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Conservation of energy using orientation 
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#1
Nov1012, 03:35 PM

P: 217

1. The problem statement, all variables and given/known data
a hockey player of mass 80 kg was skating at a velocity of 7.5 m/s [E 20 S] when he accidentally hit a linesman who was just standing on the ice. The mass of the linesman is 90 kg and his velocity after the collision was 3 m/s [ N 30 E]. Find the velocity of the hockey player after the collision and the kinetic energy lost during the collision in percentage? 2. Relevant equations I would be using the kinetic energy equation which is 1/2mv^2=1/2mv1`+1/2mv2` 3. The attempt at a solution The problem i have is that how would i get started i mean they are standing on the incline and i do not understand why the linesman's collision orientaion N 30 E. isn't that almost verticle? 


#2
Nov1012, 03:39 PM

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Kinetic energy is not conserved. (They even ask how much is lost.) But what is conserved?



#3
Nov1012, 04:20 PM

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#4
Nov1012, 04:23 PM

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Conservation of energy using orientation



#5
Nov1012, 04:31 PM

P: 217

m1v1 + m2v2 = m1v1` + m2v2` (80 kg)(7.5 m/s [E20S] = (80kg)(V1`) + (90 kg)(3 m/s[N30E]) 600 kgm/s [E20S] = (80kg)(V1`) + 270 kgm/s [N30E] 600 kgm/s [E20S]  270 kgm/s [N30E] 563.8155 kgm/s[E]+205.212086 kgm/s[S]+233.826 kgm/s[S]+135 kgm/s[W] = (80kg)(V1`) 428.8155 kgm/s[E] + 439.038086 kgm/s[S] = (80kg)(V1`) 613.44 kgm/s [S44.3E] = (80kg)(V1`) 7.668 m/s [S44.3E] But why can't i just simple use this equation 1/2m1v1^2 = 1/2m1v1`^2 + 1/2m2v2`^2 This way i don't have to use the orientation since its scaler 


#6
Nov1012, 04:47 PM

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#7
Nov1012, 05:06 PM

P: 217

after the collision wouldn't the work be transferred into some other energy??? 


#8
Nov1012, 05:14 PM

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#9
Nov1012, 05:19 PM

P: 217

why can't i use the equation for kinetic energy is it because some of the energy has been used for other energies??? 


#10
Nov1012, 05:23 PM

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#11
Nov1012, 05:39 PM

P: 217




#12
Nov1012, 07:37 PM

P: 217




#13
Nov1012, 08:32 PM

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#14
Nov1012, 08:36 PM

P: 217

Ek1 = 1/2m1v1 Ek2 = 1/2m1v1`^2 + 1/2m2v2`^2 Ek1= 1/2(80kg)(7.5m/s)^2 Ek1= 2250 J Ek2= 1/2m1v1`^2 + 1/2m2v2`^2 Ek2= 1/2(80kg)(7.668 m/s)^2 + 1/2(90kg)(3 m/s)^2 EK2= 2351.92896 J + 405 J Ek2= 2756.92896 J Ek2  Ek1 506.929 J But this does not seem right.. and how would i get this into percentage? 


#15
Nov1112, 04:36 AM

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I suspect that your calculation of the velocity is wrong. Redo your conservation of momentum work. I suggest that you write separate equations for North and East components of momentum.



#16
Nov1112, 09:45 AM

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#17
Nov1112, 11:25 AM

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Here's what I would do. Map things onto an xy axis, where east is in the positive x direction and north is in the positive y direction. Then set up two momentum conservation equations, one for the x components and one for the y components. Then you can solve for the components of the velocity of the first hockey player. 


#18
Nov1112, 11:30 AM

P: 217




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