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Collisions- Simple Algebra Problem

  • Thread starter verd
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Collisions-- Simple Algebra Problem...

Hi,

So... I think I'm having a simple algebra problem-- I was just wondering if someone could point out my error. This is the problem:

Block 1, of mass [tex]m_{1}[/tex], moves across a frictionless surface with speed [tex]u_{i}[/tex]. It collides elastically with block 2, of mass [tex]m_{2}[/tex], which is at rest ([tex]v_{i}=0[/tex]). After the collision, block 1 moves with speed [tex]u_{f}[/tex], while block 2 moves with speed [tex]v_{f}[/tex]. Assume that [tex]m_{1} > m_{2}[/tex], so that after the collision, the two objects move off in the direction of the first object before the collision.

[Image]

What is the final speed [tex]u_{f}[/tex] of block 1?


So-- I find [tex]m_{2}v_{f}[/tex] using the law of conservation of momentum:
[tex]m_{2}v_{f} =m_{1}u_{i}-m_{1}u_{f}[/tex]

And I find [tex]m_{2}v_{f}^2[/tex] using the law of conservation of kinetic energy:
[tex]m_{2}v_{f}^2 = m_{1}(u_{i} - u_{f})(u_{i} + u_{f}).[/tex]

Then I find [tex]v_{f}[/tex] using only [tex]u_{i}[/tex], and [tex]u_{f}[/tex]:
[tex]v_{f} =\displaystyle{\frac{-(u_{i}+u_{f})-(u_{i}+u_{f})}{-2}}[/tex]


Now, I have to substitute what I just found for [tex]v_{f}[/tex] into the conservation of momentum formula, and solve for [tex]u_{f}[/tex]. ...But I guess I'm having difficulty singling out the [tex]u_{f}[/tex].

This is what I've got:

[tex]m_{1}u_{1} = m_{1}u_{f} + m_{2}(\displaystyle{\frac{-(u_{i}+u_{f})-(u_{i}+u_{f})}{-2}})[/tex]
...And here, I think I messed my algebra up, but when I simplify that, I get:
[tex]m_{1}u_{1} = m_{2}(u_{f} + u_{i}) + m_{1}u_{f}[/tex]

Is that right? ...If not/if so, how do I get the [tex]u_{f}[/tex] on just one side?
 

Answers and Replies

  • #2
Doc Al
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verd said:
So-- I find [tex]m_{2}v_{f}[/tex] using the law of conservation of momentum:
[tex]m_{2}v_{f} =m_{1}u_{i}-m_{1}u_{f}[/tex]
OK.

And I find [tex]m_{2}v_{f}^2[/tex] using the law of conservation of kinetic energy:
[tex]m_{2}v_{f}^2 = m_{1}(u_{i} - u_{f})(u_{i} + u_{f}).[/tex]
OK.

Then I find [tex]v_{f}[/tex] using only [tex]u_{i}[/tex], and [tex]u_{f}[/tex]:
[tex]v_{f} =\displaystyle{\frac{-(u_{i}+u_{f})-(u_{i}+u_{f})}{-2}}[/tex]
I have no idea what you are doing here. This expression simplifies to: [itex]v_{f} = u_{i}+u_{f}[/itex], which is incorrect.

You can rewrite the momentum equation to get
[tex]v_{f} = \frac{m_1}{m_2} (u_{i} - u_{f})[/tex]
Perhaps this is what you meant? Now just plug that into the KE equation to eliminate [itex]v_f[/itex] and simplify.



...And here, I think I messed my algebra up, but when I simplify that, I get:
[tex]m_{1}u_{1} = m_{2}(u_{f} + u_{i}) + m_{1}u_{f}[/tex]

Is that right? ...If not/if so, how do I get the [tex]u_{f}[/tex] on just one side?
Amazingly, it is right. (It's equivalent to what you get when you do the "plugging in" that I suggest above.) So I suspect you made a typo earlier on. To simplify this expression, just multiply it out and move all terms containing [itex]u_f[/itex] to one side and all other terms to the other side.
 
  • #3
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woo, okay. thanks. I got it. ...Really, much appreciated.

Thanks again... (Eh, I'm having a really hard time in this class- if you couldn't tell. AND I made the mistake of taking it as a 7-week course. ...I obviously didn't know what I was in for. Really, thanks again.)
 

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