Conservation of energy using orientation

  • #1
Lolagoeslala
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Homework Statement


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?


Homework Equations


I would be using the kinetic energy equation which is
1/2mv^2=1/2mv1`+1/2mv2`


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?
 

Answers and Replies

  • #2
Doc Al
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Kinetic energy is not conserved. (They even ask how much is lost.) But what is conserved?
 
  • #3
Lolagoeslala
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Kinetic energy is not conserved. (They even ask how much is lost.) But what is conserved?

well the conservation of momentum is conserved...
 
  • #4
Doc Al
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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.
 
  • #5
Lolagoeslala
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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+233.826 kgm/s+135 kgm/s[W] = (80kg)(V1`)
428.8155 kgm/s[E] + 439.038086 kgm/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
Doc Al
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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.
 
  • #7
Lolagoeslala
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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?
 
  • #8
Doc Al
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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'.
 
  • #9
Lolagoeslala
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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?
 
  • #10
Doc Al
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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.
 
  • #11
Lolagoeslala
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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
 
  • #12
Lolagoeslala
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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?
 
  • #13
Doc Al
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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.
 
  • #14
Lolagoeslala
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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?
 
  • #15
Doc Al
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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
Lolagoeslala
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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.

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 ...
 
  • #17
Doc Al
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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.
 
  • #18
Lolagoeslala
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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 that's 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,,,,,
 
  • #19
Doc Al
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But why can't i do it like i did in post 5? I mean that's 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,,,,,
I really cannot follow what you did in post #5. Realize that momentum is a vector and that each component is separately conserved.
 
  • #20
Lolagoeslala
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I really cannot follow what you did in post #5. Realize that momentum is a vector and that each component is separately conserved.

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+233.826 kgm/s+135 kgm/s[W] = (80kg)(V1`)
428.8155 kgm/s[E] + 439.038086 kgm/s = (80kg)(V1`)
613.44 kgm/s [S44.3E] = (80kg)(V1`)
7.668 m/s [S44.3E]

so what i did was i multiplied the mass of player which is 80 kg by the initial velocity before the collision. The linesman is not moving so the velocity is 0m/s. And after the collision for the player we are trying to find the final velocity and so for the linesman we know the final velocity and the mass so i multiplied them.
I took the linesman's momentum after collision and took it to the other side where it becomes negative. I broke the north and east compoenets
Then i broke the kgm/s before and after into their separate components
 
  • #21
Doc Al
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Looks to me like you are adding up perpendicular components like ordinary numbers. You cannot do that.

Instead write two equations:
initial momentum (x components) = final momentum (x components)
initial momentum (y components) = final momentum (y components)
 
  • #22
Lolagoeslala
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Looks to me like you are adding up perpendicular components like ordinary numbers. You cannot do that.

Instead write two equations:
initial momentum (x components) = final momentum (x components)
initial momentum (y components) = final momentum (y components)

so like this?
v1 = 7.5 m/s [E20S]
V2`= 3m/s [N30E]

x components
m1v1 = m1v1` + m2v2`
(80kg)(7.047694656m/s[E]) = (80kg)(v1`) + (90kg)(1.5m/s[E])
563.8155725 kgm/s [E] - 135 kgm/s[E] = (80kg)(v1`)
428.8155725 kgm/s [E] / 80kg = v1`
5.360194656 m/s [E] = v1`

y components
m1v1 = m1v1` + m2v2`
(80kg)(2.565151075m/s) = (80kg)(v1`) + (90kg)(2.598076211m/s[N])
205.212086kgm/s-233.8268519m/s[N] = (80kg)(v1`)
439.0389379kgm/s/80kg = v1`
5.487986724m/s = v1`

combining them:
5.360194656 m/s [E] + 5.487986724m/s = v1`
7.668 m/s [S 44.3° E] = v1`

And i get the same answer...
 
  • #23
Doc Al
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OK, this time it's much clearer what you've done. (Sorry for not catching on earlier.) I don't see an obvious error in your work, so I suspect that there is a mistake in the problem statement. Double check the angles given.

(If this is from a textbook, tell me which.)
 
  • #24
Lolagoeslala
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OK, this time it's much clearer what you've done. (Sorry for not catching on earlier.) I don't see an obvious error in your work, so I suspect that there is a mistake in the problem statement. Double check the angles given.

(If this is from a textbook, tell me which.)

Hey its ACTUALLY one of the question on my worksheet. Would u like to clarify why the velocity I am getting is wrong?/ I mean.. why is the velocity incorrect?
 
  • #25
Doc Al
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Hey its ACTUALLY one of the question on my worksheet. Would u like to clarify why the velocity I am getting is wrong?/ I mean.. why is the velocity incorrect?
Given the data in the problem statement, your velocity is correct. (Unless you made an error somewhere.) But that would imply that the kinetic energy increased as a result of the collision, which I do not think was your instructor's intention in setting up the problem.
 
  • #26
Lolagoeslala
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Given the data in the problem statement, your velocity is correct. (Unless you made an error somewhere.) But that would imply that the kinetic energy increased as a result of the collision, which I do not think was your instructor's intention in setting up the problem.

Oh so ... i understand what you are trying to say.. so the kinetic energy needs to be lost rather then being increased...
 
  • #27
Lolagoeslala
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Given the data in the problem statement, your velocity is correct. (Unless you made an error somewhere.) But that would imply that the kinetic energy increased as a result of the collision, which I do not think was your instructor's intention in setting up the problem.

iven:
m1= 80 kg
m2 = 90 kg
v1 = 7.5 m/s [ E 20 S]
v1` = ?
v2= 0m/s
v2` = 3m/s [ E 30 N]

X component
m1v1 = m1v1` + m2m2`
(80kg)(7.047694656m/s[E]) = (80kg)(v1`) + (90kg)(2.598076211[E])
563.8155725 kgm/s [E] - 233.826859 kgm/s[E] = (80kg)(v1`)
329.9887135 [E] / 80kg = v1`
4.124858919 m/s [E] = v1`

y components
m1v1 = m1v1` + m2v2`
(80kg)(2.565151075m/s) = (80kg)(v1`) + (90kg)(1.5m/s[N])
205.212086kgm/s-135m/s[N] = (80kg)(v1`)
70.212086kgm/s/80kg = v1`
4.252651075m/s = v1`

4.124858919 m/s [E] + 4.252651075m/s = v1`
5.924 m/s [S 44° E] = v1`

b) Ektot = 1/2m1v1^2
Ektot= 1/2(80kg)(7.5m/s)^2
Ektot=2250 J

Ektot` = 1/2m1v1`^2 + 1/2m2v2`
Ektot` = 1/2(80kg)(5.924m/s)^2 + 1/2(90kg)(3m/s)^2
Ektot` =1403.75104 J + 405 kg J
Ektot` = 1808.75104 J

Ektot - Wdef = Ektot`
2250 J - Wdef = 1808.75104 J
Wdef = 441.24896 J

The Wdef is the energy lost during the collision

(441.248896 J / 2250 J) x 100%
19.6 %


I guess i should just re-post my answer on this thread...
 
  • #28
Doc Al
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iven:
m1= 80 kg
m2 = 90 kg
v1 = 7.5 m/s [ E 20 S]
v1` = ?
v2= 0m/s
v2` = 3m/s [ E 30 N]
Looks like you did have a mistake in the problem statement, which you've corrected. Now it should make sense.
 

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