Physics Q&A Game

  • #151
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1
The center of mass of the triangle structure just after the collision will be 2/3 of the way from the top corner to the base of the triangle. Call this point (x,y) = (0,0). So the top corner of the triangle is at (0, L/sqrt(3))

So, angular momentum about the point (0,0) before the collision is: M/2*v1*L/sqrt(3) = M*L*v1/2sqrt(3)
After the collision the angula momentum is: 3M*v2*L/sqrt(3) = M*L*v2*sqrt(3) where v2 is velocity of the incoming ball in the center of mass frame.
>>Conservation of angular momentum means these must be equal so: v2 = v1/6.

The linear momentum before the collision is M*v1/2.
After the collision, the linear momentum is 3M*vcm, where vcm is the velocity of the center of mass.
>>Conservation of momentum means that these must be equal so: vcm = v1/6

So, immediately after the collision, the two stuck together balls at the top are moving with velocity: v2 + vcm = v1/3 to the right.

The x and y components of the velocity of the bottom left ball are:

vy = v2*sqrt(3)/2 = v1*sqrt(3)/12 and vx = vcm - v2/2 = v1/12

and for the bottom right ball are:

vy = -v1*sqrt(3)/12 and vx = v1/12

Subsequently, all four balls continue to rotate at a speed of v1/6 around the center of mass. This uniform circular motion (in the cm frame) has a constant velocity component v1/6 to the right (+x direction) added to it at all times.

How's that?
 
  • #152
siddharth
Homework Helper
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Jdavel, that's the right answer. Let me post the full solution anyway for clarity.

First of all, conservation of angular momentum must be done about the center of mass of the system,ie, the center of mass of all four balls and not about the center of mass of the triangle.

This is because after collision the system will rotate only about it's center of mass (in this case, center of mass of all four balls).

As mass of each vertex of the triangle after collision is 'M' the center of mass will be at the centroid, that is, 2/3 distance from the top vertex, which is

[tex] \frac{2}{3} \frac{\sqrt{3} L}{2} [/tex]

[tex] = \frac {L}{\sqrt{3}} [/tex]

As the triangle is equilateral, the distance to each vertex is also
[tex] = \frac {L}{\sqrt{3}} [/tex]

By conservation of angular momentum
[tex] L_i=L_f [/tex]

[tex] \frac{Mv_1L}{2\sqrt{3}} = I \omega [/tex]

[tex] \frac{Mv_1L}{2\sqrt{3}} = 3(\frac{ML^2}{3} \omega) [/tex]

[tex] \omega = \frac{v_1}{2\sqrt{3} L} [/tex]

By conservation of linear momentum,
[tex] (\frac{M}{2}) (v_1) = 3M v_ \textrm(com) [/tex]

[tex] v_ \textrm(com) =\frac{v_1}{6} [/tex]

also the velocity of each vertex is
[tex] v_ \textrm(com) \vec i + \frac {L}{\sqrt{3}}\omega \vec n [/tex]

[tex]\vec v = \frac{v_1}{6} \vec i + (\frac {L}{\sqrt{3}})(\frac{v_1}{2\sqrt{3} L}) \vec n [/tex]

[tex]\vec v = \frac{v_1}{6} \vec i + \frac{v_1}{6} \vec n [/tex]


Taking components, in the x and y directions for each vertex, we get the required answer
 
  • #153
Gokul43201
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Science Advisor
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7,051
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New question :

According to macroscopic equilibrium thermodynamics, absolute zero can not be reached. However, there is a non-zero probability that the lattice of a finite body will be free from phonons. If a particle can be cooled to a temperature of about a milli-Kelvin (within reach of a good dilution refrigerator - no need for magnetic cooling even !), how small must it be so it can be said to have a lattice temperature of 0K for say, 90% of the time ?

PS : An "order of magnitude" approximation will do.
 
  • #154
10
0
I think u dont have to finish off the extinguisher.once u attain a good velocity u dont have to spray more as in space since there is no air,so u wont be slowed down.so u need to spray more only for a change of direction.
 
  • #155
Gokul43201
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Science Advisor
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7,051
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Hint for present question : Model a phonon (lattice vibration) like it were a vibrating mode of a stretched string. What is the frequency spectrum in such a case ?
 

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