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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.

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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?

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# Collisions- Simple Algebra Problem

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