Video Analysis Blog Post

Video Analysis

In this experiment, our group took a video and found items of interest about it. Our video, which is down below, show us throwing a ball up into the air and back down to get a nice curve, with some bounces at the end.

Video of experiment:

https://youtu.be/GBYTJpy1Bzs

Video Marks:

As you can see, our marks show our first, very nicely arching path that the ball travelled.  Our meter stick fell off at the very end of the experiment, but our meter mark was based on that. 

X versus T Graph:

Our x versus t graph told our group the change in position over time in both the vertical and horizontal directions. The line that arcs before falling again is the change in position in the vertical direction; this is because the ball has an acceleration of -10 m/s^2, which means it's position versus time isn't constant. However, in the horizontal direction, the ball has a constant velocity, which gives us a straight line on our graph. 

V versus T Graph:

As with our X versus T graph, the two different lines in our velocity versus time graph represent both the velocity in the vertical and horizontal directions. The yellow line represents the vertical velocity of the ball, while the red line shows us the velocity in the horizontal direction. Due to unforeseen problems with our video camera angle, our velocities are not 100 percent correct. Our vertical velocity should be closer to -10, because the ball's vertical acceleration is -10 m/s^2. Also, our horizontal velocity goes down slightly, which suggests that an unknown force pushed our ball in the opposite direction of travel. I think that it may have been caused by wind resistance, as we performed our experiment in a corridor with wind being forced through it. If we could retake our data, we would probably use a different ball, preferably one with a solid core to render wind resistance as non-factorable. 

Analysis:

A.) Our acceleration in the vertical direction was -3.82 m/s^2. It's negative because the only force acting on our ball in the vertical direction was the force of gravity.

B.) Our acceleration in the horizontal direction should have been 0 m/s^2, as it should have kept a constant velocity. However, because of an unknown force, our acceleration in the horizontal direction ended up being -.257 m/s^2.

C. and D.) The initial velocity in the horizontal AND vertical directions were both 0 m/s. This is because we started our ball from rest, without a initial velocity. 

E.) The velocity of our ball in the horizontal direction at the top of the path was a roughly constant 1.4 m/s. Even though the ball is slowing down by the top of the vertical path, there are no forces in the horizontal direction that are (supposed to be) slowing our ball down. 

F.) Our ball was traveling at 0 m/s in the vertical direction at the top of it's path, because it has to change direction and start going back down, which would change it'g velocity to a negative. 

G.) The final velocity in the horizontal direction for our ball should just be the same as it's start, which was 1.4 m/s. This is because the velocity in the horizontal direction is constant. 

H.) The final velocity of our ball in the vertical direction before it hit the ground was -3 m/s. Because the ball has an acceleration in the vertical direction, it's velocity increases by -10 m/s^2 every single second. Also, I knew it was supposed to be negative because the ball is falling back to the ground.

I.) By looking on our position versus time graph, I was able to find how high our ball went, which was roughly 3.2 meters above the ground. If I wanted to solve for this answer, I could use the equation (change in)position= Xfinal - X initial. This would ultimately give me the distance (vertically) that our ball travelled.

J.) To find the distance our ball travelled in the horizontal direction, I simply looked at our position versus time graph and found the point in which the yellow line(horizontal) matched up with the vertical (green) line. This gave me the distance the ball travelled when it was in flight, which came out to be 2.5 meters.

K.) By finding the point on our graph in which the top of our position versus time in the vertical direction was over the time, I found that it took .8 secs for our ball to get to the top of it's path.

L.) To find the total amount of time that our ball was in the air, we matched the time with when our ball finally touched the ground after one bounce. This turned out to be 2.1 secs.

 Conclusions

     In this situation, there is only an acceleration in the vertical direction because there is a unbalanced force in the vertical axis. Because the horizontal forces were balanced (nonexistent), that meant there was an acceleration of 0 m/s^2 in the horizontal direction. However, this could have been easily switched the other way around in a different situation. If our group had put a ball on a flat surface with no friction and pushed it at a constant pace, it's acceleration would've been in the horizontal direction, not the vertical. For velocity, it's quite common to see an object with a Vfinal and Vinitial in the vertical direction, while having just a constant velocity in the horizontal direction. Maybe there will be a point in the near future when this isn't the case? With regards to the velocities of our ball at the top of it's path, because the vertical and horizontal directions are split, the vertical movement has no effect on the velocity in the horizontal direction. It only has an effect in the vertical direction. This experiment, overall, gave me a deeper understanding into the understanding behind splitting the two different axises of motion, and brought together all of strategies and formulas that I've learned so far this year.    

UFPM Challenge Problem

UFPM Challenge Problem

     

     Our goal in this challenge problem was to use our learned knowledge and skills to land our weight from our Atwood machine on a cart moving at a constant velocity. However, we couldn't let the cart with our weight go, as this would've removed the 'challenge' component from our task. Below is a table with all of the variables that our group found: 

We took our data by measurements. We took our entire system (The Atwood machine) and set it on a scale to find our system mass of 1.14 kg. Our weight of the hanging force (our F pull) was very easy to find, as it told us on the actual weight. Because this was an Atwood machine, we decided to ignore friction as a factor. Using a motion sensor, we found the constant velocity of our cart, which turned out to be .31 m/s. Another thing to note is that the distance of our cart from the landing zone is actually supposed to read 1 meter, not .79 meters. The .79 meters is actually the distance of the hanging weight to our moving cart on the ground, because the distance to drop won't be to the ground; we had to include the cart's height. 

     Now that we collected all of the data we could without equations, all that was left to do was to find the acceleration of our system and the time the weight would take to hit the cart. We used the equation a = F net/ mass, plugging in .5 for F net and 1.14 for our mass. We ultimately came out with the acceleration of our system to be .438 m/s squared. Next, we wanted to use our acceleration we just found to find the time it would take for our weight to hit our moving cart. To do this, we plugged our acceleration into the equation (Change in) X = 1/2 a * (Change in) Time squared + v(initial)(change in Time). Here's our work:

.79 = 1/2 (.438)t^2 + 0

1.58 = .438t^2

t^2 = 3.607

t = 1.899 seconds or t = 1.9 seconds

With this information, we now found the time it took our buggy cart to reach our drop point. It made it in 3.2 Seconds from 1 meter away. Because these two times didn't match, our group realized that we would have to change the distance from which to start our buggy cart to match our drop time of 1.9 seconds. Once we had our correct distance, we dropped our weight and started our cart at the same time. This led to a perfect test drop, with our weight landing directly on target (We have video evidence of our drop test, but due to complications I'm not able to put it up on this blog). Thanks to our group's knowledge of everything we've learned so far this year, we were able to put it all together to solve a challenge problem such as this one. I'm looking forward to utilizing even more of my physics skills this year!