Cart Challenge - Ally and Nikki Finding the Unknown Mass

Cart Challenge - Ally and Nikki
Finding the Unknown Mass

Ally and I solved to find the mass of a variable on top of a cart. The variable we found was the weight of the soup can.

First, we weighed the mass of the cart:  .49 kg. We put the center of the cart at 110 inches and the markers on the side at 53 inches and 173 inches.


After taking 5 trials of the empty cart, the average velocity taken was .5 m/s. We then put the soup on the cart and measured the velocity. After taking 5 more separate trials, we found the average velocity with the soup can to be .25 m/s. 

Then our group needed to solve for the mass of the cart with the soup can.  We used the formula mava + mbvb = (ma + mb) (Vab).

mava + mbvb = (ma + mb) (Vab)
.49 (.5) +0 = (.49 + mb) .25
.245 = .1225 + .25mb
.1225 = .25mb
.49 kg = mb

To check our percent error, our group measured the cart with the soup can on it: 478 grams and converted it to kg: 0.478 kg

Therefore our percent error= 2.51%

To calculate the percent error:
=(.478)(.49)/ .478*100
= 2.51%

Hanging Weight UFPM post

In this lab, we had to predict where we should place a constant velocity cart so thst a falling hanging weight attached to a cart will land on a track. 

1. First, we took the masses of all of the components for the experiement.
2. Then our group calculated the time and acceleration. In order to do this, we drew our FBD's, labeling Fn, Ff, and Fg.
3. Then we calculated the sum of forces in the horizontal direction to find our Fnet.

 To find acceleration, we used A= Fnet/m


To find our mass: (then converted back into kilograms)
9.892N+.509N = 10.4N
Fnet= .509
mass= 1.04N


a = .509 / .104
a= .5 m/s^2

Then we converted the weight into kg and then into Newtons.
1kg= 100grams

buggy:4.77N
pulley: .509N
 weights: 9.892N




Then we plugged in our variables to solve for time.

x= 1/2(v)(t^2)+vi(t)
t= -1.854

In order to predict where to start the buggy, we found the velocity of the cart using a motion sensor.


To find out percent error, we took a video and tested our prediction.

The pulley landed beautifully on the cart, with a 0% error!!!

Spring Constant Practicum Challenge

Practicum Challenge: Physicists Nikki and Hank

How much does the spring have to be compressed in order for the carts to go at the same velocity?

Our group had to solve for where the two carts needed to be placed to that the carts came out with the same velocity.

Mass:

Blue= .5394 kg

Red= .55 kg

(Note: Our group weighed both carts on a scale in grams, then converted all of the measurements to kilograms)

Spring Constant:

Red= 84 N/m

Blue: 112 N/m

The following are the equations for finding the placement of where to put the blue and red carts.

Blue:

Ek=Eel

1.2mv^2= ½ kx^2

 ½ (.5394)5^2= ½ (112)x^2

.067425=56x^2

.001204m =x^2

.03469m = x

.035m or 3.5 cm=x

3.5 cm is the distance the spring is compressed for the blue and 4 cm is the distance the spring was compressed for the red. Our group solved for x by setting the Ek and Eel equations equal to each other.

Red:

Ek=Eel

1.2mv^2= ½ kx^2

½ (84)x^2=1/2 (.55)5^2

42x^2= .275

x^2= .0016369 m

x= .04045868

x= .04m or 4cm

We went to test the percent error, and the recorded velocity of both carts was .44m/s. The carts went exactly the same velocity with the distances we predicted, meaning we had a 0% trial error. :) This challenge taught us how to use formulas to solve for a variable and how to manipulate data. Hank and I solved for the distance, converted units, found the mass, and applied to Ek and Eel formulas we learned in the FINAL challenge of the year.

Unit Summary Blog

Unit Meta-cognition: 

•  impulses of two things colliding are always equal, the momentum is always equal to the impulse, but the acceleration changes
•  In order to follow Newton's 3rd law, things must move equally fast
•  when two objects end together, use the equation: MaVa+MbVb = (Ma+Mb) Vab
• When two objects end apart, use the formula p total Vi= MaVb +MbVb
•  When 2 things collide, the momentum is equal and opposite, therefore have equal impulses.
•  P represents momentum
•  J represents impulse
•  the change in J= change in P
• increasing the time will only affect the force if the J's are constant
• the impulse (J) is the same when the change in momentum is constant (smaller force= safer)
• the units of measurement for the law of conservation of matter is kgm/s
• Law of Conservation of Matter says that Cart A+ Cart B =0 (ptotal before= ptotal after)
• Why do airbags always keep you safe?
• the J's are always the same
• change in P= mass times change in velocity
• increases time because the J's are the same, by increasing the time it decreases the velocity
• the velocity goes from change in V to 0 

 What is impulse?
The simplest answer is that impulse is the change in momentum and the force upon something and how long the force is applied.
You can solve for momentum (P) using this formula: (Keep in mind if there is a negative sign in this equation, then it means the impulse was going in the negative direction.)

Δp = mΔv

Newton's Third Law:
For every action, there is an equal and opposite reaction. 
Fa= -Fb

Newton's Second Law: The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.

Conservation of Momentum Equation:

What is momentum?
Momentum is the motion of a moving body, measured as a product of its mass and velocity.

Impulse = Change in momentum
 

Example Problems:
1. Why do Otterbox cell phone cases prevent cell phones from breaking when they are dropped?

•  When a collision occurs, an objects experiences a force for an amount of time that results in mass undergoes a change in velocity, that therefore results in a momentum change. The time increases because the J's are the same. The J's are the same, because the change in velocity is the same. ( change in P= mass times change in V)

Why do Climbers use stretchy ropes?

Stretchy, nylon ropes are used by climbers because the momentum is slowed by the ropes, therefore preventing the climber from falling. The ropes are made to be stretchy so if the climber falls, it will apply a force upon the climber for a longer period of time. By increasing the time that the climber would be falling, the force exerted on the climber is reduced. Therefore the ropes increase the stopping time and decrease the stopping force.


How do air bags keep you from getting hurt?

•  Change in momentum is the same regardless of whether you stop quickly or slowly
• impulse = force times time
• The car will go from moving to not moving no matter how the car is stopped. Therefore the change in momentum is the same. The impulse will also be the same regardless of how the car stops. ( change in P= impulse ) 
• The airbag increases the time of the impulse, therefore the force on you is less. (Small force = less injury)
• If the change in P=J, the change in momentum is the same, therefore the impulse is the same too. Since the J is the same, the airbag increases the time of the impulse. This cause the force to decrease. The smaller force keeps you safe.

Impulse Momentum Theorem:
A 2kg block slides at 4 m/s. What is the momentum?

p=mv
p= 2 (4)
p= 8 m/s


A gun fires a bullet. On which object is the force greater?
 The force on both is the same because of Newton's 3rd law.

On which object is the impulse greater?
 The change in P equals the J, so the impulses are equal because the force and time is equal.

On which object is the change in momentum greater?
 The change in momentum is the same. No matter how fast the bullet moves, the change in momentum would always be the same because impulses=force times time of force applied.

Which object undergoes a greater acceleration?
 Because of Newton's 2nd law, the bullet undergoes a greater acceleration because it has a smaller mass.

Rocket Challenge: Elise, Nikki, and Will

Materials:

• rocket
• timer
• cap
• wooden stnd
• nails
• scale
• measuring tool
• wooden triangle

Variable:

• Horizontal:
• change in X= ?
• t= 3.84s 
• Vi= ?

• Vertical:
• A= -9.8 m/s
• t=3.84s
• Vi= ?
• Vf= ? 

Project Description:
 We are assigned an angle. In order to collect data, we launched the rocket at  10 degree angle difference from the angle we are trying to solve for. Our groups goal was to predict the displacement of the rocket if it was launched at a 35 degree angle.

Video:
Here is a link showing a rocket being shot vertically into the air. In our experiment, we shot the rockets on an angle horizontally. During our experiment, our video cut our rocket out of the shot but this one can be used for understanding of the experiment.

https://www.youtube.com/watch?v=x3Xh00Jn8Bs 

First we made a graph of the velocity on an angle and solved using cosign to find Vix and Viy. All of the calculations below are for a 35 degree angle.

cos 55 = adj/hyp
cos 55= 10.94/x

x= 10.94/ cos 35

Vi= 19.06

cos 35= Viy/ 19.06
Viy= 15.6

 Next, our group made calculations for  25 degree angle.

cos 65= Vix/ 19.06

Vix= 8.05

cos 25= Viy/ 19.06
Viy= 17.27

Next, we plugged in the Viy to the displacement equation to solve for time.

change in X= 1/2at^2+ Vit
0= 1/2(-10)t^2 + 17.27
-5t^2+ 17.27t + 0

Since this equation cannot be factor or simplfiied further, we plugged into the quadratic formula and got t= 3.45.

Next, we plugged in the Vix into the horizontal equation for displacement.

8.05 = change Xx/ 3.45

change in Xx= 2.78m

We found the precent error by subtracting the actual from the predicted measurement, and then dividing it by the predicted.
27.2 - 29 / 27.2 = 6.6% error

Projectile Motion Summary

Over the course of the unit, I have collected meta notes on each worksheet or activity that we did.

·      I learned that when you solve for the time, plug in Vix for initial velocity.

·      Always plug the acceleration or displacement in as a negative number.

·      When a problem does not give you the Vi in the horizontal or vertical direction, split the velocity into components and use cosign to solve for the Viy and Vix.

·      Remember:

o   How long/ how far: displacement (change in X)

o   How fst: velocity

·      The only formula that can be used in the horizontal direction is:

Vx= Change in Xx

              T

Rules:

1.     If an object starts at rest or is dropped then the Vi is 0.

2.     When something is falling down in the vertical direction, the acceleration and displacement is negative.

3.     The only force acting on an object falling is gravity, therefore the acceleration is negative because gravity is going downward.

4.     Whenever an object is at it’s highest point, before it turns around and accelerates downward, the velocity is 0 at the highest point.

5.     The only formula that can be used in the horizontal is Vxx equals change in Xx over change in time.

6.     When an object is thrown up and the returns to the ground vertically, the displacement is always 0.

7.     If the path of an object in moving upward, the displacement is positive

8.     If the path of an object is moving downwards, the displacement is negative

9.     Always use cosign to solve for the components of Vix and Viy

10. When using the quadratic formula, always use the positive answer for time.

Formulas:

o   x= 1/2at^2+viy(t)

o   vf=at+ vi

o   vf^2=vi^2+2(acceleration)(change in x)

o   a= change in v/ change in t

Key:

Here are the models the apply to the vertical or horizontal sides. When doing a goalless problem, here are all the variable to solve for:

Vertical:

o   -Unbalanced forces

o   CAPM

o   UFPM

o   Acceleration

o   vf=at+ vi

o   x= 1/2at^2+viy(t)

o   vf^2=vi^2+2(acceleration)(change in x)

Horizontal:

o   BFPM

o   CVPM

o   Constant velocity

o   Vx= Change in Xx

            T

Example:

NASA launches a rocket horizontally with a 7.0 m/s off a crater on the moon. How far will it travel horizontally before it strikes the ground outside the crater 15m below? (Disregard the lack of gravity outside of earth).

First, write all the horizontal and vertical knowns and unknown. As we solve for each variable we will go back and plug in what we found.

Horizontal:

o   Vi= 7m/s

o    Displacement= ?

o   t= ?

o   a= -10m/s

Vertical:

o   displacement= 15m

o   a= -10m/s

o   Vi= 0

o   T=?

First, use the following formula to solve for the time.

Change in X= 1/2at^2+Vit

-15= ½(-10)t^2+0t

-15=-5t^2

  - 3         -3

t= 1.73s

Now plug in the time into the horizontal equation to find the displacement.

Velocity= change in X

                    Time

7= X    

      1.73

displacement= 12.11m

Now plug in the variable solved for into the horizontal and vertical lists.

Horizontal:

o   Vi= 7m/s

o    Displacement= 12.11m

o   t= 1.73s

o   a= -10m/s

Vertical:

o   displacement= 15m

o   a= -10m/s

o   Vi= 0

o   T= 1.73s

Describe the motion of the rocket in the x-direction and explain why it moves in that manner.

The rocket moves in the x directions with no acceleration and the only force acting on it is gravity. Gravity only affects the vertical direction.

Describe the motion of the rocket in the y-direction and explain why it moves in that manner.

It drops 15m and moves downwards with acceleration of 10m/s, with only the force of gravity acting on it. There is an unbalanced force of gravity, which causes it to speed up.

The picture above is a projectile motion map, showing the projectile motion of a ball being thrown. The ball is thrown upward in its path, reaches the top, and then changes direction and accelerates downward.

Projectile Motion in Real Life:
Projectile motion can be applied in engineering designs. In oorder to design machinery, you have to have a concept of gravitational acceleration and the ability to predict the path of motion usuing mathematical formulas. By being able predict the trajectory of objects, can help students complete labs more efficiently. At the conclusions of this unit, students had many assets under their toolbelts to solve labs. We could use Pasco capstone, different kinematics formulas to solve for a specific variable, or use a force diagram and scale to solve for different forces.

 

CAPM Practicum Challenge

a. First, we drew a visual representation of the ramp. The materials we used are two wooden blocks, motion detector, ramp, ruler, cart, chalk, metronome, and the table. We used the block to prop up the table and create a ramp that is on an angle. 

b.  Then our group took a series of trials and marked the location of the back of the cart (so that the data was consistent) every second on the table with chalk. The metronome counted the seconds and indicated when we should mark. We did five trials and then entered the data into excel to make a graph. In column A, we entered the increments of time and in column B we entered the distance per second that the cart traveled.

c. In order to make the data linear, we square the B column. This also helped us find the equation of the line and to calculate our slope. The equation of the line is y=mx+b, so we plugged in the variables from our graph. Based on our graph, the equation of the line is
y=5.6958x+14.564.

Above is the table before we graphed our data. 

d. Lastly, we calculated the acceleration by using the formula slope=1/2a. In order to calculate this, we changed the units to meters. 

Our predicted acceleration is
.113916m/s^2. The prediction appears high because we calculated it in centimeters as opposed to meters.

To solve for t=4:

y=5.6958(4)+14.564

=.1056968 m/s^2 after 4 seconds

Conclusion:

I learned that by squaring the data you can make the line of the graph appear linear. You also can plug in an x,y coordinate pair to solve for b in the equation of the line. (y+mx+b)

The actual acceleration was .197m/s^2. Our group was not within 10%.