# Springlauncher And Projectile Motion

For my grade 12 physics course I have to make a spring launcher. The purpose is to aim and shoot a spring with the use of a spring launcher into a bucket. I have a spring with the k value of 28.57. I found out the k value using k=f/x. The height of the bucket is 40 cm. We can set the angle on the spring launcher. The distance of the spring launcher to the bucket will be decided on the day of the experiment. Using an equation (that I have no idea wat it is ) We have to figure out x (how far back we have to pull the spring).

So basically the distance of the springlauncher to the bucket, the angle, and x are unknown. However, on the day of the experiment, we will be given the distance to the bucket from the springlauncher.

Any help will be greatly appreciated.

basically the only equation i know of now is Fa=kx

THANK U

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^ I think you mean F=kx

unless a is some subscript your class uses.

think about the potential energy stored in the springs ;)

out of curiosity is your class calculus based?

Hey CPL.Luke,
yes a is a subscript our class uses. I think it stands for applied force. And no it isnt calculus based. Im so confuseeed. what about the potential energy stored in the springs. I am not sure what equation to used.

hmm, you learned about energy correct? and more specifically conseration of energy.

the relevant formulas are

PE=1/2kx^2

KE= 1/2mv^2

v=at
x=vt
x=1/2at^2

you need to think abuot how the energy gets transfered around in the problem, if the spring launcher is compressed than, then it stores energy, and when the spring is released it will release its energy into the ball that its launching in the form of kinetic energy.

now think about what formulas to use.

Haywire
Shouldn't you consider also gravitational potential energy?

Astronuc
Staff Emeritus
Shouldn't you consider also gravitational potential energy?
Certainly if the spring is oriented at other than horizontal. However the change in GPE over the displacement of the spring maybe negligible at small angles above horizontal, but then one would need to consider friction.

Haywire
Certainly if the spring is oriented at other than horizontal. However the change in GPE over the displacement of the spring maybe negligible at small angles above horizontal, but then one would need to consider friction.
Internal friction in the spring? Isn't it really neglegible?

Astronuc
Staff Emeritus
Internal friction in the spring? Isn't it really neglegible?
Internal friction in the spring is likely negligible. I was thing of whatever guide channel or tube might be used to direct the mass at the end of the spring, although one may not need one.

The only other equation we learnt in the course that can be applied to this is the one for elastic potential energy. The equation is E=1/2kx^2. By the way there is no mass at the end of the spring. The entire spring must be launched into the bucket.

By the way there is no mass at the end of the spring. The entire spring must be launched into the bucket.
So basically there's some device that you put your spring in, compress your spring by a certain distance, release, and watch the spring go?

The only other equation we learnt in the course that can be applied to this is the one for elastic potential energy. The equation is E=1/2kx^2.
Well, yes, the only other equation that can be applied to the compression of the spring. But when you stop squishing it, it is supposed to fly away, no? And become a projectile... which has a velocity and all sorts of fun...

So basically there's some device that you put your spring in, compress your spring by a certain distance, release, and watch the spring go?

Well, yes, the only other equation that can be applied to the compression of the spring. But when you stop squishing it, it is supposed to fly away, no? And become a projectile... which has a velocity and all sorts of fun...
Yes you hook the spring onto a device and then you pull it back a certain distance (x) and then let it go in order to launch it into the bucket. The purpose of the experiment is to find out how far back we have to pull the spriing (which is the unknown x) in order for it to travel the distance that we are given. Also, we must determine what angle we have to have it so the spring goes into the bucket.

Physics isnt exactly my strongest subject so this may take me a while :S
So basically I need to come up with a formula to figure out x (the distance I have to pull the spring back) and the angle at which I have to position the spring launcher in order for it to land in the bucket.

I know that the total energy in the spring is equal to the kinetic energy and the potential energy in the spring : Etotal= Ek + Ep

PE=1/2kx^2

KE= 1/2mv^2

I DONT KNOW WHERE TO GO FROM THERE :S

If there is no difference in height between the launcher and the bucket, I would start by looking at the range formula. You'll be given the required range the day of, so you'll need to manipulate initial velocity (which will be derived from the energy equations) and angle.

R=vi^2sin2theta/g

I now have this equation as well.

cristo
Staff Emeritus
Physics isnt exactly my strongest subject so this may take me a while :S
So basically I need to come up with a formula to figure out x (the distance I have to pull the spring back) and the angle at which I have to position the spring launcher in order for it to land in the bucket.

I know that the total energy in the spring is equal to the kinetic energy and the potential energy in the spring : Etotal= Ek + Ep

PE=1/2kx^2

KE= 1/2mv^2

I DONT KNOW WHERE TO GO FROM THERE :S
I would use the kinematic equations to obtain the velocity with which the spring must be lanuched in order for it to hit the bucket. (i.e. consider the projectile motion of the spring). Then you can use the conservation of energy equation, namely that the energy of the spring whilst compressed must equal the kinetic energy of the spring just after it is released, in order to calculate the distance the spring needs to be pulled back, by substituting the velocity obtained into this equation.

I would suggest you start by modelling the spring as a projectile and obtain the intial velocity.

Edit: If this equation is correct
R=vi^2sin2theta/g
(I've not checked it) then use this to calculate vi and then substitute into the energy equation.

Last edited:
R=vi^2sin2theta/g

I now have this equation as well.
Note that this equation is maximized when the firing angle is ...? (assuming, of course, no air resistance. I believe the maximizing angle is about five degrees larger when air resistance is not negligable.)

Now all you need is to solve for initial velocity (which would depend, of course, on how far back you pull your spring).