- #1
verd
- 146
- 0
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?
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?