Perfectly elastic collisions (proof)

  • Thread starter Phyzix
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  • #1
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An object collides elastically (perfectly) with another object (identical object) at rest. If it is not a head on collision how can i PROVE that the angle between them afterwards is 90 degrees? :confused: :bugeye:

I have NO IDEA on what to do...it's been puzzling me and my friends for a little while now...

help anyone?
 

Answers and Replies

  • #2
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In a elastic collision momentum and KE are conserved:

[tex]m_1\overrightarrow{v}_1_i=m_1\overrightarrow{v}_1_f+m_2\overrightarrow{v}_2_f[/tex]

[tex]\frac{1}{2}m_1v^2_1_i=\frac{1}{2}m_1v^2_1_f+\frac{1}{2}m_2v^2_2_f[/tex]

in this case the masses are the same so we can simplify:

[tex]\overrightarrow{v}_1_i=\overrightarrow{v}_1_f+\overrightarrow{v}_2_f[/tex]

[tex]v^2_1_i=v^2_1_f+v^2_2_f[/tex]

Consider now the first equation, look at the vectors [tex]\overrightarrow{v}_1_f[/tex] and [tex]\overrightarrow{v}_2_f[/tex], if they are added they are equal to [tex]\overrightarrow{v}_1_i[/tex], so you can picture the 3 of them as forming a triangle.

If you picture this you can see that [tex]v^2_1_i=v^2_1_f+v^2_2_f[/tex] is the Theorem of Pythagoras. [tex]\overrightarrow{v}_1_i[/tex] is the hypotenuse so the other two sides form 90º.

I hope the explanation is clear enough.
 

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