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.