Exploring PressureUnderground Sim Lab & PhET Collision Lab. PDF

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Fluid is a substance that flows easily. Gases and liquids are fluids, although sometimes the dividing line between liquids and solids is not always clear. Because of their ability to flow, fluids can exert buoyant forces, multiply forces in a hydraulic systems, allow aircraft to fly and ships to flo...

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Name: Zahid Hasan Khoka

ID:18-36652-1

Date: 20-06-21

Title: Exploring Pressure Underground Sim Lab. Introduction: Fluid is a substance that flows easily. Gases and liquids are fluids, although sometimes the dividing line between liquids and solids is not always clear. Because of their ability to flow, fluids can exert buoyant forces, multiply forces in a hydraulic systems, allow aircraft to fly and ships to float. In this lab the main objective is • Figure out the relation between pressure and depth. • Figure out how pressure and density are related. • Calculate the density of an unknown fluid.

Theory: The topic that this page will explore will be pressure and depth. If a fluid is within a container, then the depth of an object placed in that fluid can be measured. The deeper the object is placed in the fluid, the more pressure it experiences. This is because is the weight of the fluid above it. The more dense the fluid above it, the more pressure is exerted on the object that is submerged, due to the weight of the fluid. The formula that gives the P pressure on an object submerged in a fluid is: P=ρ*g*h where   

ρ is the density of the fluid, g is the acceleration of gravity h is the height of the fluid above the object

If the container is open to the atmosphere above, the added pressure must be included if one is to find the total pressure on an object. The total pressure is the same as absolute pressure on pressure gauges readings, while the gauge pressure is the same as the fluid pressure alone, not including atmospheric pressure. Ptotal = Patmosphere + Pfluid Ptotal = Patmosphere + ( ρ * g * h ) A Pascal is the unit of pressure in the metric system. It represents 1 newton/m2.

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Name: Zahid Hasan Khoka

ID:18-36652-1

Date: 20-06-21

Procedure:

Directions 1. Go to the PhET simulation “Under Pressure” https://phet.colorado.edu/en/simulation/under-pressure 2. Push the big Play arrow. a. b. c. d.

Start with the default settings. Fill the tank with water. Turn on the Grid and play with the Ruler. Use the Grid to get you data table measurements.

3. Click on the pressure gauge to move it toward the water. Measure the pressure in the water at every 0.50 m from the surface to the bottom. Page | 3

Name: Zahid Hasan Khoka

ID:18-36652-1

Date: 20-06-21

Data Table: Table 1: Measurement of pressure for water & honey by various depth.

Figure 1: Pressure vs Depth graph with respect to table I.

Figure 1 shows that the connection between pressure and depth is exactly proportional. This is because a submerged item is pushed down by a larger column of water. Pressure is lowered when items are raised and the depth of the object decreases. The graph is also shows that is linearly increasing with respect to depth. Table II: Measurement of pressure for water & honey in by various depth & gravity.

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Name: Zahid Hasan Khoka

ID:18-36652-1

Date: 20-06-21

Figure 2: Pressure vs Depth graph with respect to table II.

The above figure 2 shows that the pressure vs depth graph with respect to table II for water and honey. This graph is also shows that the linear property of pressure with depth for different fluid ( water & honey).

Figure 3: Pressure vs Depth graph with respect to table I and II.

Figure 3 shows that the pressure vs depth graph with respect to table I and II for water and honey. The link between pressure and depth is exactly proportional. Only the density of the liquid, the acceleration due to gravity, and the depth within the liquid determine the pressure within it. With increasing depth, the pressure exerted by such a static liquid grows linearly. Page | 5

Name: Zahid Hasan Khoka

ID:18-36652-1

Date: 20-06-21

4. Which variable is the independent variable (x-axis)? Answer: Depth 5. Which variable is the dependent variable (y-axis)? Answer: Pressure 6. How would your graph differ if you gathered data from Mars? Jupiter? Explain why. Answer: Because the free-fall acceleration g differs on each planet, the slopes of the graphs from Mars or Jupiter would differ from those on Earth. Because the atmospheric pressure on Mars or Jupiter differs from that on Earth, the y-intercepts of the graphs would be different. 7. What is the relationship between depth and pressure? Why do you think this happens? Answer: As depth increases, then

pressure increases. Pressure and depth have a directly

proportional relationship. This is due to the greater column of water that pushes down on an object submersed. Conversely, as objects are lifted, and the depth decreases, pressure is reduced. Pressure within a liquid depends only on the density of the liquid, the acceleration due to gravity, and the depth within the liquid. The pressure exerted by such a static liquid increases linearly with increasing depth.

8. Why might a well stop producing water? Answer: Due to a lack of electricity, well pump may have ceased operating. This can sometimes be fixed by resetting a breaker or replacing a fuse that has been damaged by a power surge or brownout. If the circuit breaker for the well pump has tripped, the well pump may be failing. Breakers or fuses can also trip or blow due to partial shorts in the wiring or motor. To be on the safe side, have a certified technician inspect the system's components to guarantee appropriate and safe operation. If electrical problems are not handled, they might become dangerous.

9. Click on the icon with the question mark on the sink to access the mystery fluid portion. Determine the density of a mystery fluid. If your last name starts with A-H, test Fluid A. If your last name starts with I-N, test Fluid B. If your last name starts with O-Z, test Fluid C. Describe your method and results bellow. Page | 6

Name: Zahid Hasan Khoka

ID:18-36652-1

Date: 20-06-21

Answer: My name is Zahid Hasan Khoka. That’s why my test Fluid C. Procedure: Turn the atmosphere off. Measure the pressure at a depth of 1.0 m. Pressure, p = p0 + ρgh. Thus, Density,ρ = (p - p0)/gh. Since I turned off the atmosphere. So, ρ = p/gh. Density for fluid C:

ρc = p/gh = 11,100 Pa/[(9.8 m/s2 )(1.0 m)] = 1,100 kg/m3 .

Discussion: In the Explore pressure under and above water lab we are focus on how pressure changes as we change fluids, gravity, container shapes, and volume. From this lab we are gather knowledge about what variables affect pressure, predict pressure in a variety of situations, determining the pressure in a fluid and so on. As we are complete all of the task which are required to fulfill our lab so we can say that our lab was successfully achieved its goal.

References: [1] https://courses.lumenlearning.com/boundless-physics/chapter/density-and-pressure/ [2] https://www.grc.nasa.gov/www/k-12/WindTunnel/Activities/fluid_pressure.html [3] https://www.choosekobella.com/well-pumps/well-pump-stops-working/

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Name: Zahid Hasan Khoka

ID:18-36652-1

Date: 20-06-21

Part –B Title: PhET Collision Lab. Introduction: Collision, also called impact, in physics, the sudden, forceful coming together in direct contact of two bodies, such as, for example, two billiard balls, a golf club and a ball, a hammer and a nail head, two railroad cars when being coupled together, or a falling object and a floor. Apart from the properties of the materials of the two objects, two factors affect the result of impact: the force and the time during which the objects are in contact. In this lab our main objective is:     

Construct momentum vector representations of "before-and-after" collisions. Apply the law of conservation of momentum to solve collision problems. Explain why energy is not conserved and varies in some collisions. Determine the change in mechanical energy in collisions of varying elasticity. Define elasticity.

Theory: Physicists use collisions to determine the properties of atomic and subatomic particles. Conservation laws play an important role in physics. If you have ever watched or played pool, football, baseball, soccer, hockey, or been involved in an automobile accident you have some idea about the results of a collision. We are interested in studying collisions for a variety of reasons. For example, you can determine the speed of a bullet by making use of the physics of the collision process. You can also estimate the speed of an automobile before the accident by knowing the physics of the collision process and a few other physical principles. Physicists use collisions to determine the properties of atomic and subatomic particles. Essentially, a particle accelerator is a device that provides a controlled collision process between subatomic particles so that, among other things, some of the properties of the target particle can be studied. In addition the study of collisions is an example of the use of a fundamental physical tool, i.e., a conservation law. A conservation law implies that something remains the same, i.e., is conserved, as you have seen in a previous module, Conservation of Energy. Conservation laws play an important role in physics. In the study of collisions in this module we are interested in one of the fundamental conservation laws, conservation of linear momentum. If the sum of the external forces is zero, then the linear momentum is conserved in the collision. This is fortunate since it provides a way around the analysis of the forces of interaction between two bodies as they collide, an otherwise formidable task. Thus the conservation-of-linear-momentum law allows one to analyze the effects of a collision without a detailed knowledge of the forces of interaction. One can deduce the converse also, as does the particle physicist in accelerator experiments, for example - some of the properties of the target particles may be deduced from the law of conservation of linear momentum and other laws of physics. Page | 8

Name: Zahid Hasan Khoka

ID:18-36652-1

Date: 20-06-21

Table III: Difference between Elastic and Inelastic Collision

Elastic Collision

Inelastic Collision

The total kinetic energy is conserved.

The total kinetic energy of the bodies at the beginning and the end of the collision is different.

Momentum is conserved.

Momentum is conserved.

No conversion of energy takes place.

Kinetic energy is changed into other energy such as sound or heat energy.

Highly unlikely in the real world as there is almost This is the normal form of collision in the always a change in energy. real world. An example of this can be swinging balls or a An example of an inelastic collision can spacecraft flying near a planet but not getting affected be the collision of two cars. by its gravity in the end.

(a)

(b) Figure 4: Collision Page | 9

Name: Zahid Hasan Khoka

ID:18-36652-1

Date: 20-06-21

Procedure: Go to the website http://phet.colorado.edu/en/simulation/collision-lab Make sure the 1-d box is checked. Part 1 Scenario #1: Elastic collision between balls of equal mass 

Make a hypothesis about initial and final momentums before playing with the sim.

Figure 5: Initial data before playing with the sim for Elastic Collision

Figure 6: Final data after playing with the sim for Elastic Collision Page | 10

Name: Zahid Hasan Khoka

ID:18-36652-1

Date: 20-06-21

Figure 5 & 6 shows that the elastic collusion of a system because after the 63.42 s collusion the kinetic energy is not change for before & after collusion.

 Make a data table for the following: mass, velocity and momentum of each ball before and after. Table IV: Data analyzing table for Elastic collision between balls of equal mass.

Ball 1

Ball 2

Position

Mass

Velocity

Momentum

Before collision

1.50 kg

1

1.50

After collision

1.50 kg

1

1.50

Before collision

1.50 kg

-0.50

-0.75

After collision

1.50 kg

-0.50

-0.75

 What is the relationship between the initial and final total momentums? Answer: The total initial momentum equals the total final momentum for a closed system. Commonly called the conservation of momentum. The total momentum of the system is the same after the collision as before it as shown by the equation initial momentum = final momentum (where final momentum is the sum of all momentums present in the system).  Describe the motion of the balls before and after the collision? Answer: In Case of the Elastic Collision, Kinetic Energy before and after the collision remains the conserved, momentum also remains conserved. On This basis, Derivation in One dimension is shown in attachment. The body of mass m₁ is moving with the velocity u₁ and body of mass m₂ is moving with velocity u₂, after the collision there velocity became v₁ and v₂. The Final Result is, v₂ - v₁ = u₁ - u₂. Scenario #2: Elastic collision between balls of unequal mass.  Make a hypothesis about initial and final momentums before playing with the sim.

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Name: Zahid Hasan Khoka

ID:18-36652-1

Date: 20-06-21

Figure 7: Initial data before playing with the sim for Elastic collision for different mass.

Figure 8: Final data after playing with the sim for Elastic Collision for different mass.

 Make a data table for the following: mass, velocity and momentum of each ball before and after. Page | 12

Name: Zahid Hasan Khoka

ID:18-36652-1

Date: 20-06-21

Table V: Data analyzing table for Elastic collision between balls of equal mass.

Ball 1 Ball 2

Position

Mass

Velocity

Momentum

Before collision

1.50 kg

1

1.50

After collision

1.50 kg

1

1.50

Before collision

2.00 kg

-0.50

-1.00

After collision

2.00 kg

-0.50

-1.00

 What is the relationship between the initial and final total momentums? Answer: For a closed system, the total initial momentum equals the entire final momentum. It's also known as momentum conservation. As proven by the equation beginning momentum = final momentum, the system's total momentum is the same after the collision as it was before it (where final momentum is the sum of all momentums present in the system).  Describe the motion of the balls before and after the collision? Answer: From the above table the motion of the bolls are same velocity for before and after collusion. After 42.62 s both balls are gain their previous velocity. In total their momentum direction is in the opposite direction of ball 1.

Part 2 Create 3 more distinct scenarios in 1-d including one totally inelastic collision. Make a hypothesis whether or not each will follow conservation of momentum. Collect some data and prove or disprove your hypothesis.

Figure 9: Initial data before playing with the sim for totally Inelastic Collision for different mass. Page | 13

Name: Zahid Hasan Khoka

ID:18-36652-1

Date: 20-06-21

Figure 10: Final data after playing with the sim for totally Inelastic Collision for different mass.

Figure 10 shows that after the 6.90 s the kinetic energy is zero. So, Its totally inelastic system. In an inelastic collision, there is always a loss of kinetic energy, i.e., the total kinetic energy of the billiard balls before collision will always be greater than that after collision. The total linear momentum of the system of balls will remain conserved even in the case of an inelastic collision. Unlike the elastic collision, where the kinetic energy is conserved, the kinetic energy in an inelastic collision is not conserved. In an inelastic collision, the kinetic energy between the colliding bodies is different at the beginning and the end of the collision.

Summary: Describe the main ideas learned in this activity regarding initial and final total momentum in 1-d collisions.

Summary: Here we are learned the details of the kinetic energy, momentum, collision, their types & importance of the effect by realistic example. In this lab we experiment with a software & also use table for judge the collusion type. In general, the precise point of contact has a large influence on the collision outcome and it is in such details that the balls decide which way they're going, and how fast. We can claim that our lab has achieved its goal because we have completed all of the tasks that are required to complete our lab.

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Name: Zahid Hasan Khoka

ID:18-36652-1

Date: 20-06-21

References: [1] https://www.britannica.com/science/collision. [2] https://digitalcommons.unl.edu/calculusbasedphysics/6/ [3]https://www.khanacademy.org/science/high-school-physics/linear-momentum-and-collisions/elasticcollisions-and-conservation-of-momentum/a/elastic-collisionsap1#:~:text=The%20total%20initial%20momentum%20equals,like%20for%20two%20colliding%20objects %3F [4] https://brainly.in/question/5875118 [5] https://byjus.com/physics/elastic-collision/

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