Rocket post

In this challenge, we were attempting to solve for how far our ball would go, after having solved for velocity at the 30& and 50 degree angles in order to solve for the 40 degree,
In order to do this, we collected the following data:

This leads to our estimates of



This leads to the math of

Video Analysis of moving objects

Have you ever wondered what the video analysis of a ball thrown through the air would be?
Prepare to be amazed, because here it is.

Here is our ball's track, the same video done twice:

These captions show the path of our ball over the course of its flight. One of the things that needs attention is that we have the meter stick, which gives a frame of reference for the flight of the ball of 1m.

This is the graph that shows both our change in position in the x direction (green), and in the y-direction (red). We can see how much distance it covers, how long it takes, and why.

When we ran our velocities, we found the following data:

This graph shows the velocities, and thus the accelerations of our graphs. In the x direction, we can see that the displacement is very low, and our graph peaks then seems to come back to almost the same as when it started. However in the y direction, our ball seems to increase its speed for the first few seconds, but then it comes to reaches a crescendo, the peak, and slows down until it hits zero, which is the point where it starts to descend.

A and B. 

Our acceleration in the y direction is 9.81 m/s^2 due to the force of gravity, and our acceleration in the x direction is 0m/s^2 as x-direction does not have an acceleration capacity, rather an ability to exist in that of constant velocity as the change in v evens out, making it constant velocity.

C & D. 

Our initial velocity in the x-direction is 5 m/s which remains constant the whole way through, but our initial velocity in the y-direction 3.09 m/s as that is the initial slope on the position vs time graph for the y direction.

E&F. 

The velocity at the top of our path in the x direction is 5 m/s, while the velocity at the top of our path in the y direction is 0 m/s as that is the point where it is about to turn around, thus reaching standstill for a moment.

G&H

The final velocity in the x direction remained the same, 5 m/s as it was travelling at a constant velocity, but had reached -9.65 on the last part of the graph.

                                                   I&J.

In order to discover the height of the ball, we can see where it was on our position graph, and it was 1.8m. The distance our ball traveled can found using x_final-x_initial=change in x. In this case, our change in x was 2-0.4=1.6 m.

K&L

It took 2 seconds for the ball to reach the top of its path (27.5s-25.5s), and it

Remained in the air (29.15-25.5) 3.65s. 

Conclusion:

1. Vertical accelerations are constant at 9.81 m/s^2 as a result of the force of gravity, but horizontal acceleration is 0 m/s^2, as it is travelling at constant velocity, not acceleration. I know this both through equations, and also the visible aspect of our data. 

2. For Vertical formula's, I can use the Δx = 1/2 aΔt² + ViΔt formula, but for the horizontal, the only formula we can use is v = Δxv/Δt, and let that solve it. In order to do so, I would solve for time through the change in x formula, and that input that into the velocity formula. These two formula's work together, not against each other. 

3. In order to solve for the displacement, I would use the Δx=xf-xi, and find it from there, but the vertical formula is a longer verison that has 0's in it to cancel it out. 

4. In order to solve for height, we would use the Δx = 1/2 aΔt² + ViΔt, input time, and then divide our x by two, as it would take equal amounts of time to fall to its peak, and to fall to the ground. 

5. At the top of its peak, there is only the horizontal velocity, as the vertical velocity is nonexistent. 

Cart and Ball post

UFPM Challenge

A modified Atwood machine:

[Image upload of diagram]

In this lab, we were attempting to calculate two variables. The constant velocity of our moving cart, and the rate at which our cart on the track was accelerating due to the hanging weight which was attached.

Our cart’s track had no angle, so all we needed to find was the rate at which our weight (50g, 0.5n) was causing it to accelerate.

Thus, the following force diagram was created.

Please note the lack of friction, thus causing our diagram to be Unbalanced.

Thus, our diagram becomes an unbalanced force particle model. In order to solve for such a situation, we use the a(acceleration)=f_net/m(kg)

Thus, our following formula became a=0.5/.6394

As a result, a=.78 m/s²

We measured how far our change in x was, we discovered that it was 90 cm which equals .9m. When we input our known variables into the Change in X=vit+ at².

The final formula to solve for time is .9=.78t². As a result, our time was 1.15 s.

Part 2 was testing the velocity of our cart on the ground, and thus we solved for the velocity of our cart, finding that it was 0.3 m/s. Thus we multiplied 0.3*1.15s ending up with 34.5 centimeters away.

In order to test our conclusion, we let our ball drop when our cart had reached 34.5 cm’s away from the meeting point.

Once we tested our predicament, it was found to be correct.

UFPM Blog

Unit Summary UFPM

This unit covered a vast amount of information in regards to unbalanced forces but was also really exciting. Examples of such forces are all balls in the air, as only gravity acts upon them, all accelerating objects, and anything really that is not maintaining a constant velocity.

The formula’s used were:

1.       vf = at + vi

2.       vf² = vi² + 2aΔx

3.       a = F_net/mass (in kg)

4.       Δx = 1/2 aΔt² + ViΔt

5.       a = Δv/Δt

We started this unit exploring what happens when there is more force going one way, than holding it back, by looking at what happened if friction, or air resistance were not factored in, or if they were, that they were not equal in force.

Examples of such scenarios are cars that are speeding up or slowing down. Part of this was learning how to solve for the frictional force, as shown below.

If a car that weighs 240 kg with an F_push  of 300n, and and F_F of 80, the following diagram looks like:

In order to solve for acceleration, we must use the a=fnet/m formula as shown below. 

We have solved for acceleration, it is .91m/s^2

If we know that a parachutist covered 1500 meters before expanding his parachute, from rest, how long did it take him?

                     It took him 17.3 seconds to fall 1500 meters.

Now, what is his final velocity right before he opens his parachute?

Thus it is 173m/s before he pulls his parachute cord open. 

This is the summary of our blog, if you have questions, please feel free to comment below. 

Collision? Where?

In this challenge, we were seeking to understand where two carts, going at different speeds would collide starting from 3 meters apart.

In order to solve the challenge, we measured the rate of acceleration (5 trials averaged) for both carts and then inserted the following data into a formula we had learnt, X = v_ot + (1/2)at^2, and set the x's equal to each other.

Our first cart had an acceleration of .0909 m/s^2, and the second cart had an acceleration of .137 m/s^2.

Thus,when you set the formula's equal to each other:

1/2(0.909)t^2 = 1/2 (-0.137)t^2=3

Which ends up being:
0.4545t^2+ 0.0685t^2=3

Which ends up with 3 meters ( our displacement) equaling 0.4545t^2 + 0.685t^2, which ends up being 0.11395^2. 

Thus, 3=0.11395t^2, divided by 0.11395 on both sides, we find t^2=25.9, wherein the square root of both sides is t=5.08s. Thus, when we let both of them run for 5.08 seconds, our prediction results in an expected collision of 1/2(0.909)(5.08^2)= collision.
Thus our collision should occur 1.2 meters from the top, 80 cm from the bottom.
When we tested it, our cart ran 1.23 meters, coming in at 77 centimeters from the bottom.
This difference, of 3.75 degrees of error, is much less than 10 percent error.

How fast can my gliding speed increase?

Well, you would think I was an ice skate racer, but instead I am a simple little cart on a simple table.

Like I said, I am only a little cart  trying to figure out how fast I can accelerate. So my proud renter took me, and put me on a track, measured every half second, and then took data to see first my change in position over time, which lead to velocity, and then solved for acceleration. 

This looked like the initial chart of:

This gave us the following graph. Note the curve upward.

We had to take the following formula, y=mx+b, and change it into a line that more closely resembles the formula's we are dealing with. The following formula is x=slope*t^2+xo. When you notice that the t is squared, it is because the time must be squared to fit into a graph to solve for acceleration.

This produced the following graph:

Note the slope of 0.070747.

In order to solve for Acceleration, we use the formula V=1/2a

Thus, as our slope is velocity, we can multiply it by 2 and solve for the acceleration. 

Our acceleration is 0.070747*2=0.141494. 

However, when we checked it with the motion detector, it found .22 as the slope. I do not know why, but will continue to ponder why I (the cart) am not accelerating at the indicated pace.

CAPM A summary of learning

Welcome to my new readers, and hello to the old.

CAPM stands for Constant Acceleration Particle Models, wherein we built from our former understanding to develop a knowledge and appreciation of what constant acceleration means.

This added acceleration graphs to our chart in contrast with velocity, showing how they intermingled. Then we worked on how we could translate from a motion graph, to a velocity graph, and then to one for acceleration. 

CAPM.1

Instantaneous velocity:

This is the velocity at any precise, given, point in time. It can be calculated in several ways from either charts or graphs. If at a graph, one can tell the velocity by using exactly where it is on a v vs t graph, such as the one below that is both constant velocity and constant acceleration.

http://www.thetrc.org/pda_content/texasphysics/e-BookData/Images/SB/57/LR/ConstAccelConstVelComparisons.png

Thus, by finding the exact position of an object which is its data points on a v vs t graph, we can see what the exact instantaneous velocity is. 

The second way to solve  for velocity is by taking mirroring in time data points on both sides of what you are looking for, and then inputting them into the formula V= Change in position/ Change in time.

Capm.2

Solving for displacement can be solved for either with formula's or with a graph. 

The formula is Change in x= xfinal-Xinitial on a x vs t graph, however on a velocity vs time graph the formula becomes change in x= 1/2*a(change in t)^2+ Vi(total time)

On a graph, we can use the position vs time graph to see where it is by subtracting the initial data points from the final coordinates.

Capm.3 

Solving for acceleration can be done in four ways.

The first is looking at a velocity vs time graph, where we can input our data into the formula a=Change in v/ change in t.

The second is solving for it in other formula's, such as solving for a in the following formula Change in x= 1/2 a (change in t)^2 + Vit.

The third is simply looking at an acceleration vs time graph and seeing where it is. 

The fourth is solving for it in the second formula:

Vf=a(change in time)+V(initial)

CAPM.4

If you look at the following position vs time graph, you may note:

This is taken from http://gradeelevenphysics.weebly.com/speed-velocity-and-acceleration.html

 You may note that it is constantly decreasing its velocity, however the acceleration is constant. It is travelling in a positive direction though the velocity and acceleration are negative as it is slowing down. 

The corresponding v vs t graph would resemble the one above, with it constantly decreasing. The acceleration is steady but nonetheless negative as it is slowing.

CAPM.5

If we take the same graph as above, we can draw a position vs time graph based on our initial understanding of what the motion means. Because our velocity vs time graph is starting positive, and never becomes negative, we know that our position vs time graph will be positive. We also know that initially it will be going faster, as our velocity graph starts from a higher position on the y (Velocity) axis. Then, as our velocity slopes downward, we know that our position vs time graph line needs to lessen its upward motion, gradually curving out until it is almost going straight.

Due to the constant nature of our velocity going towards the x (time) axis, we know that it is heading in a negative direction and thus our acceleration will be negative and a straight line.

CAPM.6

In order to construct a graph, we must see that there are 4 particles. 

The 4 particles are:

What direction is it going?

Is it constant?

What variables are we given, and what do they apply to? 

Is it possible for it to become negative?

From this, we can draw any graph we wish. 

CAPM.7

To solve for time, we can take the average formula x=1/2 a(t^2)+Vi(total time).

From this, we can solve using algebra to isolate our t. 

Please note the image below to see the process.