Investigating Unexpected Results in Elementary Experiment

[nomadreid]

Any chance we can get a picture of the student's actual slingshot?

 

[nomadreid]

By integrating the actual data up to a displacement of 3 cm, I get a stored elastic energy of 0.231 J and a launch velocity of 4.45 m/s.

The horizontal distance that the projectile went was 116 cm when fired at an angle of 26 degrees to the horizontal. If I have the formula right(?) that the range = (v*sin2A)/g, this works backwards to 14 m/s. Am I doing something wrong here?

 

[nomadreid]

I will add to the discussion my own experience when I tried to develop an experiment similar to the one described here. I didn't get very far and eventually I abandoned the idea primarily because the rubber bands I tried could not be trusted to provide the same force for the same displacement reproducibly. The cause for this lack of reproducibility was hysteresis. The force that the stretched band exerted depended on what had been done to it prior to the stretching.

A simple experiment to investigate this was a "round-trip stretching" as opposed to the "one-way" measurement described here. I suspended a series of weights of increasing magnitudes and measured the displacement then reversed the order, reducing the hanging weights and remeasuring the distances. I found that the displacements for a given weight on the way up were lower than those on the way down. Furthermore, the results from the first round-trip set were not reproduced when I repeated with a second set using the same weights and procedure. I quit when I found out that the results also varied depending on how long I waited before starting the "return" trip with the largest mass stretching the band all the while.

My point here is not to criticize this experiment but to offer a word of caution. Know thy band before comparing experimental results against an analysis that is based on Hooke's Law with a reproducible constant $k$.

Thanks, kuruman. I shall have the student do some more tests on the rubber band.

 

[haruspex]

The horizontal distance that the projectile went was 116 cm when fired at an angle of 26 degrees to the horizontal. If I have the formula right(?) that the range = (v*sin2A)/g, this works backwards to 14 m/s. Am I doing something wrong here?

You mean (v²*sin2A)/g. Also, it looks like at release the projectile is about 2.7cm above the ground, which should add about 5cm to the range.
Feeding 111cm into the formula I get 3.7m/s.
14m/s is 50kph.

 

[erobz]

Ok, so this is where the actual geometry of the slingshot could come into play. The scale on the launch pad indicates they are measuring the distance $s$, when we need $d = x-l_o$ for our formula.

Using Pythagorean theorem, that implies that:

$$ d = x- l_o = \sqrt{ \left( s + \sqrt{l_o^2 - \frac{w^2}{4} } \right)^2 + \frac{w^2}{4} } - l_o $$

We can see by using that particular metric ($s$) slingshot geometry dependence has been introduced.

In general by the triangle inequality, $s$ should overestimate $d$.

 

[nomadreid]

You mean (v2*sin2A)/g.

oops, yes. Apologies, .

Also, it looks like at release the projectile is about 2.7cm above the ground, which should add about 5cm to the range.
Feeding 111cm into the formula I get 3.7m/s.

OK, that makes a lot more sense now. Thanks very much, haruspex.

Ok, so this is where the actual geometry of the slingshot could come into play.

Thanks again, erobz. I presume it is OK for me to use your diagram (but with numbers, since an equation with that many variables is a bit much for a 13-year old) in explaining some of these ideas in my next lesson with the student?