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PeterO
#2
Nov27-11, 03:21 AM
HW Helper
P: 2,316
Quote Quote by ictsyd View Post
The question to the lab is How can the spring constant and muzzle velocity for a toy ball launcher be determined through physical measurements and energy conservation considerations?
My lab group and I took our pastic ball launchers(pretty much toy guns) and shot them straight up and measured the hieght it went. We did three trails and ended with an average hight of 2.07 m. Then we were a little confused on where to go from there.. We also know the mass: 1.5 g or .015 kg



We could find the PE of gravity from what we had so we started with that. PE=mgh. We got .305 J .. then what?
Other equations that I've seen are
KE= 1/2 mv^2 ... but how do I find the velocity?
I tried to find the velocity by doing h= V^2/2g, getting 6.373 m/s
I also have PE= 1/2Kx^2 written down, but I'm not sure what this gives me.
When I plug The PE into that equation, I get Kx= .781. What exactly is that, and what importance does it have. What do I do withit next?
Also, is there a specific equation to find the spring constand and or muzzle velocity.
Did we forget to find some important data or something?
Or do we have everything and I'm just not using the right equations?
I feel stuck. Help? Thanks!
OK, you have lots of great information and forumlae - you need direction - an overview.

When the spring in the toy ball launcher is compressed, we [theoretically] store an amount 1/2.k.x2 of energy in that spring. k is the spring constant, and x is the amount by which the spring is compressed [in metres of course]

When the launcher is "released" some of that energy is converted to kinetic energy of the ball - let's suppose all of it; I will explain later.
The ball now how kinetic energy given by 1/2.m.v2 m = mass, v = vel or speed

If that ball is released vertically in the upward direction, then as the ball rises, all that kinetic energy is converted to Gravitational Potential energy; given by m.g.h where these letters stand for the usual things.

You have most/all measurements to calculate any of those quantities and get what you want.

I hope you see that 1/2.k.x2 is effectively transformed into m.g.h

The real problem is - how efficient is the transformation of the energy from the spring.
Firstly:
If you were to compress the spring many times, with the release mechanism released, the spring will get warm. That means that during the compression/expansion of the spring some energy is converted to heat - probably not enough to worry about in a single "shot"
Secondly - and perhaps more importantly.
The spring itself has mass.
When released, one end remains stationary [against the "gun] while the other end accelerates to the same speed as the ball. If the mass of the spring is similar to the mass of the balls [or even worse; more] then the energy stored in the spring will be "shared" between the projected ball and the spring. Just how much of the stored energy is actually transferred to the ball? [I don't have an answer to that final question - it depends on the relative mass of the ball and spring]