Another way to calculate loss of energy

[Vee9]

Homework Statement

How would you calculate the loss of energy down the ramp, given only the following information:
Mass of ball = 0.0083 kg
and
http://i1097.photobucket.com/albums/g349/Physics_/Energy.jpg

Homework Equations

E = mgh
E = 0.5mv^2

The Attempt at a Solution

I already found one way:
To compare the energy at the top of the ramp, which is E = mgh and the energy at the bottom of the ramp, E2 = mgh + 0.5mv^2 and also compare E to the energy at the bottom of the projectile, E3 = 0.5mv^2.

Is this correct?
What is another way to calculate loss of energy?

 

[Azelketh]

Assuming that all the energy from falling down the ramp is converted into motion in the horizontal axis.
Then you could use the equations of motion under constant acceleration to solve for inital horizontal speed,
v^2 =u^2 +2 a s

v=u + a t

s=u t + \frac{1}{2} a t^2<br />
Dont forget to split the motion into horizontal and vertical components to find the time t that the ball is in the air.

then use
<br /> Ke= \frac{1}{2} m v^2<br /> <br />
to find the energy at the bottom of the ramp. Then compare that to the energy at the top of the ramp given by E=m g h

 

[Vee9]

Assuming that all the energy from falling down the ramp is converted into motion in the horizontal axis.
Then you could use the equations of motion under constant acceleration to solve for inital horizontal speed,
v^2 =u^2 +2 a s

v=u + a t

s=u t + \frac{1}{2} a t^2<br />
Dont forget to split the motion into horizontal and vertical components to find the time t that the ball is in the air.

then use
<br /> Ke= \frac{1}{2} m v^2<br /> <br />
to find the energy at the bottom of the ramp. Then compare that to the energy at the top of the ramp given by E=m g h

I'm sorry, but what is "u" and "s"? :S
And I think I did what you said to find v.

I just need a second method to find the loss of energy...

 

[Azelketh]

u = the the initial velocity

s= distance travelled

so from your diagram s=0.381 meters in horizontal axis
and s=0.904+0.019=0.923 meters in the vertical direction

Sorry for any confusion caused. Your method is correct and the easiest using the information you have.

 

[Vee9]

u = the the initial velocity

s= distance travelled

so from your diagram s=0.381 meters in horizontal axis
and s=0.904+0.019=0.923 meters in the vertical direction

Sorry for any confusion caused. Your method is correct and the easiest using the information you have.

lol, ohh. It's okay. :)
Thanks for your help.

The thing is, the teacher asked us to "measure as accurately as possible and in more than one method, the loss of energy."
So I was thinking of using the method above as one method.

Then comparing Ek (kinetic energy) of the ball at the end of the ramp and the Ek of the ball at the bottom of the projectile, as a second method.
Or is that wrong?

 

[Azelketh]

comparing the kinetic energy at the end of the ramp and bottom would give you the energy gained due to falling as the difference.

It wouldent allow you to find the energy lost on the ramp as the energy at the end of the ramp has already lost the energy on the ramp if that makes sense.

 

[Vee9]

comparing the kinetic energy at the end of the ramp and bottom would give you the energy gained due to falling as the difference.

It wouldent allow you to find the energy lost on the ramp as the energy at the end of the ramp has already lost the energy on the ramp if that makes sense.

Yeah, sorry. I realized that after a while.
I'm stuck.
I don't know any other way to find the difference in energy.

Also, when I find E1 = mgh, would I use (0.307+ 0.904) as "h" or would I just use 0.307 as "h?"
And when finding the time of the projectile in E2 using y = 0.5gt^2, would I use (0.019+0.904) as delta y?

 

[Azelketh]

Yeah, sorry. I realized that after a while.
I'm stuck.
I don't know any other way to find the difference in energy.

Also, when I find E1 = mgh, would I use (0.307+ 0.904) as "h" or would I just use 0.307 as "h?"
And when finding the time of the projectile in E2 using y = 0.5gt^2, would I use (0.019+0.904) as delta y?

Using
<br /> h_1= 0.307+ 0.904 = 1.211 <br />
<br /> E_1=m g h_1<br />

then E1 gives you the potential energy with respect to the ground.

Using 0.307 as h would give you the potential energy with respect to the top of the black section on your diagram, so it wouldent be useful here i think.

Using 0.019 + 0.904 as the change in height is correct.