Spring launcher equation to find x

Tags:
1. Mar 19, 2015

Siavash

1. The problem statement, all variables and given/known data
In my physics 12 class, we were given a spring and had to launch it at an angle into a target (the object launched is the spring itself). The target was on the floor level but the spring was launched from above the floor level. I need to find out how far to pull the spring back so it will travel to the target.
I also have to find out how to calculate the maximum height of the spring.
Mass of spring = 0.04012kg
k = 26 N/m (it was measured before )
Height of launch from the ground: 0.48 m
The horizontal distance from the target to the spring: 4 m

2. Relevant equations
It is clearly related to the elastic potential of the spring and kinetic energy as well.
Es = Ek
1/2mv^2 = 1/2kx^2
v^2 = x^2(k/m)
v = x√(k/m)
Beacuase we dont have the velocity, we have to work with projectile motion as well.
Δd = (v)(cosΘ)(t) ==> re-arranging => t = (Δd) / (v cosΘ)
five kinematics equations might be needed but im not sure.
3. The attempt at a solution
Because we dont have velocity, we sure have to work with kinematics as well as energy. I dont know how to relate them together to get the x (distance needed to be pulled). There are some websites that explain a bit but jumped right into the equation. such as:
http://www.geocities.ws/juliars123/springs.htm
They both got the equation: x = √((dmg)/(2k*sinΘ*cosΘ)). i used this equation and it worked. but i dont get how they got it. in specific, both of them use this equation :
d = (V*cosΘ*2V*sinΘ)/g to get the final equation, which they dont explain how they got it. Again, i dont know for sure if these two websites are right. I just would like someone to help me get the right equation or explain the equation achieved by the above websites.

2. Mar 19, 2015

AlephNumbers

How do you approach projectile motion problems?

3. Mar 19, 2015

Siavash

well obviously you right down all the things you have such as time, intital velocity, displacement, etc. and then use the appropriate kinematic equation based on what you need to find.

4. Mar 19, 2015

BvU

Well, assume the initial speed has the value v0. If you define a coordinate system with the target at (4,0) you know where the trajectory starts, namely at (0, 0.48) -- but, as always -- it's better to use symbols for as long as you can. You have an equation for horizontal movement and one for vertical movement, and there are also two unknowns: v0 and the time needed for the trajectory. And once you have v0, you go back to the energy equation to find x !

Obviously. So do just that and you'll be fine!

5. Mar 19, 2015

haruspex

It might not make much difference, but for maximum accuracy you should treat the motion in two phases.
In the first phase, the spring is sliding up the launch ramp. The spring PE is being turned into both KE and gravitational PE. The length (distance) of this phase is unknown because it depends on the initial spring compression, so that will be a variable in it. The point is that you should not assume that it leaves the launch ramp at a height of 0.48 with all its initial spring energy converted into KE.

As BvU notes, you should manipulate the equations in purely symbolic form, only plugging in numbers at the final step. This has many advantages.

6. Mar 19, 2015

Siavash

Time is not a part of this experiment. We were not allowed to record the time of flight so we only have 2 of the variables instead of the usual 3. I need a method (equation) that relates the speed to the distance needed to be stretched (x) guys. If you look at the websites you will find out what i mean !

7. Mar 19, 2015

haruspex

Sure, but in order to get enough equations you need to make use of the fact that the time is the same (though unknown) for both the horizontal and vertical motions. Therefore the SUVAT equations you select should be ones involving time.

8. Mar 19, 2015

Siavash

That is true. I did use it but i did not get the equation i mentioned above. I would like to know how to derive to that equation. I would really appreciate it if you give it a try as well to see what happens and what kind of equation you get. I used the equation x = √((dmg)/(2k*sinΘ*cosΘ)) in the class during the experiment and the spring actually went into the target. So that means its right but i dont know how to derive to it. i need to write a full lab report for this which includes full calculation :(

9. Mar 19, 2015

haruspex

Have you posted that attempt? I don't see it.
You need to use the SUVAT equations, but you have not listed any.

10. Mar 20, 2015

Siavash

Here is how far i go. when i insert t into the equation, there are still two v variables present which means you cant rearrange. maybe im making a mistake or something so please correct me or suggest new ideas.

11. Mar 20, 2015

BvU

$$v\sin\theta {\Delta d\over v\cos\theta} = v\tan\theta \Delta d\ ??$$

12. Mar 20, 2015

Siavash

isnt sin over cos equal to tan? correct me if im wrong

13. Mar 20, 2015

BvU

$${sin\theta \over cos\theta} = \tan\theta$$is correct, but $$v\sin\theta {\Delta d\over v\cos\theta} ={v\sin\theta \over v\cos\theta} \ \Delta d\ ! !$$

14. Mar 20, 2015

Siavash

are you saying that v will cancel out as well? so we are left with tan theta times delta d?

15. Mar 20, 2015

Siavash

okay so here is what i got. Can you guys check and see if it makes sense or not :)

Attached Files:

• 20150320_142246.jpg
File size:
50.9 KB
Views:
124
16. Mar 20, 2015

haruspex

Please do not post images of handwritten algebra. Take the trouble to type it in. Images might be quicker for you, but they slow down all the people trying to read your work and assist you. They also make it harder to comment on your work, identifying a particular line, for example.

17. Mar 21, 2015

BvU

Feel a bit torn up now: I second haru, that's the way it should be at PF.
On the other hand, your work is excellent: clear steps and clear comments.
It does indeed take some time to digest and check, but I did that (on a spreadsheet -- the zenith of unstructured programming ) and it's all correct.
You can now also see the difference with the expression x = √((dmg)/(2k*sinΘ*cosΘ)) in post #8 from which you said "So that means its right" .

Experiment decides in our empirical science world, but what does this mean ? Is post #15 wrong ?

Work out the numerical values now and compare. Also check post #5 by haru and see if that is an effect that should or should not be taken into accont in detail, or perhaps can be corrected for with an estimate.
(who knows, correcting in detail in the equations, it might even make the two expressions identical...)
It all comes down to accuracy considerations, which are a very important (at school often neglected), part of real physics !

Last edited: Mar 21, 2015
18. Mar 21, 2015

haruspex

The links you reference only seem to handle the case where launch and land are at the same level. They have taken g to be a positive value, whereas you are making up positive throughout, so your g will be negative. If you set $\Delta y$ to zero and switch the sign of g in your equation you get theirs.
To check whether you need to take into account the loss of KE while on the ramp, you could calculate x with the equation you have, then reduce $\Delta y$ and increase $\Delta d$ accordingly, run the calculation again, and see if x changes much. Indeed, you could iterate that process. Or you could do all the algebra again, including these effects in the relationship between x and the launch speed.

19. Mar 22, 2015

Siavash

i really appreciate your answers guys ! i only have one more question, does the gravitation potential energy of the spring increase the speed when it starts to change height? do i have to account for gravitation potential energy when im calculating the x?

20. Mar 22, 2015

haruspex

Sorry, I don't understand your question. What do you mean by "when it starts to change height"?
My point in the previous post is that you calculated x so as to provide the KE it will need when at its relaxed position. This doesn't take into account that some of the spring energy has gone into raising it from the compressed position to the relaxed position, a height of x sin(theta).