Answer:
If we had:
[tex]v_{1i}=5.3m/s[/tex]
[tex]v_{2i}=5.9m/s[/tex]
We will have:
[tex]v_{1f}=5.9m/s[/tex]
[tex]v_{2f}=5.3m/s[/tex]
Explanation:
In an elastic collision both linear momentum and kinetic energy are conserved, so we will have:
[tex]p_i=p_f[/tex]
[tex]K_i=K_f[/tex]
We will call our bumpers 1 and 2.
For the momentum equation we know that:
[tex]m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}[/tex]
Since all the masses are the same (300kg), they cancel out:
[tex]v_{1i}+v_{2i}=v_{1f}+v_{2f}[/tex]
For the kinetic energy equation we know that:
[tex]\frac{m_1v_{1i}^2}{2}+\frac{m_2v_{2i}^2}{2}=\frac{m_1v_{1f}^2}{2}+\frac{m_2v_{2f}^2}{2}[/tex]
Since all the masses are the same (300kg), they cancel out (and also the 2 dividing):
[tex]v_{1i}^2+v_{2i}^2=v_{1f}^2+v_{2f}^2[/tex]
We then must solve this system:
[tex]v_{1i}+v_{2i}=v_{1f}+v_{2f}[/tex]
[tex]v_{1i}^2+v_{2i}^2=v_{1f}^2+v_{2f}^2[/tex]
Which we will rewrite as:
[tex]v_{1i}-v_{1f}=v_{2f}-v_{2i}[/tex]
[tex]v_{1i}^2-v_{1f}^2=v_{2f}^2-v_{2i}^2[/tex]
The last of these equations can be written as:
[tex](v_{1i}+v_{1f})(v_{1i}-v_{1f})=(v_{2f}+v_{2i})(v_{2f}-v_{2i})[/tex]
But we know that [tex]v_{1i}-v_{1f}=v_{2f}-v_{2i}[/tex], so those cancel out:
[tex]v_{1i}+v_{1f}=v_{2f}+v_{2i}[/tex]
So we can write:
[tex]v_{1i}-v_{1f}+v_{2i}=v_{2f}[/tex]
[tex]v_{1i}+v_{1f}-v_{2i}=v_{2f}[/tex]
Which means:
[tex]v_{1i}-v_{1f}+v_{2i}=v_{1i}+v_{1f}-v_{2i}[/tex]
Which solving for the final velocity leaves us with:
[tex]v_{2i}+v_{2i}=+v_{1f}+v_{1f}[/tex]
[tex]v_{1f}=v_{2i}[/tex]
Grabbing any equation that relates both final velocities easily, for example [tex]v_{1i}-v_{1f}+v_{2i}=v_{2f}[/tex], we obtain:
[tex]v_{2f}=v_{1i}-v_{1f}+v_{2i}=v_{1i}-v_{1f}+v_{1f}=v_{1i}[/tex]
So we conclude that the bumpers have just exchanged velocities (something sometimes seen in billiards for example):
[tex]v_{1f}=v_{2i}=5.9m/s[/tex]
[tex]v_{2f}=v_{1i}=5.3m/s[/tex]
