ESTM Challenge Problem

ESTM Challenge Problem

     Our group's goal for this challenge was to predict where to compress two different springs so that two carts of equal mass come off with the exact same velocity. Below is a diagram of what our setup looked like, with the corresponding masses and spring constants for each cart/ spring. 

     At first glance, our group got a little bit confused as to which spring constant went with each cart, as our velocity sensors were hooked up to the opposite carts. Once we got through that small hiccup, we began working in ernest. 

     We reasoned that because we were working on a surface track with negligible friction, we could simply say that E (Elastic) = E (kinetic). We decided on setting a distance for one cart, the red spring cart, and working to find a distance for the second cart this way. Here is our work.

     All of our masses were taken using the scale. We have now found both of the starting points (x) to pull our carts back to get to the final velocity of .49 m/s. Once we had this prediction, we pulled our carts back and tested what the real velocity would be. Turns out, it was .55 m/s, which was definitely within ten percent error of our prediction. Overall, this was a very successful challenge problem that pulled together everything that we knew about the ESTM model so far. 

MTM Practicum Challenge

MTM Practicum Challenge

     Our goal for this challenge problem was to determine the mass of the unknown object that we received, without using a scale to do so. Using this information, we set about designing an experiment to figure out this problem. Hudson and I decided to use two carts for our data collection; cart A and cart B. We put our unknown object (a gatorade bottle) on cart A. We decided to use only one motion sensor, as it would make our life easier in the long run. We kept cart A stationary, and pushed cart B away from the motion sensor to hit, and stick to, cart A. This allowed us to keep our velocities positive, which would avoid confusion when calculating our bottle's mass. 

     

     As shown in the diagram above, the velocity of cart B before the collision was .705 m/s, with cart A at rest. We knew from prior knowledge that the empty carts have a mass of .48 kg, or about .5 kg. We also determined from testing with the motion sensor that the average velocity of both of the carts after they collided was roughly .195 m/s. 

     We now had all of the tools to make our prediction. Below is our thought process that we took to get our prediction:

     Unfortunately, upon weighing our bottle we found that it's actual mass was closer to .64 kg. This leaves us with a percent error of roughly 20%. This is probably due to the fact that we only has time to take a few data points, leaving us with not very accurate velocities in the first place. If we could repeat this experiment, I would make sure and take several more data points to make sure that our data is fully accurate. Overall, this was a great test of knowledge on the momentum transfer model.  

2D Motion Rocket Challenge

Rocket Challenge Blog

     The purpose of this challenge was to predict how far a rocket launched at a certain angle would travel in the horizontal direction. Our group decided to set our rocket at an angle of 50 degrees, which  we thought would launch reasonably far. Here is a diagram of what our setup looked like:

     Once we had raised enough air pressure via a bike pump, our rocket would launch. To maintain a constant initial velocity, we had to make sure and fit the pressure cap on very tightly each and every time. After a few tests with our rocket at a 30 degree angle (we weren't allowed to test within 10 degrees of 50 degrees), we collected enough data to start making our predictions for where our rocket would land. Coming into class the next morning we discovered that because our angle of launch was reversed, we had been calculating our predictions based on the completely opposite angle. Learning from this mistake, our group finally made progress.

     Starting with our 30 degree info, we found the V(initial x) to be 12.57 m/s using the formula v = (change in) x / time. We then went ahead, using cosine, and found the V initial to be 25.14 m/s. Because our rocket's initial velocity would always stay the same no matter the angle that we launch it at, we had found the piece of info crucial to making our prediction. Keeping 25.14 m/s as the initial velocity, we changed the angles to 50 and 40 degrees. Using cosine, we found both the V(initial x) and the V(initial y). These were 19.26 m/s and 16.16 m/s respectively. We could then use the equation (change in) x = 1/2at^2+v(initial)*t  to find time. This involved setting up the quadratic formula, which eventually gave us the answer of t = 3.232 seconds. Using this, we plugged it into the equation v = (change in) x / time, which gave us our final prediction of 62.25 meters. This meant that, based on our calculations, our rocket would land at a distance of 62.25 meters away from the starting point. This information is summed up below in the data table:

     Unfortunately for us, our group completely forgot about timing our test launch at 50 degrees. When we did launch it, it landed roughly 20 meters short of our predicted landing zone. I say 'roughly' 20 meters because we didn't measure our actual distance because there was only one measuring tool being shared among all the groups. However, using the rough 20 meters short estimate our group found our percent error to be about 47% off. My hypothesis for why this number is so high is because we might've taken inaccurate measurements of time when launching our rocket at a 30 degree angle. This would mean that all of our calculations from that point on got more skewed as we went along, eventually leading us 47% off-course. If we were to redo this experiment, we would take our data much more carefully than this time around. Overall, this was a fun way to visually play with the skills we've learned so far this year in physics.

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!

Fan Cart Post

Fan Cart Post

     Our group's goal in this lab was to find the acceleration of our cart, then use that to determine the position in which our cart would collide with another group's cart. To achieve this, we partnered with another group for the majority of the lab. 

Attached on the side is a diagram of what our cart roughly looked like. One important thing to note was that our wheels were frictionless, which meant we could ignore the force of friction when making our calculations. 

     Because each cart had different battery lives, and reacted differently in several different situations, each group got a different acceleration of their cart, which meant that each cart had a different slope on which they travel at. Attached below is the data that both our groups worked on to get.
     We made our prediction based on our calculations below. By finding the accelerations of both the different carts, we could then set them equal in the distance equation and find a point of intersection. However, during testing we discovered that our prediction was off by quite a margin, yet our entire group was puzzled to the reason why. In hindsight, I think this was caused by an error in finding the two accelerations of our carts. Moving forward in the future, our group will be more careful with how we handle our data. Overall, this lab helped me personally see how acceleration really connected with the previous things we have been learning, which really helped me draw that connection during my fall exam.

CAPM Model Summary

     This unit was the most elaborate one this year to me. I had the opportunity to learn many new concepts, while going back and visiting older ones from earlier this year. The CAPM model, or Constant Acceleration Particle Model, really showed and explained exactly what we mean as physicists when we want to describe speed, or acceleration. In this quest for knowledge, I connected many different parts of previous knowledge to form what I now know. The following is a brief summary of my understanding of the constant acceleration model as of now.
     The first thing that I learned was the idea of instantaneous velocity of an object, or the velocity at a certain point in time which an object is traveling at. I found that there is actually two different ways to find the instantaneous velocity of an object: by finding the slope of the tangent on a position versus time graph, or by using the mathematical equation V(final) = acceleration*time + Velocity (initial). Finding the slope of the tangent seems way harder than it actually is; all that's needed is at least three different points on a position versus time graph. Here's an example: I want to find the instantaneous velocity of an object at 6 seconds into the test. So, I'll use points 5 and 7 to test...

V(final)= a*t + V(initial)

 V= (change in)X/ (change in) T

122.5 m- 62.5 m / 7 secs - 5 secs =

60 m / 2 secs =

30 meters / sec

That's one way to find the instantaneous velocity of the object. However, if I didn't have a position versus time graph to use, I can always use the mathematical formula V(final)= a*t + V(initial).

This requires me to be using a velocity versus time graph, and simply finding the slope of the graph. I'm able to do this because the slope is also my acceleration.

     The next thing that I learned was how to find the displacement or distance an object travels in a given amount of time by using a velocity versus time graph and by using the model x = 1/2 a*t + v. I already know how to find the displacement of an object using other types of graphs, but this gives me a quicker, more efficient way to an answer. When I use the velocity versus time graph, I have to find the area under the curve, which amounts to the displacement. Here's an example problem:
Imagine a invisible triangle connecting the line to the sides of the graph. This space should look like a triangle and a rectangle. This is what you need to find for the displacement.

15 m/s * 6 secs/ 2 = 45 m/s^2+

15 m/s * 6 secs = 90 m/s^2

Displacement = 135 m/s^2 



Now, to the heart of this unit; acceleration! I actually have four different ways to find the acceleration of an object, all of them listed out below:
- Finding the slope of a Velocity versus Time graph
- Using the model a = (change in) Velocity/ (change in) Time
- Using 1/2 a * t + Velocity (initial)
- Using Velocity (final)= a * t + Velocity (initial)
I've already showed an example problem involving three of these four solutions, but I haven't touched on when to use 1/2 a *t + Velocity (initial). I should only use this equation whenever I have a Position versus Time Squared graph and need to use it to find the acceleration. However, I can only do this whenever the starting velocity is equal to 0. The trick to using this method is to not forget to remember that the slope is only equal to 1/2 of the acceleration, so I need to always add back another half to find my full acceleration.
   
      A big part of my first unit at the beginning of the year was learning all about graphs. More specifically, position versus time and velocity versus time. In this unit, I expanded on this previous knowledge by adding acceleration versus time graphs. I also found how to graphically represent an object when the surface isn't flat. This in turn changes my position versus time graph from having a straight line with a constant slope to either an exponential growth or decay graph. I developed new language to describe these new graphic ideas. Despite adding all of these new ideas, I can still find things such as the displacement the same way as before; by finding the difference in the distance from the starting point to the ending point.
   
     The same thing applies to my previous knowledge about velocity versus time graphs. I still retained all the same principles and ideas from the first unit, I just added some new knowledge to the mix. However, this time I played with the idea of a velocity versus time graph having a constant slope, meaning that the acceleration (as the slope) is staying constant as well. This meant that my velocity graph was no longer just a straight line; it began to look more like a position versus time graph from the first unit. This took on a different meaning, depending on which way the object is traveling and if it's accelerating or decelerating. Here's an example graph below:


We can tell that the object is accelerating because it's covering more distance in less time as the test runs. However, this acceleration is towards the sensor, because the position from the sensor is getting smaller. My velocity versus time graph would start down below the line, as the object is moving towards the sensor, but be moving from the origin down at a constant pace. The velocity graph is constant because the object's speed is constantly increasing, which is the acceleration (also the slope for velocity versus time). This summary has giving me a clearer picture of what I have learned going into the test, and I hope to revisit this throughout the year for additional help if needed!