Spring Cart Challenge

Challenge/goal: Predict where to compress two different springs so that the carts of equal mass come off with the same velocity.

b) What our setup looked like:

We used two of these tracks side by side during our experiment. The cart had a spring attached to the back of it.

c) What we were given:

Cart A had a spring constant that was 84 N/m.
Cart B had a spring constant that was 112 N/m.
Both carts had equal masses of .552 Kg


How we solved how far to compress the springs:

The elastic energy in the spring is equal to the kinetic energy when the cart is moving. (assuming no friction)

Eel = Ek  

Because of this, we can say that both carts originally hold the same amount of elastic energy. 

To find the elastic energy of cart A, we decided to compress the spring .06m. 

Eel = (1/2)(k)(x^2)

Eel = 1/2(84)(.06^2)

Eel = .1512

Since we know that the elastic energy of cart A will equal that of cart B, we can use the elastic energy formula to determine how far we should compress the spring. 

Eel = (1/2)(k)(x^2)

.1512 = (1/2)(112)(x^2)

x = .051m 

In order for the carts to have the same amount of kinetic energy, which would give them the same velocity, we would need to compress cart A's spring .06m and cart B's .051m. 

To test this hypothesis, we set both carts on tracks next to each other with our predicted compressions. We released both carts at the same time and used a motion sensor to determine the velocities. The velocity of both carts was .78m/s, giving us a 0% error. 

Cart with unknown mass lab

In this lab, we used our previous knowledge and the momentum conservation law to determine the mass of a small piece of metal. In our lab, we had two carts and one motion sensor. To start our lab, we measured the masses of each cart.

Ma (without weight) = .5005kg
Mb = .4882kg

After taking these measurements, we set up our experiment. We leaded one cart with the weight and pushed it until it hit another cart. Our motion sensors gave us numerical values for velocities before and after the crash, allowing us to graph the data and get an equation describing velocity before and after the crash.

Our setup looked as follows:

Our data that we collected from the motion sensor (with trials 1-7) and data when graphed looks as follows:

Velocity of A Before		Velocity of both after
0.63		0.38
0.69		0.43
0.69		0.46
0.62		0.39
0.55		0.34
0.7		0.46
0.63		0.4

The equation is: (velocity of both carts after crash) = (.7945)(velocity of cart A before crash)

This means that if the velocity of cart A before the crash was 1 m/s, then the velocity after the crash would be .7945 m/s. We can use this data, which is an average, to find the mass of the weight.


How we found the mass of the weight: Remember, Ma = the mass of the cart wight he object.We used the velocities from our equation (data above) before and after the crash in order to be as precise as possible. We plugged all of the values into our momentum conservation equation and then solved for the mass of cart A (wight he unknown object on top).

Then, to solve for just the mass of the object, we did as follows: (we weighed the mass of cart A which is how we got that value)

Conclusion: 

So, we predicted that our mass would be .87kg. We then weighed the object to see how close we were. The object actually weighed 3.4kg, a HUGE difference. I blame this on the super small number of trials with a very small variation in velocities. This data was not enough to give us accurate data, messing up our prediction. If we would have taken more data points with more variation, we probably would have gotten much closer. 

% error (I apologize for the sideways picture, it won't turn): 

Rocket Lab Post

Purpose: In the Rocket lab, we used our knowledge of constant velocity, initial velocity, and acceleration along with formulas that we previously learned in order to predict where a rocket would land at any given launch angle. Our setup looked similar to the rocket below, filled with air by a bike pump.

We used small wooden wedges to prop our rocket on an angle, which allowed us to determine how it flew. We started with a 35 degree angle. We used a timer to time how long the rocket was in the air. After launching our rocket, we used a measurer to determine how far our rocket flew. The rocket was in the air for 3.84 seconds and flew 42 meters in the horizontal direction. We used these measurements to calculate initial velocity using our horizontal component and cosine equation. It is important to remember that Initial velocity at all launch angles remains the same. Our work for this step can be seen below. 

After we found initial velocity, we were asked to launch the rocket on another angle (25 degrees) and predict where it would land. In order to do this, we drew an initial velocity diagram and then separated it in to components (step D). With the initial vertical velocity, we were able to determine how long the rocket would be in the air for (step A). After we found time, we used our horizontal displacement formula to determine how far the rocket would travel (step B). We then tested our hypotheses and our rocket landed 1.8 meters past our desired landing spot. With this data, we calculated our percent error (step C). 

                                    

Throwing a ball experiment

In this lab, we videos us throwing  a ball in a parabolic shape. We used the video and the PASCO keystone program to make graphs and learn more about the acceleration, velocity, and motion of the ball. Here is a video of our throw: 

After we plugged our video into the app, it allowed us to plot points where the ball was.

Once we plotted data points, PASCO gave us a position vs time for both horizontal and vertical movement. The green line shows the vertical position vs time while the purple line shows the horizontal movement. The green line's slope is constantly changing, and it shows us that the object has acceleration. Also, the line switches directions, showing a shift in direction of the ball. The purple line has a constant slope, indicating constant velocity. The ball had no forces acting on it in the horizontal direction and because of this it had a constant velocity.

           Time

Our second graph shows us the velocity vs time of the object in the vertical and horizontal directions. The yellow line is the vertical direction and the red line is the horizontal line.

The yellow line shows that the velocity begins at a very high rate and then begins to slow down. The line then hits the origin, showing a change in direction. The ball then begins to pick up speed in the negative direction. The line also has a slope, which is constant, which shows acceleration. The slope of our line is close to -11 (acceleration) whereas the actual acceleration of the ball should have been -9.8 m/s^2 (the true acceleration of gravity. Although the numbers weren't totally correct, they were pretty close. The red line shows no slope. This means that the velocity of the ball in the horizontal direction was constant, which is the case because of its lack of forces in the horizontal direction. 

Analysis: 

a) The only force acting on the ball after it left my hand was gravity. This means that the acceleration of the ball was -9.8m/s^2 (the acceleration of only gravity acting on anything) 

b)There are no forces acting on the ball in the horizontal direction. This means that the ball's forces are balanced in the horizontal direction, giving it no acceleration. 

c) By using our graph, we can tell that the initial horizontal velocity is very close to .6 m/s. 

d) The initial velocity in the vertical direction was 2.6 m/s according to our graph. 

e) The velocity in the horizontal direction was constant the whole time. This means that at the top of the path, the ball was still going .6 m/s. 

f) At the top of the path in the vertical direction, the ball is switching direction and momentarily has a velocity of 0 m/s. 

g) At the end of our test, the ball stops on the ground (eventually) and has a velocity of 0 m/s in the horizontal direction. 

h) At the end of our test, the ball stops on the ground (eventually) and has a velocity of 0 m/s in the vertical direction. 

i) Using the equation Change in x = 1/2(a)t^2 +vit, we can calculate the height the ball got. The ball reached its highest point after .9 seconds (according to graph)

Conclusions: What occlusions can you make about projectile motion? 

a) The vertical acceleration on our graph shows -11 m/s^2. In reality, this should be closer to -9.8 m/s^2 because of the acceleration of gravity. The only force on the ball is gravity, giving it this acceleration. The slightly larger acceleration is probably due to minor errors while plotting points on our graph. There are no forces acting on the ball in the horizontal direction. This means that the horizontal axis is balanced, and the ball has no horizontal acceleration. 

b) The equation for horizontal velocity is (change in horizontal position)/time. this velocity should stay constant throughout t ball's motion. The equation for vertical velocity is (change in vertical position)/time. This velocity will change throughout the motion of the ball. 

c) Vertical formula for displacement: Displacement = 1/2(at)(t^2) + (Vi)(t)

Horizontal formula for displacement: displacement = (.6)t

d) Displacement = 1/2(a)(t^2) + Vi(t)

e) The ball has a vertical velocity of 0 m/s at the very peak of its path, but the horizontal velocity stays constant throughout the whole motion.

UFPM Challenge

In this lab, we used a modified Atwood machine to predict the intersection of a dropping weight which had acceleration and a cart moving at constant velocity along the ground. For an extra challenge, we angled our Atwood machine slightly. The angle of our ramp was three degrees. Our apparatus looks similar to what follows:

We took measurements of the mass of each component of our contraption:

Object	Mass (kg)
Green cart (on floor)	.477
Weight hanging with string	.059
Buggy	.5883
Whole Atwood machine	1.04

Converting masses to Newtons and using our background knowledge, we created a free body diagram  to help us solve our problem.

In order to predict how to make the weight land on the cart, we calculated the time it would take for the weight to drop to the ground. To do this, we measured the total distance that the weight would fall. 

Our work is as follows:

We used our force diagram to find Fnet and used our measurements to find the mass of the Atwood machine. With this information we were able to find acceleration using the equation Acceleration=Fnet/mass. We plugged .509 into the Fnet and 1.04 in for mass, leaving us with an acceleration of .489m/s^2

Because we knew the distance the weight would drop and the acceleration, we used the following method to solve for the amount of time the weight would drop.

We then used a motion sensor to detect the velocity of the cart that was moving along the ground. Once we found the velocity of that cart (which was .119 m/s), we multiplied it by the time in order to find out how far back to start the cart for the intersection to work. We used the equation 

distance=(velocity)(time) to find out that the cart would travel 25.2 cm. We then placed the cart 25 cm away from where the weight would land. We then let both the cart on the floor and the dropping weight go at the same time and videos the result. Our predictions were correct and the weight landed perfectly on our cart! A video of the experiment is posted below. 

Physics Challenge- Wind cars

In this lab, we used motion sensors to get two equations of different wind powered cars. These cars both accelerated from rest. In the lab, we tried to predict where the cars would have to start in order to intersect at the origin. Our work is as follows:

Constant Acceleration Particle Model

Position vs Time Graphs: Previously we have known that when a position vs time graph has a straight line on it, the object is moving at a constant velocity. With the CAPM model, we learned that these graphs can also have a curved line, which indicates acceleration. Also, we know that steeper slopes indicate that an object is moving fast, and that when an object moves in the positive direction it is going forwards and when it is going towards the negative side of the graph that it is going in the negative direction. We can also apply this information to our new curved lines.

As the line moves down as the vector
indicates, the velocity is increasing.
As the vector going up indicates, the
velocity is also increasing. 

If a velocity line is going up in the direction
 of the top vector, the object is moving in
the positive direction, and if the line is going
downward like the bottom vector, then the
object is moving in the negative direction. 

If the line moves toward the origin over
 time line the top vector does, the object
is slowing down. Also, if the line moves
towards the origin like the bottom vector,
the line is also slowing down. 

If the line on the graph moves towards
the origin like the top vector, the
object is still moving in the positive
direction. Similarly, on the bottom vector,
the object is moving in the negative
direction. 

In this example, the object starts with
a velocity of 10m/s. From seconds 0-2, the
object is slowing down. Then, as it crosses
the origin, it changes direction and
begins speeding up in the negative direction.

In this graph, the object is increasing in speed in the positive direction. To calculate displacement,
we can calculate the area between the line and the x-axis. This area is in the shape of a triangle, so
we can use Displacement = 1/2 (Base)(Height) to find displacement. By substituting values in from
the triangle that the line makes with the x-axis, we create an equation that should look like this:
1/2(10)(10) = displacement. From there, we can conclude that the object finished 50m from where it started.
To take a different approach, we can also use the equation ΔX = 1/2aT^{2 + initial velocity} to find the displacement. Initial velocity is the velocity the velocity at zero seconds. In this case, initial velocity is zero. Also, we can find the acceleration by using the equation (ΔV)/(Δt) = a. We can only use this equation for acceleration when our velocity vs time graphs are a straight time, indicating constant acceleration. In this case, the acceleration is 1m/s^2. After we find acceleration, we can plug in our numbers and solve for the displacement. ΔX = 1/2(1)(100) + 0 which comes out to equal 50m, the same as our first calculation.

Average Acceleration:

Formula: 

Average acceleration is the average rate at which an object is speeding up. Average acceleration may be very useful for many other equations and formulas in physics. 

In this graph, to find the instantaneous velocity at five seconds we can
find the slope between two pints that will make 5s the midpoint.
 For example, we can find the slope between the points at 4s and 6s.
This line is also tangent at five seconds. The fourth second falls at 1m and
the sixth at 2m. Therefore, we can use the equation (change in position)/(change in time) 
to find the line tangent. This tangent line will be the instantaneous velocity at that point. 

Example: 

In the position vs time graph above, the object has four distinct motions. From point A to B, the object is moving in the positive direction. Also, since the line is curved, we can determine that it is either slowing down or speeding up. The way to differentiate between the two is look at the slope of the line. At the beginning of the line, the slope is very steep. It gradually becomes less steep until the slope becomes zero at point B. With this information, we can tell that the object is initially going very fast and then slows down. From point B to C, the object is moving towards the negative side of the graph, telling us that the object is moving backwards. The slope closest to point B is not very steep, but gets steeper as it gets closer to point C, indicating that the object is speeding up. At point C something interesting happens. The object crosses the origin with a very steep slope (high velocity) and then begins to slow down again. This can be seen with the gradual decline of slope. At point D, the object switches direction again and begins to travel in the positive direction again. Starting with a small slope, the object begins traveling slowly and then speeds up until it reaches the origin, finishing at the same position  that it started at. 

Velocity vs Time Graphs: Constant velocities are shown on a velocity vs time graph as a horizontal line. When a velocity vs time graph has a diagonal line, the velocity is either increasing or decreasing. An object with an increasing or decreasing speed has an acceleration. As seen on the velocity vs time graphs below, as the line moves away from the origin, the object is increasing in speed. Also, if the line is going in the positive side of the graph, it is moving forwards (above the origin) and if it is in the negative side of the graph, it is moving backwards (under the origin) 

Example: 

 Displacement and Velocity vs  Time graphs: We can use two methods to find the displacement from a velocity vs time graph. The first is to find the area of the space between the line and the x-axis. This space represents how far the object has traveled. Also, we can use the equation ΔX = 1/2aT^2 where ΔX = displacement, a = acceleration, and t = time. 

Example problem: 

Acceleration vs Time Graphs: Acceleration vs Time graphs do not tell us much other than the slope of the velocity vs time graph. a vs t graphs are directly related to the slope of v vs t graphs. For example, if the slope of a velocity vs graph is negative, the acceleration line will also be on the negative side of the graph. Keep in mind that there is no acceleration when the velocity vs time graph has a straight horizontal line with no slope. 

Examples: 

In this velocity vs time graph, the line has a positive slope from 0-5 seconds, no slope from 5-15 seconds, and then a negative slope from 15-25 seconds. Because of this, we can conclude that the acceleration for the first five seconds is positive, there is no acceleration for the next ten seconds, and then there is a negative acceleration for the last ten seconds. The acceleration vs time graph would look as follows: 

Motion Maps: We have now learned how to make both acceleration and velocity motion maps. In velocity motion maps, the distance between the points represents how far the object has traveled and the vector represents the velocity. The larger the vector, the larger the velocity. In acceleration motion maps, the distance between dots represents distance traveled, but the vectors represent acceleration. Constant acceleration is represented by vectors of the came length as seen below. In the image below, the object's velocity is increasing with constant velocity. 

Average Velocity: Average velocity is calculated by the equation (change in position)/(change in time) and represents the average velocity that an object travels at over a given amount of time. 

Instantaneous Velocity: Instantaneous velocity is the velocity at a given time. To find instantaneous velocity on a position vs time graph, you can find a line tangent to the point at which that time is located (also the same as making that time the midpoint of two other points on the line). 

Real World Examples: Physics is ever present in the real world. One example of a real world situation where acceleration is present is as follows: A dog runs down his driveway with an initial speed of 5m/s for 8s and then uniformly increases his speed to 10m/s in the next five seconds. How long is his driveway? 

How to solve: So, we are given both time and velocity. Since we are trying to solve for displacement, we can use the equations ΔX = 1/2aT^{2 + initial velocity} and x = Vt. In the first eight seconds of the run, the dog has a constant velocity. Because of this, we use the exertion ΔX = Vt. After substitution, this becomes ΔX = (5m/s)(8s), simplified to 40m. Keep in mind that this is only the first half of the dog's journey. The second half can be calculated with the equation ΔX = 1/2aT^{2 + (initial velocity)(time)}

. We can find acceleration with the equation ΔV/Δt which comes out to be 1m/s^2. After substitution, this equation becomes ΔX = 1/2(5)^2 + (5)(5) This can be simplified to ΔX = 37.5. So, now that we have both the first part of the journey and the second, we can add them together in order to find the total displacement (length of the driveway). 40m + 37.5m = 77.5m, the length of the driveway.