- #1

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My question is how do I calibrate this properly? Our teacher said we need to use projectile motion equations and energy equations. I am not entirely sure how to do this so can someone help me please?

- Thread starter muddy_waters
- Start date

- #1

- 14

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My question is how do I calibrate this properly? Our teacher said we need to use projectile motion equations and energy equations. I am not entirely sure how to do this so can someone help me please?

- #2

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- #3

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I thought of that, but the question arises as to how to fire the marble at that specific velocity?

- #4

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Then you can calulate energies set the zero-point for potential energy when the catapult is loaded and then use energy conservation:

[itex] E_{initial} = 1/2kx^2\\

E_{final} = mgy + 1/2mv^2 = 1/2kx^2\\

1/2mv^2 = 1/2kx^2 -mgy \\

v = \sqrt{\dfrac{k}{m}x^2-2gy}[/itex]

- #5

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Elastic energy = gravitational potential + kinetic energy

Ee = Eg + Ek

If so how would I calculate the elastic constant for my specific bungee cord?

- #6

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- #7

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[itex] mg - k\Delta x = 0 \\

k = mg/\Delta x[/itex]

So you would need to measure the initial length of the bungee.

- #8

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ΔDx = (V1sin2θ)/g

- #9

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[itex]

x = v_0 \cos(\theta)t \\

y = y_0 + \sin(\theta) t -1/2gt^2

[/itex]

Where y0 is the height of the catapult just before shooting. Now the task is to eliminate the time from the above i.e. isolate the time in x and insert into y. then you would solve your quadratic equation for x with y = 0. Can you see why?

- #10

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Also in your derived equation, the velocity you solved for is initial right? Just making sure

- #11

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Sorry I don't follow, could you explain why I need the y as well as the x?

- #12

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[itex] y = y_0 + \sin(\theta) v_0 t -1/2gt^2[/itex]

But yeah that is right the v0 is the velocity from #4.

Edit: You need the time to be eliminated, because you don't know how many seconds your marble will fly.

So the only way is to define the ground as y=0 and therefore the y is needed aswell as the x.

- #13

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You dont need the y it is only interesting when set to zero.

- #14

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- #15

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Edit: With your suggestion the y_0 should be zero, and we would get the same result.

- #16

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y = -0.48 + xsinθ/vcosθ - 1/2g(x/vcosθ)^2

Is this correct? Now all I have to do is set the equation to zero, and plug in the x which is the given distance, and solve for v. Then from there I work backwards and use the energy eq'ns to solve for how much to pull it back?

Also is θ the angle at which the arm is to the ground?

- #17

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Make a sketch with your definition of the coordinate system and see.

the theta you got right.

Hmm you could do that, or you could derive a equation which only need to take the desired firing range x? It could be done from our discussion and eq.s.

But i get the same eq. except from the sign on y0

- #18

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When doing that you would insert the v from #4 as v0 and arrive at the following result:

[itex] 0 = y_0 + \dfrac{\sin(\theta)}{\cos(\theta)} x -1/2 g x^2\dfrac{1}{\cos(\theta)^2 (\dfrac{k}{m}\Delta x^2-2gy)}[/itex]

Which you would need to solve for delta x :)

Edit: That is good physics: Obtain a finished equation and plug in your numbers at last then you could also optimize it by varying the catapult height and so on.

EditEdit: The y is now the catapult height. sorry.

[itex] 0 = y_0 + \dfrac{\sin(\theta)}{\cos(\theta)} x -1/2 g x^2\dfrac{1}{\cos(\theta)^2 (\dfrac{k}{m}\Delta x^2-2gy)}[/itex]

Which you would need to solve for delta x :)

Edit: That is good physics: Obtain a finished equation and plug in your numbers at last then you could also optimize it by varying the catapult height and so on.

EditEdit: The y is now the catapult height. sorry.

Last edited:

- #19

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[itex] \Delta x = \sqrt{\dfrac{gx^2m}{2k(y_0+\tan(\theta ) x)\cos(\theta )^2}+\dfrac{2gym}{k}}[/itex]

- #20

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- #21

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- #22

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- #23

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Total need to be stretched and the x is the total fired dist.

- #24

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Oh my God thank you so much. I really appreciate your help! :)

- #25

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Np. but the y and y0 are the same i just realized, and i like the problem

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