MTM Challenge

MTM Challenge

Determine the mass of the unknown object in the cart.

First, we determined that the mass of the cart was 0.488 kg. We decided that, in order to predict the mass of the object, we had to divide momentum by velocity and then subtract that from the mass of the cart.
We needed to find the momentum and velocity, first.

We recorded the velocity of the cart and, after a few trials on the track with the cart, we determined that, without the object on top, the average cart's velocity was .524 m/s.
With the object, the average velocity was -.282 m/s.

We used the Law of Conservation of Momentum to plug in our data.

mava + mbvb = (ma + mb)vab
(0.488)(0.524) + mb(-.282) = 0
0.255712 + mb(-.282) = 0

mb = 0.907

If we subtract the mass of the cart from mb, we can find the mass of the unknown object.

0.907 - 0.488 = 0.419kg

So, we predict that the mass of the unknown object is 0.419kg!

Impulse Force Model Review

Impulse-Momentum Force Model

The Impulse-Momentum theorem is Fnet(change in t) = change in(mv). This formula can be used to calculate the momentum of an object in certain situations.

For example, if a 1000kg sports car is traveling at 30.0 m/s, you can plug those numbers into the equation to solve for the momentum.

P represents momentum. The equation can be represented as P = mv. So, if m = 1000km and v = 30.0 m/s, multiplying them will give you the P - value: 30,000 kg. 

Impulse is also relevant in this model. Impulse = Change in Momentum. (Change in P = J). 

Impulse is represented as J. So, the formula we can use is Change in P = m(change in v).

So, if we return to the sports car scenario, we can calculate the impulse needed to increase the car's speed from 30.0 m/s to 35.0 m/s.

Change in P = m(change in v)

Change in P = (1000)(v final - v initial)

Change in P = (1000)(34 - 30)

Change in P = 5,000 kg

Using a new formula, J = F(change in t), we can determine the forces acting upon the sports car in the scenario. If the car were to crash, we could find the amount of force the object the car crashes into exerts on the car.

If the car, traveling 35 m/s, crashes into a rock wall and comes to rest at 0.25 seconds, we can plug the numbers into the equation to find the force exerted on the car by the wall.

* The force becomes negative depending on which direction it is exerted In this situation, the J will be negative because the car's force on the wall is positive. So, the wall's force back on the car will be negative.

J = F(change in t)

J = Change in P = mv = 100(35) = 35000

- 35,000 = F(0.25)

- 35,000/.25 = F/.25

- 140,000 N = F

Newton's 2nd Law is also applied in this scenario. A common misconception is that the wall would exert more force on the car because it is bigger. However, according to N2L, the forces are the same. Because the wall has a greater mass, the effect of the crash will be greater on the car. 

Diagrams can also be used to help visualize Impulse Force Model scenarios. 

For example, if the scenarios is that of a car and truck crash, we can examine the situation and create a momentum conservation diagram. If the truck is 5000 kg and rear-ends the 1200 kg car that had been traveling at 13 m/s, the truck would be slowed from 14 m/s to 12 m/s and the car would speed up.

Initial object/mass/velocity:

Final object/mass/velocity:

The momentum conservation equation would be mava + mbvb = mava + mbvb. Because we are comparing the initial event with the final event, we set them equal to each other.

mava + mbvb = mava + mbvb

(500)(14) + (1200)(13) = (500)(12) + (1200)(v)

70,000 + 15,600 = 60,000 +1200v

85,600 = 60,000 + 1200v

25,600 = 1200v

v = 21.3 m/s

When the momentum is larger, the change in velocity will be smaller and, when the momentum is smaller, the change in velocity will be larger.

M(change in v) = m(CHANGE IN V)

When comparing the truck and the car, we can determine that the change in momentum and the force of impact are equal because of N2L: 

"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."

F = Ma and F = mA

The acceleration can be found using the formula a = Fnet/m.

Energy Challenge!

Energy Challenge

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

We began by determining the spring constants for each spring.

Red: 84 n/m

Blue: 112 n/m

From there, we made an LOL chart to find the mathematical equation to use in order to find the amount of energy in the system. 

Then, we determined that Eel = Ek would, in relationship to the formula, be equal to 

1/2kv^2 = 1/2mv^s

The Eel and mass for each spring was:

Eel = 1/2kv^2

Eel = 1/2(84)^2 = 3528J

Mass = 550g = .55kg

Eel = 1/2kv^2

Eel = 1/2(112)^2 = 6272J

Mass = 539.4g = .5394kg

With this new information, we were able to plug everything into the Ek formula. We gave the system a predicted velocity of .55. 

Ek = 1/2mv^2

Ek = 1/2(.55)(.55)^2 = .0831875J

Ek = 1/2mv^2
Ek = 1/2(.5394)(.55)^2 = .08158425J

Now, to predict the amount each spring must be compressed, we re-tried the Eel formula and plugged in the new numbers we found in the previous formulas.

Eel = 1/2kx^2
3528 = 1/2(.0831875)x^2 
X = 69.42935806 = 69.4m


Eel = 1/2kx^2
6272 = 1/2(.08158425)x^2
X = 79.19570195 = 79.2m


From this information, I can say that the red spring should be compressed 69.4m while the blue spring should be compressed 79.2m. Unfortunately, I was unable to test the springs on the track but the calculations seem a bit too high.

Energy Storage Model - Unit Summary

Energy Storage Model

Unit Summary

The Energy Storage Model is used to observe how energy is used and transferred. Energy is a conserved, substance-like quantity with the capability to produce change. It does not come in different kinds but, rather, different forms. 

Eg - Gravitational Energy

Eel/Esp - Elastic Energy/Spring Energy

Echem - Chemical Energy

Ek - Kinetic Energy

Etherm - Thermal/Heat Energy

Ediss - Dissapated Energy

To represent the transfer of energy in a graph, we use pie charts and LOL charts. 

Pie charts are qualitative representations of the energy in a system. The size of the each piece of the pie chart indicates how the energy is distributed throughout the event. The size of the circle itself represents the total energy of the system. 

For example, if an object rests on a coiled spring and, then, is launched upwards, you can represent the event with three pie chats.

For an LOL chart, you can use energy bar graphs to display a quantitative analysis of the event. The initial energy is represented with a bar graph showing the amount of energy stored in the beginning of the event. Then, in the middle, a circular flow chart indicates which items are inside and outside the system. The last bar graph represents the energy transfer at the end of the event. By using the information from the bar graphs, you can come up with a mathematical expression to represent the event. 

For example, if a moving cart travels 5.0 m/s collides with a spring, you can use an LOL chart to discover the maximum compression of the spring. 

If you choose to exclude the spring from the system, the chart will look a lot different. An outside force would be called a work force.

Hooke's Law

Concentrating on Eel, you can find the Eel stored in a spring with Hooke's Law. 

Eel = 1/2KX^2

The spring constant can be found when observing a graph measuring force vs. displacement. The rise/run equals the spring constant. N/M can also give you the spring constant because rise/run is, on a graph, measuring the N and M. 

Other important equations are:

Ek = 1/2mv^2 to find the kinetic energy 

Eg = mgh to find the gravitational energy 

Eth = FfrX to find the thermal energy 

W = Fx to find the work force when a force is acting outside the system 

P = Etrans/t to find the amount of energetic power

J/S = Watts

Power is different than energy. In terms of calories, Power is what calories are converted into in order for you to be active. It can be found by, first, determining the Ek in an event. 

For example, if someone eats a 700-calorie lunch and radiates about 100 J per second, you can determine how long it will take them to radiate away their lunch with the Power equation. 

If 1 food calorie = 4186 J, then the person transferred 2,930,200 J of energy. Dividing this by 100 (J/s) will give you the amount of time it will take to radiate away the lunch. 29, 302 seconds or 8 hours.

By using the graphical representations as well as the equations, you can solve numerous word-problems and physics situations to find the values of different energies. 

Rocket Challenge

Rocket Challenge! 

Move the target to predict where the rocket will land.

The greatest challenge my group faced in calculating the rocket's landing was not in the setup but, rather, the actual mathematic calculations. Initially, we placed the angels of the rocket on the diagram in the wrong places. The rocket's stand was inclined backwards rather than forwards. So, we had to go back and adjust our calculations and re-test the launch. Once we re-calculated, we got a more realistic prediction. 

Initial Data: 

Change in Xy = 0m (Starting from the ground and ending on the ground)

a = 10m/s (Constant in the vertical)

Fg = 9.8

We had to solve for Vi and 25 degrees, which was the angle of launch we were predicting. 

We began testing at 40 degrees.

Average: t = 3.74 and x = 49.05

With this information, we could see that the change in Xx = 49.05 and t = 3.74.

Then, we were able to plug this information into an equation and find Vx.

V = Change in Xx/Change in t

V = 49.05/3.74

Vx = 13.11 m/s

Next, we created a graph to calculate the velocity.

Cos = adj/hyp

Cos 50 = 13.11/hyp

hyp = 13.11(cos(50))

hyp = 20.40

V = 20.40

Then, once we found the velocity, we could create a new graph to find the initial velocity of x and y.

Cos 65 = Vix/20.40

Vix = 20.40 (cos(65))

Vix = 8.6

Cos 25 = Viy/20.40

Viy = 20.40 (cos(25))

Viy = 18.5

Vix = 8.6

Viy = 18.5

Once we had all the variables, we were able to plug them all into an equation to find the change in time. However, my calculations seemed to turn out incorrect.

Change of X = 1/2a (change in t^2) + vi (t)

0 = 1/2(-9.8) (change in t^2) + (18.5)(t)

-18.5 = -4.9 (change in t^2) + t

*-13.6 = change in t^2 + t

Initially, with our incorrect calculations, we predicted the rocket would land at 23.13 meters. However, it actually landed at 36.5 meters. 

UFPM Challenge - Atwood Machine

UFPM Challenge!

Atwood Machine Experiment

Mission: Land the mass from the modified Atwood Machine on the constant velocity cart.

We started off by finding the total weight of the system: 

The weight of the cart + the 300 g added = 838.1 g, the total weight of the system.

We then converted grams to newtons and created a free-body diagram to better visualize the forces acting upon the Atwood Machine. 

838.1/1000 = .8381 kg x 10 = 8.381 N

 

If we use the formula a = fnet/m, we can find ft as well as the acceleration because fnet = ft since ft is the only force moving the cart. 

a = .507 N/.8381 kg

a = .605 m/s^2

ft = 50.7 g/1000 = .0507 kg x 10 = .507 N

Next, we took data on the constant velocity cart to see exactly how fast the cart travels.

We found:

Sec  |  M

0       .22

1       .53        

2       .84

3       1.16

4       1.47

5       1.83

Using this information, we were able to calculate the constant velocity: .30 m/s^2

v = change in x/t

.30 = .53 - .22

(.31) .30 = .31/t (.31)

t = .093 m/s^2

Vertical distance = 81 cm

Vi = 0 cm

Change in position = 81 cm/100 = .81 cm

Using another kinematics formula, we can find t.

Change in x = 1/2a (change in t) + Vi(change in t)

.81 = 1/2(.605)t^2 + 0(change in t)

.81/.3025 = .0325t^2/.0325

Take the square root of 2.67768595 = The square root of t

t = 1.64 s

So far, we know:

a = .605 m/s^2

Change in position = .81 cm

t = 1.64 s

Vi = 0

Since t shows the amount of time it will take the cart on the ground to travel .81 m, we multiplied 1.64 by .30, the velocity, and found that it will take .492 m for the cart to align with the dropping weight!

UFPM Unit Summary

UFPM 

Unbalanced Force Particle Model

Let's talk about the Unbalanced Force Particle Model! UFPM!
This model will help us to solve for unbalanced forces, even when we don't know the acceleration.

First, let's take a look at Newton's 2nd Law. It tells us that acceleration is directly proportionate to force and inversely proportionate to mass. Using this info, we can find the acceleration, or "a", of any moving object by using the formula a=fnet/m.

Right about now, you might be wondering what an fnet is. Well, fnet stands for net force. You can find the net force by adding together all the forces acting upon your object.

For example: If an 80 kg skier is standing still, waiting for the ski lift, there will be 800 N (Convert kg to N by multiplying the kg amount by 10) of negative gravitational force pulling down on the skier and 800 N of positive normal force pulling up on the skier. So, the net force will be 0.

-800 N + 800 N = 0 N

Now that we know the net force of the skier, we can calculate their acceleration!
Using the formula a = fnet/m, just plug in the amount of force in the fnet and the amount, in kg, of the skier.

a = 0 N/80 kg
a = 0 m/s^2

In this scenario, the force is vertical. However, if the skier were to be observed while they were skiing down a hill, the net force would change.

If the ski slope our skier is racing down a 20 m slope slanted at a 65 degree angle, we can use trigonometry to find the net force and then find the acceleration. In this scenario, the fgx is the fnet because it is the only force acting upon the skier as they go down the slope.

COS (25) = 80 kg/fgx
fgx = 80 (COS (25) )
fgx = 72.50 kg (10) = 725 N

Now that we know the the fnet, we can find the acceleration using our previous formula.

a = fnet/m
a = 725 N/ 80 kg
a = 9.06 m/s^2

Since we know the basics now, we can move on to solving for many other variables. Using kinematics formulas and Newton's 2nd Law, we can solve for acceleration, mass, net force, initial velocity, final velocity, change in time, and displacement.
Here are the equations:

vf = a(change in t) + vi
change in position = 1/2a(change in t^2) + vi(change in t)
vf^2 = vi^2 + 2a(change in position)
ff = coefficient of kinetic friction(fn)

If the skier is traveling at 4 m/s when they begin to descend the slope, how fast will they be going when the reach the bottom of the slope 20 m later?
We know:
vi = 4 m/s
change in position = 20 m
a = 9.06 m/s^2
m = 80 kg

If we use the kinematics formula vf^2 = vi^2 + 2a(change in position)  we can find the value of vf.

vf^2 = 4^2 + (2(9.06))(20)
vf^2 = 16 + 362.4
vf^2 = 378.4
Take the square root of 378.4 to get a vf of 19.5 m/s

We can also find the amount of time it took the skier to descend the mountain using another kinematics formula.

vf = a(change in t) + vi
19.5 = 9.06(change in t) + 4
15.5 = 9.06(change in t)
1.71 = t... approximately

That's a fast skier!
Unless, of course, the snow they are skiing on has friction.

If the coefficient of friction is 0.08, we can find the amount of friction acting upon the skier!
But, first, we have to return to trigonometry to find the fgy which is equal to the fn.

COS (65) = 80 kg/fgy
fgy = 80 (COS (65) )
fgy = 33.81 kg (10) = 338.1 N = fn

ff = coefficient of kinetic friction(fn)
ff = 0.08 (338.1)
ff = 27.05

*BONUS
Atwood Machines!

Atwood Machine Pro-Tips:

Weight = Pull (dangle-y weight)

Mass = Whole System (dangle-y weight + weight on the table)

a = fnet/m

fnet = weight of the hanger (dangle-y weight)

m = mass of system (dangle-y weight + weight on the table)

Everything else you can solve just like a normal object! Just be sure to make free-body diagrams for both weights to help visualize the forces acting upon the weights. 

LA FIN~