Finding the velocity after a collision?

  • #26
Dick
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Ah yes, staying up till 6 in the morning for exams takes its toll on me :)

But that was for the question about finding ##v_{driving}##

My final question was regarding finding the energy lost due to deformation. What was wrong about my expression for ##W_{deformation}##?

Nothing. But you don't need to find that right now. My comment about summing the loses was more conceptual. You should now have three equations in the three unknowns ##v_{driving}, v_{impact}, v_{collision}## you just need to eliminate the other two velocities and get an expression for ##v_{driving}## alone in terms of the other values.
 
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  • #27
Nothing. But you don't need to find that right now. My comment about summing the loses was more conceptual. You should now have three equations in the three unknowns ##v_{driving}, v_{impact}, v_{collision}## you just need to eliminate the other two velocities and get an expression for ##v_{driving}## alone in terms of the other values.

My assignment consists of two questions. One is to find an expression for ##v_{driving}## and the other question is to find the energy that was used for the deformation of the cars. :)

I feel pretty confident about the expression for ##v_{driving}## where I use the 3 equations with the 3 variables.

That only leaves me with the the second question that I was given regarding the energy used for the deformation.
Was the above method that I used for finding ##W_{deformation}## correct?
 
  • #28
Dick
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My assignment consists of two questions. One is to find an expression for ##v_{driving}## and the other question is to find the energy that was used for the deformation of the cars. :)

I feel pretty confident about the expression for ##v_{driving}## where I use the 3 equations with the 3 variables.

That only leaves me with the the second question that I was given regarding the energy used for the deformation.
Was the above method that I used for finding ##W_{deformation}## correct?

Saying ##W_{deformation}=\frac{1}{2}m_1v^2_{impact}=\frac{1}{2}(m_1+m_2)v^2_{collision}## would be correct. No, I don't think the expression you gave is correct. Both of the equations you gave before that had sign errors. And ##v_{rest}## is just 0, right? Probably don't need a separate symbol for it.
 
  • #29
Saying ##W_{deformation}=\frac{1}{2}m_1v^2_{impact}=\frac{1}{2}(m_1+m_2)v^2_{collision}## would be correct. No, I don't think the expression you gave is correct. Both of the equations you gave before that had sign errors. And ##v_{rest}## is just 0, right? Probably don't need a separate symbol for it.

Could it really be this simple:

Keeping in mind that "driving" is the initial velocity of the first car,
"impact" is right before the collision
"collision" is right after the collision

[itex]KE_{lost}=KE_1-KE_2[/itex]

Where ##KE_1## is before the collision and ##KE_2## is after the collision.

Before the collision:
[itex]KE_1=\frac{1}{2}mv_{impact}^2=\frac{1}{2}mv_{driving}^2-\mu mgl_1[/itex]

After the collision:
[itex]KE_2=\frac{1}{2}(m_1+m_2)v_{collision}^2=\mu (m_1+m_2)gl_2[/itex]

So the kinetic energy lost can be found by either two equations:
[itex]KE_{lost}=(\frac{1}{2}mv_{impact}^2)-(\frac{1}{2}(m_1+m_2)v_{collision}^2)[/itex]

Or:

[itex]KE_{lost}=(\frac{1}{2}mv_{driving}^2-\mu mgl_1)-(\mu (m_1+m_2)gl_2)[/itex]

Does that work?
 
  • #30
Dick
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Could it really be this simple:

Keeping in mind that "driving" is the initial velocity of the first car,
"impact" is right before the collision
"collision" is right after the collision

[itex]KE_{lost}=KE_1-KE_2[/itex]

Where ##KE_1## is before the collision and ##KE_2## is after the collision.

Before the collision:
[itex]KE_1=\frac{1}{2}mv_{impact}^2=\frac{1}{2}mv_{driving}^2-\mu mgl_1[/itex]

After the collision:
[itex]KE_2=\frac{1}{2}(m_1+m_2)v_{collision}^2=\mu (m_1+m_2)gl_2[/itex]

So the kinetic energy lost can be found by either two equations:
[itex]KE_{lost}=(\frac{1}{2}mv_{impact}^2)-(\frac{1}{2}(m_1+m_2)v_{collision}^2)[/itex]

Or:

[itex]KE_{lost}=(\frac{1}{2}mv_{driving}^2-\mu mgl_1)-(\mu (m_1+m_2)gl_2)[/itex]

Does that work?

Yes, that works. And it's easy to see why. You are just subtracting the two frictional losses from the original kinetic energy. The difference is what is lost in the collision itself.
 

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