Conservation of energy using orientation

In summary: I would map things onto an x-y axis, where east is in the positive...In summary, the hockey player lost 596.92896 j in kinetic energy when he collided with the linesman.
  • #1
Lolagoeslala
217
0

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

well the conservation of momentum is conserved...
 
  • #4
Lolagoeslala said:
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
Doc Al said:
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
Lolagoeslala said:
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
Doc Al said:
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
Lolagoeslala said:
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
Doc Al said:
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
Lolagoeslala said:
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
Doc Al said:
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
Doc Al said:
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
Lolagoeslala said:
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
Doc Al said:
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
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
Doc Al said:
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
Lolagoeslala said:
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
Doc Al said:
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
Lolagoeslala said:
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
Doc Al said:
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
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
Doc Al said:
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
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
Doc Al said:
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
Lolagoeslala said:
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
Doc Al said:
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
Doc Al said:
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
Lolagoeslala said:
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.
 

1. What is conservation of energy using orientation?

Conservation of energy using orientation is a principle in physics that states that the total amount of energy in a closed system remains constant over time. This means that energy cannot be created or destroyed, but it can be transferred or converted from one form to another.

2. How does conservation of energy using orientation apply to everyday life?

Conservation of energy using orientation applies to everyday life in many ways. For example, when you turn on a light switch, you are converting electrical energy into light energy. When you ride a bike, you are converting chemical energy from your body into kinetic energy. Understanding this principle can help us make more efficient use of energy in our daily activities.

3. Can energy be lost during conservation of energy using orientation?

No, energy cannot be lost during conservation of energy using orientation. It can only be transferred or converted into other forms. This is because energy is a fundamental property of the universe and cannot be destroyed.

4. What role does orientation play in conservation of energy?

Orientation plays a crucial role in conservation of energy, as the transfer or conversion of energy depends on the orientation of the objects involved. For example, when a ball rolls down a hill, the potential energy is converted into kinetic energy due to the orientation of the slope.

5. How does conservation of energy using orientation relate to the laws of thermodynamics?

Conservation of energy using orientation is closely related to the laws of thermodynamics, particularly the first law which states that energy cannot be created or destroyed. This principle also helps to explain the second law of thermodynamics, which states that energy tends to dissipate and become more disordered over time.

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