Robin Seibel

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8 years, 128 days

MaplePrimes Activity

These are replies submitted by PsiSquared

Ok, I was stupid.  I inserted an image that was a PNG (screenshot). When I deleted that image and instead used a jpeg version of that screenshot, the document printed as a pdf without error.

@Rouben Rostamian  


I inserted the base image into my Maple Document.  Then I added to the image using Maple drawing tools.  To create the pdf, I used the standard Mac technique: seclect Print for the  file, and then in the print dialogue box, select "Save As PDF".  Attached is the PDF.Soln_for_2_Particles_in_a_Potential_for_Jordan.pdf

@Rouben Rostamian  The base image was from a file, but then I added to the image with a lot of other elements, adn those added elements didn't remain in place when the doc converted to pdf.

Problem 2:  At the moment they come to rest, where is the center of mass of the 2 particle system?



Looking at the figure above, we can see in the energy diagram where the kinetic energy for each particle goes to zero.  For Particle A, that is at x = -10, and for Particle B, that is at x = 5. I am assuming that each square has a width of 1-unit.  In general the center of mass for a 2 particle system at any given time will be:

x[cm] = (m[A]*x[A]+m[B]*x[B])/(m[A]+m[B])

So substituting in what we know so far:

x[cm] = (-10*m[A]+5*m[B])/(m[A]+m[B])

But we can find out more.  If our particle's come to rest at the same time, that means that Particle A has to travel a distance `Δx` = 10 in the same distance that Particle B travels a distance `Δx` = 5.  Now, we know that:

-(diff(U(x), x)) = F


In other words, the negative of the slope of the potential is the force. In this case, that means the force applied to each particle on either side of the origin is constant.  Therefore the magnitude of the force to the left ofx = 0 is 1, while the magnitude of the force to the right of the origin is "2 ."  Let's use those forces to find the speed of each particle at x = 0.  Particle A is originally at rest at "x=5,"while Paricle B is originally at rest at x = -10.  Their velocities at the origin will then be:

"v[A]=sqrt(2 a[A]Deltax[A])=sqrt(2(2/(m[A]))5)=2sqrt(5/(m[A]))"
"v[B]=sqrt(2 a[B]Deltax[B])=sqrt(2(1/(m[B]))10)=2sqrt(5/(m[B]))"

Since both particles come to a rest at the same time and pass each other at the origin at the same time, and since t = `Δx`/v, we get:

`Δx`[A]/v[A] = `Δx`[B]/v[B]


5/(2*sqrt(5/m[A])) = 10/(2*sqrt(5/m[B]))``


``sqrt(5/m[B])/sqrt(5/m[A]) = (2/5)*(10*(1/2))


sqrt(m[A]/m[B]) = 2


m[A]/m[B] = 4


m[A] = 4*m[B]

Now we can complete our center of mass calculation.

From above:

x[cm] = (-10*m[A]+5*m[B])/(m[A]+m[B])


x[cm] = (-10*(4*m[B])+5*m[B])/(4*m[B]+m[B])


x[cm] = -35*m[B]/(5*m[B]) and -35*m[B]/(5*m[B]) = -7



Note: I left values for forces, energy, position, and velocity unitless in my calculations beccause they were't specified in the problem.  











Thanks.  That works pretty well, except that my output is "count + xxxx", with "xxxx" of course being some random number.  Note that I am using Maple 11.

Would it be worth putting randomize() inside the proc dice?

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