# Conservation of energy using orientation

1. Nov 10, 2012

### Lolagoeslala

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. Nov 10, 2012

### Staff: Mentor

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

3. Nov 10, 2012

### Lolagoeslala

well the conservation of momentum is conserved...

4. Nov 10, 2012

### Staff: Mentor

Yes, momentum is conserved. Use that to figure out the velocity of the first guy after the collision.

5. Nov 10, 2012

### Lolagoeslala

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. Nov 10, 2012

### Staff: Mentor

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. Nov 10, 2012

### Lolagoeslala

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. Nov 10, 2012

### Staff: Mentor

The kinetic energy 'lost' in the collision is transformed to other forms of energy, such as 'heat'.

9. Nov 10, 2012

### Lolagoeslala

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. Nov 10, 2012

### Staff: Mentor

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.

11. Nov 10, 2012

### Lolagoeslala

oh ok ... thanks :D

12. Nov 10, 2012

### Lolagoeslala

Umm how would i find the kinetic lost during the collision in percentage?

13. Nov 10, 2012

### Staff: Mentor

Calculate the total kinetic energy before and after the collision. Subtract to find the change.

14. Nov 10, 2012

### Lolagoeslala

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. Nov 11, 2012

### Staff: Mentor

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. Nov 11, 2012

### Lolagoeslala

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. Nov 11, 2012

### Staff: Mentor

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. Nov 11, 2012

### 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,,,,,

19. Nov 11, 2012

### Staff: Mentor

I really cannot follow what you did in post #5. Realize that momentum is a vector and that each component is separately conserved.

20. Nov 11, 2012

### Lolagoeslala

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