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Momentum & kinetic energy

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Hi, I was reading Andre LaMothes book Tricks of the Windows Game Programming Gurus and got into this. On page 813 he combines conservation of momentum and conservation of kinetic energy equations and gets that:
vaf = (2*mb*vbi + vai * (ma - mb)) / (ma + mb)

and

vbf = (2*ma*vai - vbi * (ma - mb)) / (ma + mb)
      
I was wondering how did he do that? I tried for few hours with no result... (I know how to combine equation groups) Could someone tell the steps how to get to those equations? Then he does some calculation with those:
ma = 2kg
mb = 3kg
vai = 4m/s
vbi = -2m/s

And he gets:

vaf = 1.6 m/s
and
vbf = 2.4 m/s
      
If I do that with those exactly same numbers I get -3.2 m/s and 3.6 m/s. So are equations wrong or has he calculated wrong or what's the problem? Thanks. Edit: missing some brackets...

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To answer the first part of your question, this equation is dealing with perfectly elastic collisions. We know that energy and momentum are conserved, so we get the following equations:

.5*m1*v12+.5*m2*v22 = .5*m1*v1''2+ .5*m2*v2''2

We can cancel out the .5 on each term, and move the v1 expressions on one side, move the v2 expressions on one side, and then factor using the difference of perfect squares to get m1*(v1-v1'')(v1+v1'') = m2*(v2''-v2)(v2''+v2)

From conservation of momentum m1*v1 + m2*v2 = m1*v1'' + m2*v2''
once again, move all of the v1 terms to one side, v2 terms to the other, and you get the following:

m1(v1-v1'') = m2*(v2''-v2).

Divide this into the other equation and you get this result:

v1 +v1'' = v2 + v2''

From here, you should easily be able to derive the expressions that you showed earlier.

Brendan

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quote:

To answer the first part of your question, this equation is dealing with perfectly elastic collisions. We know that energy and momentum are conserved, so we get the following equations:

.5*m1*v12+.5*m2*v22 = .5*m1*v1''2+ .5*m2*v2''2

We can cancel out the .5 on each term, and move the v1 expressions on one side, move the v2 expressions on one side, and then factor using the difference of perfect squares to get m1*(v1-v1'')(v1+v1'') = m2*(v2''-v2)(v2''+v2)

From conservation of momentum m1*v1 + m2*v2 = m1*v1'' + m2*v2''
once again, move all of the v1 terms to one side, v2 terms to the other, and you get the following:

m1(v1-v1'') = m2*(v2''-v2).

Divide this into the other equation and you get this result:

v1 +v1'' = v2 + v2''


This is clear.
quote:

From here, you should easily be able to derive the expressions that you showed earlier.


Could you please show few more steps how to get there?

It''s summer vacation, my brains aren''t working...

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