Angular momentum and collisions

In summary, the conversation discusses the concept of angular momentum and collisions, specifically in the context of a wooden block attached to a rigid rod being hit by a bullet. The discussion covers the calculation of the fraction of kinetic energy lost in the collision, using equations for conservation of momentum and kinetic energy. It is also mentioned that Latex can be used to write equations in a clearer format.
  • #1
XxBollWeevilx
78
0
[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 traveling 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.
 
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  • #2
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
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
XxBollWeevilx said:
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
cepheid said:
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
XxBollWeevilx said:
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!
 

Related to Angular momentum and collisions

1. What is angular momentum?

Angular momentum is a measure of the rotational motion of an object. It is the product of an object's moment of inertia and its angular velocity.

2. How is angular momentum conserved in collisions?

In a collision, the total angular momentum of a system remains constant, meaning that the sum of the angular momentum of all objects involved before the collision equals the sum after the collision.

3. How does the law of conservation of angular momentum apply to objects in space?

The law of conservation of angular momentum applies to objects in space because there is no external force to change their rotational motion. This means that their angular momentum will remain constant unless acted upon by an external force.

4. Can angular momentum be negative?

Yes, angular momentum can be negative. This typically occurs when the direction of rotation is opposite to the direction of the chosen axis of rotation, resulting in a negative value for angular momentum.

5. How is angular momentum related to torque?

Angular momentum is closely related to torque, as torque is the rate of change of angular momentum. In other words, torque is the force that causes an object to rotate, and the resulting change in angular momentum is directly proportional to the applied torque.

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