Angular momentum and collisions

  • #1
[SOLVED] Angular momentum and collisions

Homework Statement



A wooden block of mass M resting on a frictionless horizontal surface is attached to a rigid rod of length [tex]\ell[/tex] and of negligible mass. The rod is pivoted at the other end. A bullet of mass m and travelling parallel to the horizontal surface and normal to the rod with speed [tex]\vec{v}[/tex] hits the block and gets embedded in it. What fraction of the original kinetic energy is lost in the collision?

Homework Equations



Fraction = (K[tex]_{f}[/tex]-K[tex]_{0}[/tex]) / K[tex]_{0}[/tex]
v[tex]_{f}[/tex] = (mv / M+m)

The Attempt at a Solution



The main thing that confused me here is the way the collision works and what the energy is before and after the collision. I said that before the energy before would simply be (1/2)mv^2, the kinetic energy of the bullet. Then after, the kinetic would be (1/2)(M+m)v^2 where the masses are combined and v is the final speed. I got the equation for final velocity above from my collision formulas for inelastic collisions, but each time I try to work it out, I cannot get the correct answer, which is M / (M+m). Also, I am confused by how I find the fraction of the original energy lost...is my fraction formula above correct, or is it simply final energy over initial energy? I'm just having trouple conceptualizing it. Thanks so much.
 

Answers and Replies

  • #2
Doc Al
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The energy lost would be: KEi - KEf
Expressing this as a fraction of the initial energy is just: (KEi - KEf)/KEi
 
  • #3
OK, thanks. But I don't feel confident that my substitution for final velocity would be correct. I'll try and rework the problem and see if I can come up with it.
 
  • #4
cepheid
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Everything you have is correct. From conservation of momentum (which we're only applying to the magnitudes of the momentum, which I guess is ok at the instant of the collision?), you get:

[tex] v = \frac{m}{m+M} v_0 [/tex]

The kinetic energy lost is equal to:

[tex] -\Delta K = K_0 - K = \frac{1}{2}mv_0^2 - \frac{1}{2}(m+M)v^2 [/tex]

[tex] = \frac{1}{2}mv_0^2 - \frac{1}{2}\frac{m^2}{m+M}v_0^2 [/tex]

[tex] = \frac{1}{2}m\left(1-\frac{m}{m+M}\right)v_0^2 [/tex]

[tex] = \frac{1}{2}mv_0^2 \left(\frac{M}{m+M}\right) = \left(\frac{M}{m+M}\right)K_0 [/tex]
 
  • #5
Doc Al
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OK, thanks. But I don't feel confident that my substitution for final velocity would be correct. I'll try and rework the problem and see if I can come up with it.
You don't have to rework anything. Just calculate the initial and final KE.
 
  • #6
Everything you have is correct. From conservation of momentum (which we're only applying to the magnitudes of the momentum, which I guess is ok at the instant of the collision?), you get:

[tex] v = \frac{m}{m+M} v_0 [/tex]

The kinetic energy lost is equal to:

[tex] -\Delta K = K_0 - K = \frac{1}{2}mv_0^2 - \frac{1}{2}(m+M)v^2 [/tex]

[tex] = \frac{1}{2}mv_0^2 - \frac{1}{2}\frac{m^2}{m+M}v_0^2 [/tex]

[tex] = \frac{1}{2}m\left(1-\frac{m}{m+M}\right)v_0^2 [/tex]

[tex] = \frac{1}{2}mv_0^2 \left(\frac{M}{m+M}\right) = \left(\frac{M}{m+M}\right)K_0 [/tex]
Thanks for the help. The only thing I don't understand is between the last two lines of what you did. How did M get on the top of the fraction, each time I calculate it I work it out I seem to get m on the top. And once I simplify KEi-KEf, I need to divide by KEi to get the fraction, correct? Thanks so much.
 
  • #7
Nevermind about that last question, I see how KEi will cancel anyway. But how does M get on the top of the fraction?
 
  • #8
Doc Al
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Nevermind about that last question, I see how KEi will cancel anyway. But how does M get on the top of the fraction?
Hint: 1 = (m + M)/(m + M)
 
  • #9
Doy! Can't believe I didn't consider that. Thanks so much for the assistance!

For future reference...how do I make my equations like cepheid did above, is there a certain button I can click in the post screen to do so? It would make my equations look much clearer.
 
  • #11
Thank you!
 

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