Conservation of energy using orientation


by Lolagoeslala
Tags: conservation, energy, orientation
Lolagoeslala
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#1
Nov10-12, 03:35 PM
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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?
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Doc Al
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#2
Nov10-12, 03:39 PM
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Kinetic energy is not conserved. (They even ask how much is lost.) But what is conserved?
Lolagoeslala
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#3
Nov10-12, 04:20 PM
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Quote Quote by Doc Al View Post
Kinetic energy is not conserved. (They even ask how much is lost.) But what is conserved?
well the conservation of momentum is conserved...

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#4
Nov10-12, 04:23 PM
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Conservation of energy using orientation


Quote Quote by Lolagoeslala View Post
well the conservation of momentum is conserved...
Yes, momentum is conserved. Use that to figure out the velocity of the first guy after the collision.
Lolagoeslala
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#5
Nov10-12, 04:31 PM
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Quote Quote by Doc Al View Post
Yes, momentum is conserved. Use that to figure out the velocity of the first guy after the collision.
so...

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
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#6
Nov10-12, 04:47 PM
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Quote Quote by Lolagoeslala View Post
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
You can't use it because it's not true. In general, kinetic energy is not conserved in a collision. (When it is conserved, it's called a perfectly elastic collision.) But momentum is always conserved.
Lolagoeslala
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#7
Nov10-12, 05:06 PM
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Quote Quote by Doc Al View Post
You can't use it because it's not true. In general, kinetic energy is not conserved in a collision. (When it is conserved, it's called a perfectly elastic collision.) But momentum is always conserved.
Im confused..... so for the kinetic energy it is the work being done...
after the collision wouldn't the work be transferred into some other energy???
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#8
Nov10-12, 05:14 PM
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Quote Quote by Lolagoeslala View Post
Im confused..... so for the kinetic energy it is the work being done...
after the collision wouldn't the work be transferred into some other energy???
The kinetic energy 'lost' in the collision is transformed to other forms of energy, such as 'heat'.
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#9
Nov10-12, 05:19 PM
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Quote Quote by Doc Al View Post
The kinetic energy 'lost' in the collision is transformed to other forms of energy, such as 'heat'.
yes.... so
why can't i use the equation for kinetic energy
is it because some of the energy has been used for other energies???
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#10
Nov10-12, 05:23 PM
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Quote Quote by Lolagoeslala View Post
yes.... so
why can't i use the equation for kinetic energy
Because kinetic energy is not conserved. (Total energy including everything is always conserved, but you have no way of using that fact.)
is it because some of the energy has been used for other energies???
That's right.
Lolagoeslala
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#11
Nov10-12, 05:39 PM
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Quote Quote by Doc Al View Post
Because kinetic energy is not conserved. (Total energy including everything is always conserved, but you have no way of using that fact.)

That's right.
oh ok ... thanks :D
Lolagoeslala
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#12
Nov10-12, 07:37 PM
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Quote Quote by Doc Al View Post
Because kinetic energy is not conserved. (Total energy including everything is always conserved, but you have no way of using that fact.)

That's right.
Umm how would i find the kinetic lost during the collision in percentage?
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#13
Nov10-12, 08:32 PM
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Quote Quote by Lolagoeslala View Post
Umm how would i find the kinetic lost during the collision in percentage?
Calculate the total kinetic energy before and after the collision. Subtract to find the change.
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#14
Nov10-12, 08:36 PM
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Quote Quote by Doc Al View Post
Calculate the total kinetic energy before and after the collision. Subtract to find the change.
so...

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?
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#15
Nov11-12, 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.
Lolagoeslala
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#16
Nov11-12, 09:45 AM
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Quote Quote by Doc Al View Post
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.
I am still getting the same answer...... i mean what do you mean by separate components i do break them apart.. as i did in the calculation above .....
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#17
Nov11-12, 11:25 AM
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Quote Quote by Lolagoeslala View Post
I am still getting the same answer...... i mean what do you mean by separate components i do break them apart.. as i did in the calculation above .....
I did not quite follow the calculation you did in post #5.

Here's what I would do. Map things onto an x-y 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.
Lolagoeslala
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#18
Nov11-12, 11:30 AM
P: 217
Quote Quote by Doc Al View Post
I did not quite follow the calculation you did in post #5.

Here's what I would do. Map things onto an x-y 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.
But why can't i do it like i did in post 5? I mean thats how we have been thought in class to break the components in the equation itself, i have never seen of breaking them apart and then using two different equations,,,,,


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