# Spring Gun

1. Mar 7, 2014

### Steven60

I have a question about a spring gun. Suppose the barrel of a spring gun is placed horizontally at the edge of a horizontal table. You put say a marble in the barrel and compress the spring x cm and after releasing the marble it travels a horizontal distance of y cm before hitting the floor (so motion is of a projectile). My question is whether or not the horizontal distance traveled and the amount the spring is compresses make a linear relationship? If so, then how can I prove this? This is not homework.
Thanks!

2. Mar 7, 2014

### Windadct

Seems to be purely a math problem. Perhaps sketch the system and write the relevant equations needed to determine this?

3. Mar 7, 2014

### UltrafastPED

This is a nice physics exercise - there are several physical considerations, and then some simple math.

You have two forces acting on the marble ... the spring force, which launches the marble, and gravity.

Once the marble leaves the launch tube it will have a constant "horizontal" speed - ignoring air resistance - and an initial vertical speed of zero. Call this initial horizontal speed V.

The vertical speed will increase with time due to the constant gravitational acceleration - and will hit the floor at a definite time which depends only on the height of the table. Call this duration T.

Then the distance from the table to the point of contact will be D = V x T.

The time T does not depend upon the spring force, only on the height of the table and local value g=9.8 m/s^2.

Thus you only need to determine if the speed V is proportional to the spring force; by Hook's law we know that a "good" spring obeys F = -k * X, where X is the compression/extension distance and k is the spring's constant.

If we switch to energy we have work done on marble is W = Integral[F dx] over the interval x=[0,X]. Note that the force is changing as the spring moves! So W = Integral[ k*x dx] = 1/2 k*X^2.

But this work has been converted into kinetic energy of the marble. For a marble of mass=M, and given that it is NOT rolling or spinning, then the kinetic energy is KE=1/2 M*V^2 = 1/2 k*X^2=W.

Thus V = k/M Sqrt[X]. xx Correction: xxx Make that V = Sqrt[k/M] * X.

Thus the hypothesis is true!

Thanks to dauto for noticing the mistake at the end! :-)

Last edited: Mar 7, 2014
4. Mar 7, 2014

### DrewD

The equations involved will be $d=vt$ for constant $v$ (and assuming that the initial point when exiting the spring gun is defined as 0 distance), $U_{spring}=\frac12k\Delta x^2$ and $K=\frac12mv^2$. $\Delta x$ is the amount the spring is compressed, and $v$ is the velocity of the object as it leaves the spring. This approximation assumes that the object does not stick which is a good assumption for a spring gun. Solve for $v$ to find the relationship between $\Delta x$ and $d$.

5. Mar 7, 2014

### dauto

You made a mistake at the very end. In fact, after correcting the mistake, you proved that the hypothesis is true.

6. Mar 9, 2014

### Steven60

Thanks for your replys. I actually worked this out myself and actually did the same exact steps as UltrafastPED.

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