Spring launcher equation to find x

[Siavash]

Homework Statement

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

Homework 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 don't 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 I am not sure.

The Attempt at a Solution

Because we don't have velocity, we sure have to work with kinematics as well as energy. I don't 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://04adams.tripod.com/physics/labs/term2/spring_lab/
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 don't 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 don't explain how they got it. Again, i don't 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.

 

[AlephNumbers]

How do you approach projectile motion problems?

 

[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.

 

[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 !

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

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

 

[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.

 

[Siavash]

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!

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 !

 

[haruspex]

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 !

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.

 

[Siavash]

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.

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 don't know how to derive to it. i need to write a full lab report for this which includes full calculation :(

 

[haruspex]

I did use it but i did not get the equation i mentioned above.

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

 

[Siavash]

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

Here is how far i go. when i insert t into the equation, there are still two v variables present which means you can't rearrange. maybe I am making a mistake or something so please correct me or suggest new ideas.

 

[BvU]

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

 

[Siavash]

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

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

 

[BvU]

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

 

[Siavash]

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

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

 

[Siavash]

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

 

[haruspex]

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

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.

 

[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 !

 

[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.

 

[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 I am calculating the x?

 

[haruspex]

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 I am calculating the x?

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).

 

[Siavash]

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).

I meant that at the time of stretching the spring, it has both elastic potential energy and gravitational energy and during its flight all of them convert into kinetic energy. The velocity that we are calculating right now is only from the elastic potential energy. As it changes its height, gravitational energy affects the speed and makes it to increase. Isnt that right? and if it is, doesn't it cause the object to miss the target maybe?

 

[BvU]

No, this isn't about the flight phase. The trajectory calculation done for that is completely correct and deals with kinetic and potential energy perfectly well.

It's about the stretching phase only. From letting go to to when the center of mass is at 0.48 m the spring stretches from "compressed by x" to "uncompressed length". That means its center of mass will move upwards by $x\sin(30^\circ)$. An energy of $mgx\sin(30^\circ)$ is needed to achieve that, and that energy can only come from the spring compression energy.

Now, depending on the confuguration at launch, that may require a correction. The deciding criterion is: was the top end of the compressed spring at y=0.48 m or was the top end of the uncompressed spring at 0.48 m (so before the compression).

If I'm not mistaken, for the latter case that is a considerable extra compression required. In fact so much that that extra compression also justifies a correction.
And then your answer will deviate even more from the value that gave a hit.

I don't think you have provided enough information for your kind helpers to make that choice, so it's up to you.

-----

Next issue: the required accuracy. If your drawing is to scale, the target distance is (4 $\pm$ 0.4) m. So you can check if that can distinguish between a correct calculation (yours) or an incorrect one (the website expression you used, which, as haru made clear in post #18, isn't really appropriate). Perhaps you've just been lucky !

 

[haruspex]

I meant that at the time of stretching the spring

I imagined it as compressed then released, not stretched. If stretched, it requires some kind of hook at the top of the ramp.

it has both elastic potential energy and gravitational energy

Before release, yes it has elastic PE and, relative to the height of the target, some gravitational PE.

during its flight all of them convert into kinetic energy

Yes, by the time it lands it will only have KE.

As it changes its height, gravitational energy affects the speed and makes it to increase.

Your flight equations already take into account that its speed will increase after it passes the highest point.

In most questions like this, there is a mass (placed or attached) at the top end of the spring and the mass of the spring itself is largely irrelevant. In the present case, am I to understand that it is just a simple uniform spring, with no other masses involved? If so, the expansion stage (I'm assuming a compressed spring here) gets a bit tricky. At the point where the spring becomes relaxed (i.e. just as it loses contact with the backstop on the ramp), the KE is not uniformly distributed along the spring. If the front of the spring is moving at speed v, the mid point will be moving at v/2, etc. In respect of the subsequent trajectory, it's the speed of the mid point that matters, i.e. it will be as though the whole spring is moving at v/2.
If the spring has length L and mass m, the total KE in the spring at this point is therefore $\frac 1{2L^3} m \int _{x=0}^{L}(xv)^2.dx = \frac 16 mv^2$, while the effective KE for the flight is $\frac 12 m (\frac v2)^2 = \frac 18 mv^2$. I.e. only 3/4 of the original PE is converted into useful KE. The rest goes into oscillations in the spring.
If it is a matter of stretching the spring, a corresponding issue arises, though the details may look different.

 

[Siavash]

I imagined it as compressed then released, not stretched. If stretched, it requires some kind of hook at the top of the ramp.

Before release, yes it has elastic PE and, relative to the height of the target, some gravitational PE.

Yes, by the time it lands it will only have KE.

Your flight equations already take into account that its speed will increase after it passes the highest point.

In most questions like this, there is a mass (placed or attached) at the top end of the spring and the mass of the spring itself is largely irrelevant. In the present case, am I to understand that it is just a simple uniform spring, with no other masses involved? If so, the expansion stage (I'm assuming a compressed spring here) gets a bit tricky. At the point where the spring becomes relaxed (i.e. just as it loses contact with the backstop on the ramp), the KE is not uniformly distributed along the spring. If the front of the spring is moving at speed v, the mid point will be moving at v/2, etc. In respect of the subsequent trajectory, it's the speed of the mid point that matters, i.e. it will be as though the whole spring is moving at v/2.
If the spring has length L and mass m, the total KE in the spring at this point is therefore $\frac 1{2L^3} m \int _{x=0}^{L}(xv)^2.dx = \frac 16 mv^2$, while the effective KE for the flight is $\frac 12 m (\frac v2)^2 = \frac 18 mv^2$. I.e. only 3/4 of the original PE is converted into useful KE. The rest goes into oscillations in the spring.
If it is a matter of stretching the spring, a corresponding issue arises, though the details may look different.

First of all, yes the spring is the only object and there is no other mass involved. second of all, yeah what your saying makes absolute sense. As i said before this was an experiment and i have to write a lab report for it including the errors in the experiment and the loss of energy. Do you think i should add this? if so, what other things i should add for errors as well as loss of energy? I also want to thanks for all of your responses

 

[Siavash]

No, this isn't about the flight phase. The trajectory calculation done for that is completely correct and deals with kinetic and potential energy perfectly well.

It's about the stretching phase only. From letting go to to when the center of mass is at 0.48 m the spring stretches from "compressed by x" to "uncompressed length". That means its center of mass will move upwards by $x\sin(30^\circ)$. An energy of $mgx\sin(30^\circ)$ is needed to achieve that, and that energy can only come from the spring compression energy.

Now, depending on the confuguration at launch, that may require a correction. The deciding criterion is: was the top end of the compressed spring at y=0.48 m or was the top end of the uncompressed spring at 0.48 m (so before the compression).

If I'm not mistaken, for the latter case that is a considerable extra compression required. In fact so much that that extra compression also justifies a correction.
And then your answer will deviate even more from the value that gave a hit.

I don't think you have provided enough information for your kind helpers to make that choice, so it's up to you.

-----

Next issue: the required accuracy. If your drawing is to scale, the target distance is (4 $\pm$ 0.4) m. So you can check if that can distinguish between a correct calculation (yours) or an incorrect one (the website expression you used, which, as haru made clear in post #18, isn't really appropriate). Perhaps you've just been lucky !

For the last thing that you mentioned, yeah i think i got lucky because the difference between the answers that the equations give are not that much and probably one of my friends stretched it a bit more just to account for the energy loss at the time of the experiment
Now for ur point, the top end of the spring was at 0.48 m when not stretched. And i personally don't think it makes a considerable difference ( Do you think it does?) but it was a very good point that you mentioned. Thanks for all of your responses ! I really appreciate it

 

[haruspex]

Do you think i should add this?

It would be an excellent thing to add, but had I thought of your doing that I would not have laid it out so completely. Instead, I would have nudged you along the path to figuring this out for yourself. A compromise would be to include it but acknowledge the assistance you had.

 

[BvU]

And i personally don't think it makes a considerable difference ( Do you think it does?)

Do the calculation, then you'll know.

 

[Siavash]

Do the calculation, then you'll know.

I did and it doesn't !

 

[BvU]

Strange, I did and it did. Did I make a mistake ?

 

[Siavash]

It would be an excellent thing to add, but had I thought of your doing that I would not have laid it out so completely. Instead, I would have nudged you along the path to figuring this out for yourself. A compromise would be to include it but acknowledge the assistance you had.

for sure !

 

[BvU]

My compliments to teacher for coming up with such a simple but physicswise rich experiment !