ESTM Challenge 

Challenge: We predicted the compression of two springs with different spring constants so that the two  carts come with the same velocity. 

Procedure: The only data we took was the mass of the two carts (.544 kg for the cart with the blue spring and .552 kg for the cart with the red spring). We were given the spring constants of each spring (112 N/m for the blue spring and 84 N/m for the red spring). We than calculated for the compression of the spring needed. After we calculated the compression for each spring, we tested our predictions using the motion sensors to detect the velocity of each cart.

Diagram:

Calculations:

To solve for the compression we used the energy conservation equation Eel = Ek. We did not consider friction so this equation was appropriate to use in this situation. Our work is shown below.

Conclusion:

Once we calculated the compression of the two springs, we tested our predictions using motion sensors and PASCO Capstone. We analyzed the velocity vs. time graphs to examine the velocities of the two carts. The cart with the red spring traveled at around 0.47 m/s and the cart with the blue spring traveled at around 0.41 m/s. Our prediction was off by about 6.8%.

Unknown Mass Challenge

Challenge: Find the mass of an object placed on a cart using the conservation of momentum principle. We used two carts, a can (unknown mass), and motion sensors. 

Procedure: First, we knew the mass of an empty cart was 0.5 kg. We then measured the velocity of the cart in order to find Va. After we measured the velocity of the first cart, we then measured the velocity of the second cart with the can (Vab).

Diagram:  

Data:                                                  

Velocity – Cart 1	Velocity – Cart 2
0.56 m/s	- 0.28 m/s
0.57 m/s	- 0.29 m/s
0.54 m/s	- 0.27 m/s
0.57 m/s	- 0.28 m/s
0.51 m/s	- 0.28 m/s
0.57 m/s	- 0.29 m/s

                                                         

Calculations:

 We then used the law of conservation of momentum to calculate the mass of the cart. Pi = Pf or mava + mbvb = (ma + mb)(vab)

We were 4.4% off when we tested our prediction which was around 500 kg and the actual mass was around 478 kg.

Rocket Challenge

Challenge: Launch a rocket and predict where it would would land. In order to do this, we had to use kinematic formulas to calculate and predict the horizontal distance traveled by the rocket. We initially set the rocket at 40° in order to calculate and predict the distance the rocket would travel at a greater launch angle. 

Procedure: First, we collected data from a launch at 40°. We used a timer in order to know the time the rocket was traveling in the air and we used a contractor measuring wheel to find out the horizontal displacement of the rocket (in meters).

Diagram:  

Data:                                                  

Time (s)	Distance (m)
3.73	52.2

                                                         

Calculations:

 We then used the equation Vx= ∆x/t to solve for the horizontal velocity, Vx. Vx = 13.99 m/s. Using trigonometry, we can calculate cos(50°)= 13.99/vi 

So, the actual velocity = 21.76 m/s

After we solved for the actual velocity, we now could apply this known velocity to our final test, since the launch velocity for the rocket is always the same. 

We launched our rocket at a 25° angle for our test. 

cos(65°)= vix / 21.76

vix = 9.2 m/s

cos(25°)= viy/ 21.76

viy = 19.72 m/s

∆x = 1/2at^2 + vit (∆xy = 0)

acceleration = -9.8 m/s^2 (acceleration due to gravity)

We can now plug in the known variables: 

0 = -4.9t^2 + 19.72*t

Now we can use the quadratic formula: 

t = -b + or - (√b^2-4ac)/(2a) Where a = -4.9, b = 19.72; c = 0

Solving the Equation: We calculated the time to be around 3.3 seconds. We then use Vx = ∆x / t (9.2 for Vx and 3.3 for t). Now we must solve for the horizontal Vx. We calculated the Vx (Vertical Displacement) to be 30.36 meters.

We set our target 30.36 meters from our launch site, and shot the rocket at an angle of 25°. Our rocket landed 32.3 meters from our launch site, which was exactly 1.94 meters off. So our percent error was 6.4%, which is higher than expected.  We concluded that the wind most likely influenced the trajectory of the rocket.

Unit 5 Summary: Particle Models in Two Dimensions

In this unit, we applied previously learned concepts to a new model. This model particularly focuses on projectile motion. We used the kinematic equations to calculate the unknowns of different scenarios.

Motion Maps:

The motion maps to the left represent the motion of a ball being thrown forward. The left side represents the ball's motion in the horizontal direction, while the right side represents the ball's motion in the vertical direction. As you can see with the motion maps, the horizontal and vertical motions of a projectile are completely independent of one another. The ball moves with constant velocity in the horizontal direction since there are not any forces acting upon it in the horizontal direction. However, the ball accelerates in the vertical direction downward due to the force of gravity that is acting upon it. The force of gravity is the ONLY force acting upon the ball in this direction.

Formulas and Units to know:

Horizontal:

Δx=vi Δt

vi = vf

Vertical:

vf =gΔt+vi

y = 1/2aΔt2 + viΔt vf2 = vi2+ 2gy y=1⁄2(viy +vy)Δt 

Concept Practice:

1. 

A ball is thrown downward with an initial speed of 15 m/s.
What is the acceleration of the ball? Calculate the displacement of the ball during the first 3 seconds.

Now calculate the time required to reach a speed of 40 m/s. Also, calculate the speed of the ball after falling 80 meters.

2. John's dog is chasing a cat across a table surface 1.3 m high. The cat quickly jumps off the table as John's dog slides off of the table at a speed of 4.2 m/s. Where will John's dog land on the floor?

3. A baseball is thrown straight upward and return to the thrower's hand after 4 seconds in the air. A second ball is thrown at an angle of 40 degrees with the horizontal. At what speed must the second ball be thrown so it reaches the same height as the one thrown vertically?

4. A group of students are visiting the Empire State Building while on a trip to New York City. A student becomes quite angry at another student and decides to throw him off from the observation deck, which is 371 meters high, with an initial downward speed of 10 m/s (pretend that the steel barriers that surround the observation aren't there). How long does it take the student to hit the street below? How fast is the student going at the moment of impact?

5. An angry student hurls an apple out of a window with a velocity of 14 m/s at an angle of 25 degrees above the horizontal. If the launch point is 8 meters above the ground, how far from the building will the apple hit?

Conclusion:

In this unit, I learned about projectile motion. I used previously learned concepts and applied it to models in two dimensions. I learned that the kinematic equations that apply to the the vertical component have to be associated with CAPM and UFPM. The formulas that apply to the horizontal have to be associated with BFPM and CVPM. At the beginning of the unit, I discovered that motions that occur on the horizontal and motions that occur on the vertical are completely independent of each other. I further used trigonometry to calculate the Viy and the Vix, which we applied to the latest challenge in class, the rocket challenge.  

Sources Cited:

http://paynephysics.wikispaces.com/file/view/Particle+Models+in+Two+Dimensions.pdf

http://w3.shorecrest.org/~Lisa_Peck/Physics/All_Projects/photojournal/shelby/shelby.html