Spring-loaded Toy Gun, with Friction

[Mudbird]

The answers to these problems are in my book, but I need a little help understanding just how they fit together. The spring is compressed 5 cm from equilibrium, and the projectile travels 15 cm from the compressed point to the opening of the gun barrel. They want the velocity at exit (which I got right), and the MAXIMUM velocity, which apparently occurs just BEFORE, not AT, spring equilibrium or x=0.

 

Variables:

kinetic friction = .032 N (constant for length of gun barrel, 15 cm)

mass of projectile = 5.30 g = .0053 kg

spring constant = 8.00 N/m

starting point x = -5.0 cm = -.05 m

ending point xf = 10.0 cm = .10 m = (xi + .150 m)

 

I need to find the x-value where velocity is maximized, and the max velocity itself.

 

I would have thought that was just where the ball separates from the spring which has been pushing it - at that point the only force in effect would be friction within the gun barrel, slowing it down from there on. (We're not looking at what happens after the ball exits the barrel.)

 

For exit velocity I got v^2 = [2(.032N)(.150m)]/.0053kg = 1.9622 (m/s)^2

Taking the square root of that I got v = 1.40 m/s, which agrees with the book.

 

Now for max velocity I assumed that the spring would be pushing the ball right up to x=0, at which point the ball would begin slowing down due to friction. That is, the ball would have traveled 5.00 cm to reach max v, which I then calculated as 1.78 m/s. However, the book says max v = 1.79 m/s and occurs after 4.60 cm of travel, or at x = -.40 cm.

 

All I can figure is that the spring's force is diminishing as it gets closer to 0, so the spring itself is slowing down while the ball is speeding up, and that the ball actually leaves the spring behind just before reaching x = 0. I reasoned that acceleration of the ball from the spring then ceases (drops to zero) and since acceleration is the derivative of velocity I should be able to find a max value of v by setting a=0. It's at this point that my brain freezes up and I stare at the wall.

 

A) Am I on the right track, or is something else going on here?

 

B) How do I set up the equation for finding the maximum velocity?

 

Thanks.

[Mr Skeptic]

At what point does the force the spring is applying become less than the force of friction on the projectile?

[Mudbird]

Okay, thank you... I actually had arrived as far as setting 1/2kx^2 = f(k)x, but when the distance traveled worked out to 4.2 cm (or x= -.008) instead of 4.6 (or x=-.004) I thought I must be doing something wrong and started over. I've worked it through this time.

 

Taking the initial spring potential energy 1/2k(.05m)^2 as the total energy of the system, I then subtracted f(k)(.008m) from that to get 1/2mv^2 = .008656. This yields a velocity of 1.807 m/s after the spring has moved 4.20 cm.

 

Once again, the book says the max velocity should be 1.79 m/s, at a distance of 4.60 cm from the firing point, or -.4 cm from equilibrium.

 

Was I supposed to divide that .008 by two somewhere along the line?

 

Thanks for taking the time to help me.

[swansont]

All I can figure is that the spring's force is diminishing as it gets closer to 0, so the spring itself is slowing down while the ball is speeding up

 

You need to rethink this. Diminishing force means accelerating at a smaller rate, but it still means accelerating. Slowing down implies an acceleration of the opposite sign.

[Mudbird]

I'm sorry, I just feel so stupid about this. I've covered pages with calculations and wasted hours that I needed for even harder homework. Can someone please patiently explain to me where, in the spring's path, I should expect to find the highest velocity. Near or at the starting point (-5cm)? Near zero (equilibrium)? Past zero? Surely the spring must be slowing down by then?

 

Presumably the ball moves at the same speed as the spring as long as they are in contact (there will be some deformation of the ball, which is described as soft rubber, but we're not asked to calculate that). Is the speed of the ball increasing or decreasing while it is in contact with the spring? At some point the spring reverses direction while the ball keeps going forward. To me it seems logical that, before this happens, there would be a gradual lessening of the forces exerted by the ball and spring against each other. There must be a point, BEFORE the spring bounces back, when the spring's effect on the ball becomes negligible, and I would guess that that is the point when the ball's speed is highest, as friction will slow it down thereafter.

 

The maximum speed according to my book is about 26% higher than the speed at exit, and occurs before the ball passes x=0.

 

What I can't figure out is HOW to find out where along the gun barrel this maximum speed occurs. Or, conversely, how to find the speed itself so I can work out the distance. Do I need the derivative of acceleration? Can someone please tell me what's happening here? Thank you so much.

[Mr Skeptic]

I assume you don't know calculus. It would be convenient for this problem but you can do it with constant force and knowledge of potential energy instead.

 

The force of a spring is -kx where x is the displacement from equilibrium and k is the spring constant. The friction of the barrel is constant, so at the start you have both the force of the spring and the friction acting in opposition. The energy loss from friction is F*d. The potential energy of the spring is (1/2)kx^2.

 

First, find the displacement at which the force of the spring equals the force of friction. Next, take the starting potential energy and subtract the potential energy at this point, and the losses from friction at this point. That's your maximum kinetic energy, from which you get the max speed. Then add to this the remainder of the potential energy and subtract the remainder of the friction losses (or subtract the whole friction loss from the whole initial potential energy), and that gives you the kinetic energy at the exit.

[Mudbird]

Mr. Skeptic, you are a lifesaver.

 

So all I had to do was set -kx = f(k) itself, without multiplying or adding anything else. Where I went wrong was in confusing the Spring Force with the Spring Potential Energy, and the Friction Force with the "work" done by friction.

 

The problem was enormously more simple than I thought. It took me a minute and a half once I got the right variables in place.

 

Thank you ever so much for explaining step-by-step, in complete sentences that my word-oriented brain can follow. Your assumption that I don't know calculus is half right: I've taken it over and over but have yet to master it. I wish there were more explication in the physics class of EXACTLY how the concepts of calculus apply to the formulas we are memorizing. I try to take apart all the units and quantities to see what makes them tick, so to speak, but I can't always put them back together right.

 

Thanks again for your help.

 

When you have a chance, would you mind posting the convenient, calculus-based way of solving this type of problem? If it's not too much trouble.

 

Cheers,

E.R. "Mudbird"

[Mr Skeptic]

Well if you know calculus, you don't need to remember how to get energy from the force for each particular circumstance. You just need [math]W = \int_{x_0}^{x_f}F \cdot dr[/math]. That will give you the energy regardless of what the force is or how it changes. For the above problem, you are essentially given the result of the integration for the force of friction and the spring force.

Edited April 6, 2010 by Mr Skeptic