# Homework Help: Ball fired out of a spring gun differential equation

1. Nov 19, 2011

### Catalytical

1. The problem statement, all variables and given/known data

You're firing a ball of mass 10.0g out of a spring gun, and the spring is the entire length of the barrel, which is .05m long. The force constant for the spring is 500 N/m. There is a constant resisting force of 10.0N acting on the ball as it travels down the length of the barrel. What is the speed of the ball as it leaves the gun?

2. Relevant equations

mx'' + cx' + kx = 0
Work total = ΔK
Work = F*d
K = 1/2mv^2
Potential energy of a spring = 1/2kx^2

3. The attempt at a solution

So basically, I'm wondering if it's possible to solve this problem using the above differential equation. I know that the cx' term refers to the damping force, which I assume to be the resisting force. So I tried my equation to be:
.01x'' + 10 + 500x = 0
I let time t= 0 correspond to when the spring has been fully compressed, so x(0) = -.05 (I let x = 0 be at the end of the barrel), and the block is not moving when the spring is compressed, so x'(0) = 0.
I then solve the differential equation, and find the time it takes for the spring to reach it's uncompressed distance, when x = 0.
After I find the time taken, I plug that into x'(t) to find the velocity of the block when it leaves the spring. However, the answer I get from doing this doesn't match the answer I get when using conservation of energy techniques.

For conservation of energy, I said that:
Work total = ΔK, where the total work done is due to the resisting force, and the spring's potential energy.
(1/2)(500)(.05)^2 - (10)(.05) = 1/2mv^2
solved for v, and got 5 m/s.

So, unless I'm doing something wrong in my energy approach, the answer should be 5m/s.
Basically what I'm asking is, is it possible to model this situation with the mass-spring differential with the information given, and if so, what would be the proper way to handle the cx' term?

2. Nov 20, 2011

### cepheid

Staff Emeritus
You realize that x' is the velocity of the object, right? So the cx' term would be correct if the resisting force were proportional to the object velocity. But you're told explicitly in the problem that the resisting force is CONSTANT. So the correct term in the equation would just be 'c', where c = 10 N.

You don't try and apply some canned equation of motion to describe a system, and hope that it's the right one. You use Newton's 2nd law to derive the equation of motion: In this case:

Fnet = ƩF = ma

Fspring + Fresistance = ma

-kx - c = mx''

mx'' + kx + c = 0

...or something. Make sure you get the signs right. The "x" in kx is the displacement of the spring from its equilibrium position, which may or may not be the same thing as the position of the mass.

3. Nov 20, 2011

### Catalytical

Thanks. It actually turns out that my equation was right, but I had found the wrong root using my calculator, so I was getting the wrong answer. I Feel kind of stupid =P

4. Nov 20, 2011

### cepheid

Staff Emeritus
No, I mean, your starting differential equation was wrong for the reasons I stated. But then you plugged in a constant of "10" for the cx' term (completely ignoring the x' part) thereby turning it into the correct equation. You see why this makes no sense, right?

5. Nov 20, 2011

### Catalytical

I guess it makes it seem like I had some faulty logic going on.
I knew that cx' was equal to the damping force, and that the damping force was 10N, so I just substituted 10N for cx' in that equation.

But yeah, I probably should have tried deriving the equations first, instead of just blindly guessing.

6. Nov 20, 2011

### cepheid

Staff Emeritus
It seems like you're still not getting it. x' is the velocity. cx' is ONLY an accurate model for the damping force IF the damping force is proportional to the velocity. In this case, the damping force is NOT proportional to the velocity, because it is CONSTANT. So the damping term is NOT of the form cx' in the first place (and it is WRONG to even write it as cx' initially). "Substituting 10 N for cx' " is precisely the step that I was saying was completely nonsensical. When you did that, did you not ask yourself, "what happened to the x' ???"