The last time I looked at this projectile motion lab, I was confused. My different methods for measuring the launch speed of the ball were not even close to being consistent. So, I am bringing out the big guns – video. I made a video of the ball shot both horizontally off the table and vertically. No point posting the whole video (unless you really need it), but here is a screen shot of what the setup looked like.
These videos were made with my flip video camera, it doesn’t have adjustable shutter speed so that there is some blur. Also, notice the carbon paper on the floor. This is so I can also measure where it landed along the floor. Ok, but first the analysis from the video. Here is the trajectory (x vs. y) for the horizontally shot ball.
Looks parabolic. Now for the vertical motion.
From the parabolic fit, the acceleration is -9.93 m/s2. And, I can get the launch speed from horizontal motion.
I did not use the first data point to fit the function because it was a little difficult to see exactly when the ball came out of the launcher. However, the rest of the points fit fine. From the vertical motion, I assume the scaling is close enough. The slope of the linear fit gives a launch speed of 3.24 m/s. Ok, how does this agree with the carbon paper data? From that same setup, I measured the initial height at 0.849 +/- 0.005 m and a horizontal distance of 1.30 +/-0.01 m (yes, I didn’t measure this very well for lots of reasons). As I have done before, the time can be found from the y-direction – remember the initial y-velocity is zero m/s.
Now for the x-direction where the x-velocity is the initial launch velocity (and the acceleration is zero).
So, with the measurements I made, I get an initial velocity of 3.12 m/s (not worrying about uncertainty just yet). Anyway, it is in the same ball park with the other measurement. Now what about the video of the ball launching straight up? Here is the data:
I can use this data two ways. First, I can use it to just find the initial velocity from the fit of the parabolic equation. Or, I could use this to find the height of the ball. From the fitting equation, the ‘b’ parameter is the y-velocity at t = 0 seconds. That doesn’t help here since I messed up. Look at the graph. It starts at t = 12.4 seconds (t = 0 is at the start of the video where I am walking from the camera to the launcher). Ok, I can fix this. I can use that parabolic fit to get the y-velocity as a function of time by taking the derivative of position with respect to time. I get:
Where a and b are the parameters from the fit (a is not acceleration). If I put in t = 12.375 seconds, I get a y-velocity of 1.81 m/s. Oh no. This is pretty different. Ok, what about the height measurement? I eyeballed this in the last method, but now I can get something better. From the video, the ball goes 0.22 meters high. I will use the work energy principle to find the initial speed. The only thing that does work on the ball is gravity, so I can write: (note that I am using work done by gravity instead of potential energy for no real reason)
Putting in a height of 0.22 meters, I get an initial velocity of 2.08 m/s. Again, I have not looked at uncertainty yet but this is fairly close to the other value from the video.
Er? Why are they different?
Two methods for horizontal shot give about the same values and two methods for vertical shots give about the same value (but different than horizontal). The only thing I can think of to account for the different is the gravitational force on the ball while it is being shot vertically. During the “shot” there are two forces that can do work on the ball, the gravitational force from the Earth and the force from the spring. Here is a diagram.
Note that for a horizontal shot, gravity does no work since it is perpendicular to the direction of the displacement (also, the floor of the launcher pushes up on the ball and does no work). Looking at just the ball and the spring, I can write down the work done. I am looking at the spring as part of the system since that is a non constant force. This will let me use the spring potential energy.
Here, s is the amount the spring is compressed. I had been going under the assumption that the work done by gravity over this short distance didn’t really matter, but clearly it does. What would the spring constant have to be to give me these different values for the initial velocity? I will call the one vh for the initial horizontal velocity and vv for the vertical velocity. Here is the same expression for horizontal velocity (in terms of m, k and s):
Let me take the difference in the square of the velocities (vh2 – vv2):
I can easily measure the mass of the ball. This will give me a value for s that I can calculate and compare to reality. The mass of the ball is about 16 grams. This would give a spring compression of:
No way. I tried to measure the compression of the spring and I get somewhere around 0.035 meters. I can only think of one reason (well, two if I count the possibility that I screwed up somewhere). Maybe there is a significant mass on the end of that spring. This would mean that the spring has to accelerate both the ball and the mass and that I would need to consider work done on the extra mass in the vertical case (but not in the horizontal case). Ok. I can’t stop now. I am going to get a rough value for the spring constant.
To get a value for the spring constant, I stood the launcher on its end (so it was pointing up). I put a stick in the launcher (without the ball) and added masses on top. I recorded the mass as the amount it was compressed. Here is the data (I only collected 4 data points because I was in a hurry to find the answer).
This is a great example of why the graph is better than just one data point. What if there is mass on the end of that spring (that is hidden)? The graph and the slope don’t change because there is some hidden mass (well, the graph might, but not the slope). Anyway, this data shows that 1/k = 0.005 m/N or k = 200 N/m.
So, how far would I have to compress this to shoot the ball horizontally with a speed of 3 m/s?
I measured a compression of 0.036 meters. What if there is an “extra mass” in there? I can solve for this using the measured compression and mass of the ball.
With this, I get an “extra mass” of 0.7 kg. That seems really high. I really don’t know what is going on here. These are my final thoughts (still need to look at this more)
- Maybe there is ‘extra mass’ on the spring or even the mass of the spring is important
- Maybe there is some significant friction force