value, more knowledge regarding velocity and position of the object can be obtained.The system studied in our experiment has two masses connected by a string, one moving freely vertically, while the other mass is a smart cart moving along a track. In this following experiment, we have simplified by assuming the pulley is massless and frictionless. Hence, tension on both sides of the string should be equal.Methods:Equipment:10g stackable massesHangerBraided stringSmart cartPulley with clampDynamics track feetElastic bumperDynamics trackProcedure:Use PASCO software and connect the cart to the software by using Bluetooth.Clamp the pulley to the end of the track and place it at the edge of the table.Place the cart on the track and hang a string to one side of the cart.Hang a mass at the other end of the string, where its mass is known and hold it at the edge of the table.Before recording the actual data, use samples to ca position vs. time and force vs. time is shown for each condition.0 g Smart cart:20 g of hanging mass:The considered range can be found through the two graphs shown above. It is the range where the position of the cart increases as well as where the force is remains fairly constant.Within the range considered, the gradient of the position-time graph can be calculated. From the calculated gradient, velocity-time graph is then plotted.This is the velocity-time graph of 0 g smart cart with 20 g of hanging mass. By following similar procedure, the graph can be plotted for the remaining data. However, for the simplicity of the data representation, only the 0 g smart cart will be shown as a demonstration. For more information, refer to the appendix of the following lab report.Discussion:As shown from the graphs in results section, the velocity-time graph of the experiment displays linear relationship.Referring back to Newton’s second law, F=ma, the relationship between the hanging mass and te hanging mass, the mass of the smart cart and acceleration. The gravitational force should remain at a fixed constant (9.81m/s2). Thus, the factor which contributes in altering the acceleration is .Increase in the mass of mass of the cart contributes to decrease in acceleration whereas increase in the mass of the hanging mass should lead to slight increase in acceleration due to increase in the value of numerator in the fraction show above.From the collated data in this experiment, the findings support the relationship explained theoretically. This is shown from the gradient of the velocity-time graph presented below:0g cart mass:Hanging mass (g)204060Gradient of v-t curve0.3940.5610.664250g cart mass:Hanging mass (g)204060Gradient of v-t curve0.3510.5460.550500g cart mass:Hanging mass (g)204060Gradient of v-t curve0.3340.4520.519As shown from the gradient of the velocity-time curve, the gradient of the curve increases when the hanging mass increases. Furthermore, the general gradientis arguable that the experimental data shows acceleration which is slower than theoretical findings.Moreover, the stopper present in the pulley also influences the findings. During the experiment, the stopper results in deceleration of the cart as it prevents the cart from falling. However, such presence results in exclusion of the very end point which we should have recorded. The exclusion of the data can possibly influence calculation such as the gradient of the curve to find the velocity. If the velocity is not accurate due to this factor, this results in affecting the findings of acceleration as acceleration is found by the gradient of the velocity-time curve.Conclusion:To conclude, the following example successfully demonstrates how the mass of either the hanging mass or the smart cart can influence the velocity and the acceleration of the cart. This is examined through Newton’s second law which states that F=ma.Increase in the hanging mass provided force onto the cart, which, in :

자연과학| 2021.01.05| 9페이지| 2,500원| 조회(97)

물리보고서, 일반물리, 일반물리실험, 영어레포트, 실험레포트, 물리실험레포트, 뉴턴..

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[일반 화학 실험 영어 레포트] Molar Solubility; Common-ion effect

Exp. 9 Molar Solubility, Common-Ion EffectIntroduction:The following experiment aims to explore the common ion effect of calcium hydroxide by comparing molar solubility of two solutions: saturated calcium hydroxide and calcium hydroxide solution with common-ion present.Common-ion effect is the effect resulted from the presence of ion which is involved in the equilibrium. The effect of the common-ion is explained by LeChatelier’s principles. As the concentration of the common-ion increases, the equilibrium will shift towards the direction which will reduce the concentration of the common-ion. As the equilibrium shifts, it will affect the solubility of molecules such as calcium hydroxide as the dissolution of the substance is also a dynamic equilibrium.As shown from the equation above, the dissolution of calcium hydroxide is an equilibrium system. The molar solubility constant is calculated by the following equation:As ,The molar solubility of solution is equivalent to the concentration olor changes from yellow to orange. Record the volume of HCl used in titration.Repeat the titration for 3 times.From the data obtained, calculate molar solubility and solubility product.Molar solubility of calcium hydroxide in the presence of a common ionUsing a 25 mL pipet, add 25 mL of the solution with present to 125 mL flask with drops of methyl orange added.Titrate the solution with standardized HCl solution until the color changes from yellow to red-orange. Record the volume of HCl used in titration.Repeat the titration for 3 times.Calculate molar solubility of the solution using the data obtained.Results:Standardization of HCl SolutionThe moles of solution is obtained by the following calculation.For trial 1, the initial buret was 29 mL and once it was read to 44 mL, we have added additional HCl to the buret as we believed that there would not be enough for the titration. Hence, the second initial reading is 3 mL and second final reading is 11 mL. Volume is calculated from the tf saturated solution (mL)25Concentration of standardized HCl solution (mol/L)0.0548Initial buret reading (mL)1.58.614Final buret reading (mL)24.83236.9Volume of HCl added (mL)23.323.422.9Moles of HCl added (mol)0.0012770.0012820.001255Moles of in saturated solution (mol)0.0012770.0012820.001255equilibrium (mol/L)0.05110.05130.0502equilibrium (mol/L)0.02550.02560.0251Molar solubility of (mol/L)0.02550.02560.0251Average Molar solubility (mol/L)0.0254of0.0000670.0000670.000063Average0.000066Standard deviation of%RSD of3.52%Molar Solubility of Calcium Hydroxide in the Presence of a Common IonThe calculation for the following section is similar to part B. Trial 1 will be shown as sample calculation.As moles of HCl = moles of in solution, the calculation proceeds as followings.The summarized findings will be presented in the table below.Trial 1Trial 2Trial 3Volume of saturated with added solution (mL)25Concentration of standardized HCl solution (mol/L)0.0548Initial buret reading (mL)2.221.31 in the findings are due to accumulation of minor errors which are presented in the experimental procedure. Such errors involve the error in obtaining the correct amount of volume using the pipet due to observing from the wrong point of view, or the error resulted from the determination of end-point of titration from the color change. Especially in the determination of color change, as the color-change from yellow to orange and then to red-pink is done by only a small amount of volume difference, it may be challenging to determine the exact color that we aim to expect. Moreover, the temperature of the solution may also affect the findings according to LeChatelier’s principle. If the temperature is too high, the solubility of the solution would have decreased. However, I believe temperature did not play the major role in influencing the following experiment as 3 trials were obtained around the same time, and the experiment was conducted in a constant environment.Molar solubility of calcfindings.Conclusion:To sum up the following experiment, there has not been any major errors which greatly impacted the findings. Hence, we can possibly conclude that the findings from the experiment are highly accurate. However, to reduce minor errors and human errors from the experimental procedure, it is better to repeat the experiment and obtain the average value. From the comparison between the molar solubility of the solution in part B and part C, it has successfully showed the common-ion effect.The chemical equation above shows the chemical equilibrium for the dissolution of calcium hydroxide. As the common-ion was present in part C, the concentration of the product has increased. Hence, according to LeChatelier’s principle, the equilibrium would then shift to the left, resulting in decrease in molar solubility. This is evident from the results where the molar solubility of saturated was 0.0254 mol/L, but it was reduced to 0.0198 mol/L when the common ion was present.Lab Question

자연과학| 2020.06.11| 7페이지| 2,500원| 조회(113)

용해도, 일반화학실험, 일반화학, 영어레포트, 실험레포트, 화학실험레포트, 화학보고..

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[일반 화학 실험 영어 레포트] Galvanic Cells, the Nerst Equation 평가A+최고예요

Introduction:The following experiment aims to explore one of the electrochemical cells – galvanic cells. Electrochemical cells are based upon redox (reduction-oxidation) principle. In the case of galvanic cells, reactions happen spontaneously. In other words, without the need of external energy, the redox will continuously occur. Reduction and oxidation can be explained by the transfer of electrons, where reduction is the gain of electrons and oxidation is the loss of electrons. Often, the reaction is presented by half-equation (redox couple).<중 략>Conclusion:The following experiment explored galvanic cell. The first experiment explored different redox couples, and how potential difference is resulted from different tendencies for metal ions to be reduced. The errors in the experiment was relatively high, and we could speculate that such errors are mainly due to the instruments used.

자연과학| 2020.06.11| 7페이지| 2,500원| 조회(122)

일반화학실험, 일반화학, 영어레포트, 실험레포트, 갈바닉 전지, 네른스트 식, 화학..

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