Projectile Motion 

Video Analysis

Experiment & Purpose: The purpose of this was to study the properties of projectile motion. In this experiment I simply threw a tennis ball forward, and took a video. From there, I uploaded the video to the PASCO Capstone application in order to analyze the motion of the ball. I tracked the ball's motion using the tracking feature on the application. I was than able to examine some key characteristics of projectile motion. Once I had created the graphs and examine them, I realized that motion in the horizontal direction is independent from the motion in the vertical direction.

The image above shows the path of the tennis ball after leaving my hand. The green crosses represent the position of the tennis ball.  The yellow symbol represents the measurement tool that was necessary for the subsequent calculations. 

The graph above shows the position of the ball over time. The purple line represents the ball's horizontal displacement, and the green (highlighted yellow) line represents the ball's vertical displacement.

The graph above shows the velocity of the ball over time. The red line represents the ball's horizontal velocity, and the blue line represents the ball's vertical velocity.

Vf  - Final Velocity
V0 - Initial Velocity 
a - Acceleration
t - Time
Δx - Displacement
xi - Starting (initial) position

Analysis:

With the data collected so far, I can calculate a number of unknowns. 

• Vertical Acceleration: -9.8 m/s² (force from Earth's gravity)
• Horizontal Acceleration: 0 m/s²  (no unbalanced force in the horizontal direction)

Vertical Initial Velocity: 0.21 m/s (Δx = Vit + .5a*Δt²)

Horizontal Initial Velocity: -11 m/s (negative due to being thrown to the left)

Velocity at the top of ball's path (vertical) is 0 m/s because it is changing direction.
Velocity at the top of the path (horizontal) is -11 m/s (constant velocity)

Final Velocity (vertical):  -3.8 m/s      last data point on v vs. t graph
Final Velocity (horizontal): -13.1 m/s last data point on v vs. t graph

Highest point reached by ball: 1.6 meters (vertex of parabola on graph)
Distance ball traveled: 4.51 meters  (using Δx = .5*a*Δt²+ Vi*Δt)

Conclusion: After this experiment, I learned that motion experienced on the the vertical component does not correlate with the motion experienced on the horizontal component this is due to the object's inertia, since no external horizontal force is needed to maintain the horizontal motion. So the V = Δx/Δt formula may only be applied to the horizontal component due to the object experiencing constant velocity only on the horizontal component, not the vertical component. On the vertical component the only two models that can be applied are the CAPM and UFPM models, however the only two models that can be applied to the horizontal component are BFPM and CVPM models. We also learned that that the only force that acts primarily on the object is gravity. 

UFPM Challenge

Challenge: Land the mass from the modified Atwood machine on the constant velocity cart. 

First we calculated the mass of the system. Mass of system = 0.6 kg. Then we solved for acceleration. 

a=Fnet/m

a=.5 N/.6 kg

a=0.833 m/s²

We than measured the distance from the starting position of the weight to where it would land on the cart, which was .78 m.

We than solved for time...

x=1/2at^2+Vit

(.78)=1/2(.833)(t)²

t=1.37s

By using: V=Δx/Δt, we calculated where we should position the cart. We calculated the velocity of the cart to be .27 m/s. 

V= .27 m/s, so .27 = (Δx)/(1.37)  

Δx = .37 m 

Once the calculations were finished we decided to test our hypothesis. So we positioned the car .37 meters from the position of where the weight hit the floor. 

We successfully proved our hypothesis correct after a few tests (we had to angle the cart slightly to the left due to its tendency to travel a bit to the right).  Our prediction was quite close and we made it within 5%.

Unit 4 Summary: Unbalanced Force Particle Model

In this unit, I learned about Newton's Second Law of Motion and applied it to a multitude of scenarios.  I also learned about unbalanced forces and how to calculate the net force acting on an object. 

Newton's Second Law: 

• The second law says that the force acting on an object is equal to the mass of that object times its acceleration. 
• This is represented by the formula: F = ma (ΣF = m*a)
• F is force, m is mass and a is acceleration. 
• The acceleration of an object is directly proportional to the net force acting on it, and is inversely proportional to its mass. 
• The direction of the acceleration is in the direction of the net force acting on the object. 
• So, the net force is the sum of all the forces acting on an object. A net force is capable of accelerating a mass. 

  
  

  The bobsled accelerates because the team exerts a force.

Friction and Coefficients of Friction

• F_f = Mk * Fg       
• F_f = frictional force (N)
  M (Mu) = frictional coefficient
  Fg = Weight of object
  
  
  

  Units and Equations for this Model:

  Vf  - Final Velocity

  V0 - Initial Velocity 

  a - Acceleration

  t - Time

  Δx - Δ is displacement

  
  

  
  

  x0 - Starting (initial) position

  
  

  

  
  

  In this model, we began to apply the kinematic equations to scenarios in order to calculate unknown variables. 

  
  
  
  
  
  
  
  
  

Concept Practice: 

1. Above is a modified Atwood Machine, we can calculate the net force on the system using the hanging block's mass. Let's say the hanging block's mass is 100g. 

I drew the free body diagram and calculated the net force on the system.

2.

Above is a person pushing a 15 kg box with 20 N to the left. The positive direction is the right and the coefficient of friction between the surfaces is .11. What is the acceleration of the box?

I drew the quantitative free body diagram and than proceeded with calculating the acceleration. As you can see, it was a multi-step process. First, I had to calculate the Friction (F_f = Mk * Fg) which was 15 Newtons. I than had to calculate the ΣF, which was -5 Newtons. The final step was to calculate the acceleration (a = ΣF/m), which was .33 m/s².


3.

Michael and Angela are riding the elevator up to the 95th floor of the building. The elevator is traveling up at a constant velocity of 17 m/s. Michael and Angela have a combined mass of 130 kilograms. Construct a force diagram for the situation and than calculate the force the floor exerts on the passengers. 

Since the elevator was traveling at constant velocity, it is quite easy to calculate the force exerted by the floor (or the ΣF). The acceleration is force of Earth's gravity and and the mass is 130 kilograms (or 1300 N). By knowing the mass and the acceleration, I can calculate the ΣF. Which the ΣF equals 12,740 Newtons. 

3b. The same elevator soon begins to accelerate upward at 4 m/s². Now what is the force exerted by the floor on the passengers? 

Since the elevator is now accelerating, I can calculate the ΣF using the a = ΣF/m formula. The elevator is accelerating upward at 4 m/s². The mass of the elevator is 130 kg, so we can plug these known variables into the equation (4 = ΣF/130). ΣF = 520 Newtons. Since I know the ΣF, I can now solve for the normal force. I can do this by using the ΣF = Fg + Fn formula. So 520 = -1300 + Fnormal, and 1300 + 520 = 1820. So the normal force is 1820 newtons! 

4. 

 In the image above, a 75 kg skier hits the slope and is accelerating down the slope. Calculate the net force of the skier and his acceleration as he travels down the slope. 

The skier is traveling down an inclined slope which measures 30 degrees. When drawing the free body diagram for problems involving inclined slopes, always place the given between the Fg and Fgy. The Fgx equals the ΣF, so once I solved for the ΣF using Cos(60) = Fgx/750 = (750)(Cos60) = 375, I had also calculated the Fnet. To calculate for the acceleration, I plugged in the known variables into the a = ΣF/m formula (a = 375/75). 375 divided by 75 equals 5, so the acceleration is 5 m/s².

5.

The rollercoaster, Kingda Ka, is the world's tallest rollercoaster at a height of 139 meters (456 feet). As the passenger cars are launched from rest at the start they accelerate uniformly to a speed of 57 m/s (128 mph) in just 3.5s. While accelerating, friction exerts  450N on the train. Loaded with passengers, the train has a mass of  8325 kg. Calculate for all the unknowns. 

The train goes from rest to 57 m/s in just 3.5 seconds. With this information, I calculated the acceleration using the  a = Δv/t (a = 57 - 0/3.5) formula, which equals around 16.2 m/s². Once I calculated the acceleration I could now calculate for the Δx using Δx = Vit + .5at²  (Δx = 0*3.5 + .5*16.2*3.5²). I calculated the Δx to be 85.85 meters. I than calculated the ΣF using a = ΣF/m (16.2 = ΣF/8,325).  I calculated the ΣF to be 135,865 Newtons. Once I had calculated the ΣF, I could calculate the Fpush using the ΣF = Fpush + FF (134,864 = Fpush - 450). I collocated the Fpush to be 135,315 Newtons. 

Conclusion:
In this unit, I learned about Newton's Second Law of Motion and the Unbalanced Force Particle Model. Newton's Second law says that the force acting on an object is equal to the mass of that object times its acceleration. I also learned that acceleration t is directly proportional to the net force, but is inversely proportional to mass. I learned that the net force is the sum of all the forces acting upon an object and that the direction of the acceleration is in the direction of the net force acting on the object. Unbalanced forces must act upon an object in order to accelerate and an object at rest is being acted upon by balanced forces. In this unit, I began to use the kinematic equations in order to calculate for the unknowns. When dealing with the motion of elevators, I learned that when an object is speeding up, the acceleration is in the same direction as the velocity. However, when the acceleration and velocity are in opposite directions, the object is slowing down. 

Sources Cited:

http://images.hellokids.com/_uploads/_tiny_galerie/20100102/bobsled-source_ckr.jpg

http://s3.amazonaws.com/siteninja/site-ninja1-com/1404085954/large/pushing_heavy_box_10368-300x196.png

https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEga2Ox_q5J2IdzhRXvj3Oi6Ws0mKx3K-r6kAy_Blx46ITVpaO60cZ-qZbM_KMLDPvMzzi0H7KUcMTi_gSXvfVuv5VH3TfvgABSnpFyOIZjm3hyphenhyphen05oQv5U-rb2E1vPfjyROA7OpOoOmAuEo/s1600/notfreefall.jpg

https://www.sixflags.com/sites/default/files/kingda_ka_3.jpg

CAPM Final Challenge

Challenge: Predict where your fan cart and other group's fan cart will collide. Your prediction needs to be within 3 centimeters rather than within 10% error. 

With the data recorded, we calculated the average acceleration of the fan cart using the motion sensor to detect the velocity of the cart. This sensor provided us with a velocity versus time graph of the object's motion along with the equation of the graph: y= mx + b or v = at + initial v. The slope of this line is equivalent to the acceleration. We used the data to create a graph and calculated the average acceleration using the formula: a = ΔV/ΔT.

a = (.61 - 0) / (4 - 0) = .1525 m/s²

^{We recorded this data using Cart A and compared the acceleration of Cart A} 

^{to Cart E. We calculated the acceleration of Cart E to be} .0213 m/s²

^{After we had completed this part, compared our results with with another group to calculate the location in which the two carts would collide.} 

Δx = (1/2) * a * (Δt)^2 + V_it

Cart A:

Δx = (1/2) * (.1525) * (4)^{2 + (0)(4) = 1.22 meters}

^{Cart B:}

Δx = (1/2) * (.0213) * (2)^{2 + (0)(4) =  .426 meters}

^{We calculated that the carts would collide at 29 centimeters. When we actually put the carts to the test, we were only 4 centimeters off.} 

Conclusion: In this experiment we used acceleration to calculate where two carts would collide (traveling in a direct path). We plugged the two formulas into a graphing calculator and calculated where they would collide by calculating the point of intersection. Our prediction was pretty close to our actual results, only 4 centimeters off.