## Conservation of energy using orientation

 Quote by Doc Al 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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 Quote by Lolagoeslala 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,,,,,
I really cannot follow what you did in post #5. Realize that momentum is a vector and that each component is separately conserved.

 Quote by Doc Al 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[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]

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

 Mentor Blog Entries: 1 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)

 Quote by Doc Al 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[S]) = (80kg)(v1) + (90kg)(2.598076211m/s[N])
205.212086kgm/s[S]-233.8268519m/s[N] = (80kg)(v1)
439.0389379kgm/s[S]/80kg = v1
5.487986724m/s[S] = v1

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

And i get the same answer...

 Mentor Blog Entries: 1 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.)

 Quote by Doc Al 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 im getting is wrong?/ I mean.. why is the velocity incorrect?

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 Quote by Lolagoeslala Hey its ACTUALLY one of the question on my worksheet. Would u like to clarify why the velocity im 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.

 Quote by Doc Al 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.....

 Quote by Doc Al 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[S]) = (80kg)(v1) + (90kg)(1.5m/s[N])
205.212086kgm/s[S]-135m/s[N] = (80kg)(v1)
70.212086kgm/s[S]/80kg = v1
4.252651075m/s[S] = v1

4.124858919 m/s [E] + 4.252651075m/s[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 %