Conservation of energy using orientation

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?

Doc Al

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

Lolagoeslala

well the conservation of momentum is conserved...

Doc Al

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

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[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

Doc Al

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

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???

Doc Al

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

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???

Doc Al

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.

Lolagoeslala

oh ok ... thanks :D

Lolagoeslala

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

Doc Al

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

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?

Doc Al

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

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 .....

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.