anonymous
  • anonymous
A car traveling north at 10.0 m/s crashes into a car traveling east at 15 m/s at an unexpectedly icy intersection. The cars lock together as they skid on the ice. The two cars have the same mass. What is their combined speed after the collision? Am I wrong in assuming I cannot come to a proper answer without both the mass and velocity of each vehicle? Normally I'd just split the momentum of each car into x and y components of the combined force, and use the combined force with the combined mass to find the final combined velocity. Or is there another way to do this?
Physics
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SOLVED
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katieb
  • katieb
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apoorvk
  • apoorvk
you are almost right. but the problem is we dont have the mass right?? so, assume the mass 'm' for each body (they have same masses its given). now solve with mass as m.. you will see that when coming to the end of the calculations, the 'm' will get cancelled, leaving your answer in absolute numerical terms!!
amistre64
  • amistre64
m1v1+m2v2 = (m1+m2)v3 is the formula for this if i recall it right
amistre64
  • amistre64
but since m1=m2 we can factor them out as: (m)(v1+v2) = mv3

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anonymous
  • anonymous
So, Pf= 10mne(kg*m/s)+15mne(kg*m/s) And Pf=mf*vf So Vf=25mne(kg*m/s)/mne(kg*m/s) Then cross off "mne (kg)" And you get Vf=25m/s?
amistre64
  • amistre64
|dw:1332803233884:dw|
anonymous
  • anonymous
Well if I'm doing this correctly amistre, the final force vector, and the hypotenuse of this triangle is 150, but if you try to find the final velocity of that force vector, you're just dividing by the variable, which will cancel out the mass, but still leave you at 150m/s no? Vf=Pf/mf | Vf=150kg*m/s / "x"kg? Apoorvk's way got me to a much more realistic 25m/s?
amistre64
  • amistre64
im just going by what little i recall. with the masses out of the picture we would add the force vectors to obtain the resultant vector force. since its the hypotenuse of a right triangle the speed would be the magnitude of that vector
amistre64
  • amistre64
I get 18.03 as an approximation
anonymous
  • anonymous
Ah, I see, thanks for helping!
amistre64
  • amistre64
youre welcome
anonymous
  • anonymous
Be careful in your use of the words 'force' here. There are no EXTERNAL forces acting here. Therefore, since from Newton's Second Law, the sum of the external forces is equal to the time rate of change of the momentum\[\sum F = {dp \over dt}\] Additionally, this is a perfectly inelastic collision because the masses stick together. There are several things we can assume. First, momentum is conserved (as established by Newton's Second Law and the lack of external forces) and energy is not conserved. As amistre64 correctly listed, since momentum is conserved, \[m_1 v_1 + m_2 v_2 = (m_1 + m_2) v'\]Keep in mind that velocity is a vector quantity and should be analyzed as such.

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