# Collisions- Simple Algebra Problem

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 $$m_{1}$$, moves across a frictionless surface with speed $$u_{i}$$. It collides elastically with block 2, of mass $$m_{2}$$, which is at rest ($$v_{i}=0$$). After the collision, block 1 moves with speed $$u_{f}$$, while block 2 moves with speed $$v_{f}$$. Assume that $$m_{1} > m_{2}$$, 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 $$u_{f}$$ of block 1?

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

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

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

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

This is what I've got:

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

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

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verd said:
So-- I find $$m_{2}v_{f}$$ using the law of conservation of momentum:
$$m_{2}v_{f} =m_{1}u_{i}-m_{1}u_{f}$$
OK.

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

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

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

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

Is that right? ...If not/if so, how do I get the $$u_{f}$$ 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 $u_f$ to one side and all other terms to the other side.

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.)