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Question: What would be the final velocity of the coupled railroad cars?

Answer: For this scenario, we have

v2 = (ma*va1 +mb*vb1)/(ma + mb), or

v2 = (ma*10)/2*ma = 5 m/s

The final velocity of the pair of coupled cars is half the initial velocity of the car that was moving.

Scenario #2: Now assume that due to loading, Car-A has twice the mass of Car-B, the initial velocity for Car-A is 10m/s and the initial velocity forCar-B is 0.

Question: What would be the final velocity of the coupled railroad cars?

Answer: For this scenario, we have

v2 = (ma*va1 +mb*vb1)/(ma + mb), or

v2 = (ma*va1)/(ma + 0.5ma), or

v2 = ma*10/1.5*ma = 6.67 m/s

When the car at rest is less massive than the car in motion, the final velocity is a little higher than when the two cars have the same mass.

Scenario #3: Finally, assume that due to loading, Car-A has twice the mass of Car-B, the initial velocity for Car-A is 10m/s and the initial velocity forCar-B is 5m/s.

Question: What would be the final velocity of each railroad car?

Answer: For this scenario, we have

v2 = (ma*va1 +mb*vb1)/(ma + mb), or

v2 = (ma*10 +0.5ma*5)/(1.5*ma), or

v2 = (10/1.5) + (0.5*5/1.5) = 8.33 m/s

When both cars are already moving in the same direction, the final velocity in that direction is greater than when one of the cars is stationary.

Center of mass examples

This section contains solutions to problems involving the center of mass.

Two objects

Two objects are located on a flat lawn with the following mass values and locations:

  • obj1m = 15 kg
  • obj1x = 1 m
  • obj1y = 5 m
  • obj2m = 3 kg
  • obj2x = 4 m
  • obj2y = 2 m

What are the coordinates of the center of mass?

Solution:

A JavaScript script that will solve this problem is shown in Listing 2 .

Listing 2 . Center of mass for two objects.
<!---------------- File JavaScript02.html ---------------------><html><body><script language="JavaScript1.3">document.write("Start Script</br>"); //Create arrays for mass,xCoor, and yCoor values;var mass = new Array(15,3); var xCoor = new Array(1,4);var yCoor = new Array(5,2); //Declare a counter variable.var cnt = 0; //Use a loop to compute total mass by summing the individual// mass values. var massTotal = 0;for(cnt = 0;cnt<mass.length;cnt++){ massTotal += mass[cnt]; }//end for loop//Use a loop to compute x-coordinate of the center of mass // by summing the normalized sum of products.var cmX = 0; for(cnt = 0;cnt<mass.length;cnt++){ cmX += mass[cnt]*xCoor[cnt]/massTotal}//end for loop //Use a loop to compute y-coordinate of the center of mass// by summing the normalized sum of products. var cmY = 0;for(cnt = 0;cnt<mass.length;cnt++){ cmY += mass[cnt]*yCoor[cnt]/massTotal}//end for loop //Display the resultsdocument.write("massTotal = " + massTotal + " kg</br>"); document.write("cmX = " + cmX + " meters</br>"); document.write("cmY = " + cmY + " meters</br>"); document.write("End Script");</script></body></html>

The comments describe how the problem is solved.

The output is shown in Figure 4 .

Figure 4 . Center of mass for two objects.
Start Script massTotal = 18 kgcmX = 1.5 meters cmY = 4.5 metersEnd Script

The x-coordinate for the center of mass is 1.5 meters, and the y-coordinate for the center of mass is 4.5 meters.

Three objects

Three objects are located on a flat lawn with the following mass values and locations:

  • obj1m = 5 kg
  • obj1x = 1 m
  • obj1y = 1 m
  • obj2m = 5 kg
  • obj2x = 3 m
  • obj2y = 1 m
  • obj3m = 10 kg
  • obj3x = 2 m
  • obj3y = 3 m

What are the coordinates of the center of mass?

Use the code from Listing 2 but make the change shown in Listing 3 in order to populate the world withthree objects having different mass values and different coordinates.

Listing 3 . Change to add a third object.
//Create arrays for mass,xCoor, and yCoor values; var mass = new Array(5,5,10);var xCoor = new Array(1,3,2); var yCoor = new Array(1,1,3);

The output is shown in Figure 5 .

Figure 5 . Center of mass for three objects.
Start Script massTotal = 20 kgcmX = 2 meters cmY = 2 metersEnd Script

Do the calculations

I encourage you to repeat the calculations that I have presented in this lesson to confirm that you get the same results. Experiment with the scenarios, making changes, and observing the results of your changes. Make certain that you can explain why your changes behave as they do.

Resources

I will publish a module containing consolidated links to resources on my Connexions web page and will update and add to the list as additional modulesin this collection are published.

Miscellaneous

This section contains a variety of miscellaneous information.

Housekeeping material
  • Module name: Force and Motion -- Momentum, Impulse, and Conservation of Momentum for Blind Students
  • File: Phy1160.htm
  • Revised: 10/02/15
  • Keywords:
    • physics
    • accessible
    • accessibility
    • blind
    • graph board
    • protractor
    • screen reader
    • refreshable Braille display
    • JavaScript
    • trigonometry
    • conservation of momentum
    • momentum
    • Newton's cradle
    • impulse
    • action and reaction
Disclaimers:

Financial : Although the openstax CNX site makes it possible for you to download a PDF file for the collection that contains thismodule at no charge, and also makes it possible for you to purchase a pre-printed version of the PDF file, you should beaware that some of the HTML elements in this module may not translate well into PDF.

You also need to know that Prof. Baldwin receives no financial compensation from openstax CNX even if you purchase the PDF version of the collection.

In the past, unknown individuals have copied Prof. Baldwin's modules from cnx.org, converted them to Kindle books, and placed them for sale on Amazon.com showing Prof. Baldwin as the author.Prof. Baldwin neither receives compensation for those sales nor does he know who doesreceive compensation. If you purchase such a book, please be aware that it is a copy of a collection that is freelyavailable on openstax CNX and that it was made and published without the prior knowledge of Prof. Baldwin.

Affiliation : Prof. Baldwin is a professor of Computer Information Technology at Austin Community College in Austin, TX.

-end-

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
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20/(×-6^2)
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Commplementary angles
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Differences Between Laspeyres and Paasche Indices
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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Source:  OpenStax, Accessible physics concepts for blind students. OpenStax CNX. Oct 02, 2015 Download for free at https://legacy.cnx.org/content/col11294/1.36
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