anonymous
  • anonymous
Micah and Delanie are in the a soapboc cart derby. Thay are about to crash into each other. They both are buckled into their respective carts and when they have their inelastic collision, they all go at a speed of 0.40 m/s in the direction frm which Micah was originally coming. If Micah's total cart mass (including himself) was 120. kg and Delanie's total cart mass (including herself) was 85.0kg, what would be the speed of each cart if the total energy "lost" in the collision was 1672.8J?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
@amistre64 please helpppp
amistre64
  • amistre64
|dw:1359734933000:dw|
amistre64
  • amistre64
hmm \[120v1+85v2=285(.4)+1672.8\] is what my idea is

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amistre64
  • amistre64
of course making sure the energy lost is put in proper relation
anonymous
  • anonymous
i know im pretty lost. i have to got to class. ill let you know aht the answers is and how my techer did it later. thanks for the help.
amistre64
  • amistre64
k, ill see what i can do on my own to see if my idea has any merit :)
amistre64
  • amistre64
and thats spose to be 205, not 285 ... typo :)
amistre64
  • amistre64
the energy in the system as it stands is: 205(.4)^2/2 = 16.4 J add that to what was lost; 1672.8 + 16.4 ------ 1689.2 kg m^2/s^2 = Mv^2/2 =102.5kg v3 m^2/s2 1689.2/102.5 = (v3)^2 sqrt(16.48) = v3 v3 is about 4.06 m/s if im thinking this correctly :)
amistre64
  • amistre64
this at least gives me an equation: 120 v1 + 85 v2 = 205(4.06) if i could see another way to determine a second equation in v1 and v2 we would have 2 eqs in 2 unknowns
amistre64
  • amistre64
im thinking that maybe: 120/2 v1^2 + 85/2 v2^2 = 1689.2 kg m^2/s^2 60 v1^2 + 42.5 v2^2 = 1689.2 but i think im just going in circles .... oh well
amistre64
  • amistre64
v2 = sqrt((1689.2 - 60x)/42.5) 120 v1 + 85 sqrt((1689.2 - 60x)/42.5) = 205(4.06) the calculations give me v1 as about 2.69 v2 as about 5.99 ifn i did it correctly to begin with :)
anonymous
  • anonymous
yeah you did you got it right!
anonymous
  • anonymous
help

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