# Conservation of Energy Question

1. Dec 30, 2005

### Rock00

Question is:
------------

Two children are playing a game in which they try to hit a small box on the floor with a marble fired from a spring loaded gun that is mounted on a table. The target box is 2.20 m horizontallyl from the edge of the table. Bobby compresses the spring 1.10 cm, but the marble falls 27.0 cm short. How far shuld Rhoda compress the spring to score a hit?

--------

Diagram:

_______ projectile |
_|_spring_|-------------___ v
| | marble<--- ---O_
| | --__
| | --__
| |__________________________|box|_
|<------------2.20 m ---------->|

Given
-----
dx = 2.20 m
Sc (spring compressed) = .011 m
(falls .27 m short)

Sc=?
(hits the targe)

so basically what should be the compression of the spring before it can hit the target?

-----------------
I didn't know how to start. I wrote down F=-ks. I figured that the initial velocity is 0. PE at start is 0. Total change in energy is 0. But I do not know what else to do. Please help!!!

2. Dec 30, 2005

### Rock00

Clear diagram of previous post!

Attached is a clearer diagram of the previous posted diagram!

#### Attached Files:

• ###### diagram.jpg
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3. Dec 30, 2005

### LeonhardEuler

Are you sure that the PE at the start is 0? What about the compression in the spring or the hieght of the object?

4. Dec 30, 2005

### Homer Simpson

with kinematics you should be able to find the horizontal velocity that got the projectile to go 220 - 27 cm. Knowing that velocity should let you solve for the spring constant. With this knowledge, find out what horizontal velocity will get the projectile to go where you want it to go. Then using spring equations again, figure out how much spring compression will make that initial velocity.

Does this seem to help?

5. Dec 30, 2005

### LeonhardEuler

However, since he does not know the hieght of the table, he will not be able to find the spring constant. It doesn't matter though. Solve for the height in terms of the spring constant and both terms will cancel out in the end.

6. Dec 30, 2005

### marlon

Ahh, this is a classic one.

You will be needing :
1) split up the problem in two parts (one part with the spring on tha table, the second part when the object has left the table and fall down because of gravity)

2) Nexton's second law (determin all the forces in the horizontal and vertical direction when the object is falling down : gravity)

3) conservation of total energy when the particle just leaves the table. This will be the "connection" between the particle just leaving the spring (calculate the energy it gets from the spring) and the motion under gravity.

good luck

marlon

edit : Both k and m will not be important for your calculation. You do not need to know them, because they will cancel out

Last edited: Dec 30, 2005
7. Dec 30, 2005

### Homer Simpson

Oh yeah, I missed that detail. That makes it much more tricky. Build a model.

8. Dec 30, 2005

### marlon

Guys, just use k and m for springconstant and mass. You will see that both quantities will cancel out (if you solve this question correctly :) ). You don't need them...

marlon

9. Dec 30, 2005

### Rock00

Hi, do you think you can show me the process of how you started the problem. I would greatly appreciate it.

Thanks

10. Dec 30, 2005

### LeonhardEuler

The way I did the problem was to write equations for the x- and y-coordinates as functions of time. Then I solved for time in the x equation, and plugged the expression I got for t into the y equation. This gives y as a function of x. The expression for x has an initial velocity, v_0 in it. Use conservation of energy to solve for v_0 in terms of k, the spring constant, m, the mass, and s, the displacement of the spring. Then it is just a matter of using the information given to solve for h in terms of k, m, and s. k and m cancel away in the end.

11. Dec 30, 2005

### marlon

First forget about the 1.73 cm. That's the solution of another question. Sorry...

well, start with conservation of energy : $$\frac{kx^2}{2} = \frac{mv^2}{2}$$ ; x is the distance the spring is compressed.

From this, you get a velocity (just bring on k and m...don't mind them) when x = 0.011m

When leaving the table, the object moves under the influence of gravity.
Applying Newton's second law gives you

$$y = \frac{gt^2}{2}$$
$$x' = vt$$ ; x' is the horizontal distance travelled by the object. You know that x' = 1.93 m if the spring is compressed 0.011 m.

From the above two equations, write y as a function of x' and v. Calculate the y-value with the 0.011m and 1.93m data.

Then, using this y-value (k and m are still not known) you can get the x-value (distance that the spring is compressed) when you know that x' = 2.2m

marlon

12. Dec 30, 2005

### marlon

Correctamundo

marlon

13. Dec 30, 2005

### Rock00

Where are you getting the 2 from and what is the t for?

is 2 part of the equation and t is for time and if t is time then how would you get time or 1.93 m.

Thank you

14. Dec 30, 2005

### Rock00

does anyone know how marlon got $$y = \frac{gt^2}{2}$$

any help appreciated

15. Dec 31, 2005

### marlon

Don't you know the equation for the vertical position of an object moving under the influence of gravity ?

In general, the formula is :

$$y = y_0 + v_{0y}t + \frac{a_y t^2}{2}$$

$$y_0$$ is the initial position ; for you it is 0

$$v_{0y}$$ is the initial velocity component in the y direction ; for you this is 0 because the v after energy conservation only has a component along the x direction.

$$a_y$$ is the acceleration in the y direction ; for gravity this is g

regards
marlon

16. Dec 31, 2005

### Rock00

Thank you marlon. I still do not know what to do after I get the Y.

I got Y=301686.116/(k*m)

Help would be appreciated.

17. Jan 2, 2006

### marlon

read my first post carefully. it's all in there

marlon

18. Jan 2, 2006

### Rock00

trust me I have been looking at it for days, but I can't seem to get an answer. I literally have five pages of work and to no avail. I need a solution. As always, appreciated.

19. Jan 2, 2006

### marlon

conservation of energy yields (for the spring pressed over 0.011m)
$$(1/2)k(0.011)^2 = (1/2)mv^2$$
$$v^2= (k/m)(0.011)^2$$

conservation of energy yields (for the spring pressed over x' m)
$$(1/2)k(x')^2 = (1/2)mv'^2$$
$$v'^2= (k/m)(x')^2$$

Newton's second law yields:
$$h = gt^2/2$$
$$x = vt$$

thus

$$h = \frac{g(1.93)^2}{2v^2}$$

and ofcourse also

$$h = \frac{g(2.2)^2}{2v'^2}$$

Calculate this h for x = 1.93m and replace the v by the first equation for the velocity coming from conservation of energy. This gives you height h.

Use this h in

$$v'^2 = \frac{g(2.2)^2}{2h}$$

Above formula will give you v' and use this v' in

$$v'^2= (k/m)(x')^2$$

to solve for x'

marlon

edit : g=9.81 m/s²

Last edited: Jan 2, 2006
20. Jan 2, 2006

### Rock00

Thank you!

Just out of curiousity, is the coefficient we get for V' the answer to X'.

I got $$V' = (.017) (sqrt (k)/sqrt(m))$$

and when I plug in for $$v'^2= (k/m)(x')^2$$

I get x=.017. Is it suppose to be that way, I am really confused. Well if it is thank you Marlon and everyone else who has posted on this topic. I couldn't have done this problem alone. I hope to back for more help. I need help in succeeding in physics. I have a passion for physics, but I am not good at it right now, but I hope to be from this forum.

Thank you again.

Last edited: Jan 2, 2006