Momentum Review

Hello readers,

Momentum-
  1. a property of a moving body that the body has by virtue of its mass and motion and that is equal to the product of the body's mass and velocity; broadly :  a property of a moving body that determines the length of time required to bring it to rest when under the action of a constant force or moment
--Webster definition

Momentum=mass (kg)*velocity(m/s)
                 p =m*v
Whenever we write momentum, we must remember that it is on a vector, meaning we are required to indicate the direction, in both a sketch, which will say which way is positive and negative, and in our equations through the use of - to indicate going down or backwards on the x and y scale.

Any object in motion has momentum, however if it is at rest, it has none. This is clear on a free body diagram, where despite an object having a force due to gravity, and a normal force, if it is stationary, there is none due to momentum. The unit of momentum is kg per m/s.

Sample problem:
A bowling ball weighing 8 Kg is rolling along at 2 m/s, what is its momentum?
      p=m*v
      p=8*2
      p=16kg/(m/s)
Momentum = 16Kg/(m/s)
The next step in momentum is impulse. Impulse is derived from Newton's second law, which states that:
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.
--Physics classroom 
Mass is always measured in Kg.
Acceleration is: (Δv/ΔT)
 Forcenet=mass*acceleration
 wherein force and acceleration are directly proportional, and mass and acceleration are inversely proportional.
 This means that if acceleration increases, so does force directly. And if the force stays the same, but the mass increases, acceleration decreases in a direct manner.

 The unit of impulse (J) is newtons, and it is the force times the time. Impulse=change in momentum.
 Thus we have the formula:
Impulse=(forcenet*change in time=(mass*acceleration))*change in time.)
Thus, j=p
which means fnet*Δt=m*(a=(Δv/Δt)).
 So our final impulse formula is:
j=fnet*Δt

 Sample problem:
What is the impulse of a 1800 kg car that collides with a fencepost weighing 20 Kg for 3 seconds?
First, to find fnet I multiply 1800*10 which is 18000
-----------> j=fnet*Δt
                  j= 18000*3
                  j=54000N

Now that we know what momentum and impulse are, we can learn the new formula's which take advantage of this understanding in collision. However there is a mathematical model that we base these formula's on.
This is p(total before)=p(total after) as energy must be accounted for throughout.
There are 2 types of collisions:
Unstuck--> unstuck  =  mava+mbvb=mava+mbvb
 Unstuck-->stuck      =  mava+mbvb=(ma+mb)*(Va*b)

We must remember that the force in which two objects hit each other is always equal due to newton's third law, however the repercussions on each other in terms of momentum can be different.

Sample problem:
Unstuck to stuck
If a 150 kg skateboarder travelling at 15 m/s hits a 275 kg marble ball at rest, what is the final velocity of the two travelling forward after they collide?
mava+mbvb=(ma+mb)*(Va*b)
150*15+275*0=(150+275)*(vab)
2250= 475*Vab
2250/475= (475*Vab)/475
Vab=4.7 m/s.

Unstuck to Unstuck 
A 2500 kg blue car is travelling forward at 40m/s, and collides with another 1500 kg red car traveling 16 m/s, what is the red car's speed after rebounding off each other if the blue car slows down to 8 m/s?
Formula: mava+mbvb=mava+mbvb
 (2500*40)+(1500*16)=(2500*8)+(1500*vb)
                         124000=20000+1500VB
              124000-20000=20000+1500VB-20000
                         104000=1500Vb
                104000/1500=1500Vb/1500
                     69.33 m/s=vb
The red car speeds up to 69.33 m/s.

Using energy to predict velocity

This lab was designed to see if we could cause two carts to travel at the same velocity while only being given the spring constant of both springs (red and blue).
We labeled our data according to spring color.

The two formula's we used were:
Ek=1/2*m*v^2
Eel=1/2Kx^2

Red:
K=84
M= 0.55 Kg

Blue:
K=112
M= 0.5422 Kg


Once we had this data, we set a compression to red, of 5 cm, 0.05 M, and then input it into the formula to solve for energy:
Eel=1/2Kx^2
Red=0.5*84*0.05^2
       = 0.105j
and then we set the energy equal to solve for the Blue compression:
blue=0.5*112*x^2
0.105=56x^2
x= 0.043m

To then solve for velocity, we input this data into the formula:
Eel=Ek because inital energy= final energy
so
Both red and blue Eel=0.105 j which means 0.105j=EK
Thus to solve for velocity we input the energy (0.105j) into the Ek position.
 0.105j=1/2*m*v^2
Red:
0.105j=0.5*0.55*v^2
0.105j=0.275*v^2
0.105j/0.275=(0.275*v^2)/0.275
0.381=v^2
v= 0.617m/s

Blue:
0.105j=0.5*0.5422*v^2
0.105j=0.2711*v^2
0.105j/0.2711=(0.2711*v^2)/0.2711
0.387=v^2
       v=0.622m/s


To double check our blue compression data, we input the velocity of red (0.61 m/s) into the formula to find our Ek:

Ek=0.5*0.5422*0.617^2
Ek=.103j

Which when used as part of the Eel=Ek formula means that Eel=.101.
Thus, when we input this into the Eel formula, we solved for the compression of x for blue.
 Eel=1/2*k*x^2
.103=0.5*112*x^2
.103=56*x^2
.103/56=(56*x^2)/56
 .0018=x^2
 0.0424=x

When we tested the carts, we pulled the cart with the red spring 0.05m (5 cm) back, and the cart with the blue spring back 0.042m (4.2 cm).
We used pasco as our testing software, and our red cart came in at 0.56 m/s for velocity, while the blue cart came in at 0.61 m/s for velocity.
This resulted in a percentage of error for our red cart of (error %=(0.61-0.56)/0.61) 8.1% error.

Energy Transfer model

Welcome, one and all to energy transfer principles.

The formula's used are
Ek: 1/2 m*v^2
Eel: 1/2kx^2
Eg: m*g*h
W: fnet*deltax

Across charts, we use the idea that energy is a closed system, except for Ediss, where once it disappears to, it is gone.
This results in formula's such as
Ek=Eg, where one variable needs to be solved for, or a time concept, or a force present can be used.

The other concept is joules, these principles of energy that when indicated by time(seconds) become watts, and in hours kwhr's.
Across energy transfer, we can use pie charts which are conceptual, to LOL charts which are more precise. One of the necessary factors to remember is that in the system which defines the energy entering or leaving, if we are using earth's gravitational force, it must be included.

If we have a 150 kg skateboarder going at 3 m/s, when friction is negligible, what is his Ek?
Ek=1/2 m*v^2
     = 0.5*150*3^2
     = 675 j

This same skateboarder accidentally goes off a cliff whose height we do not know.
To solve for Height we can use:
Ek=Eg
675j=M*g*h
675j=150*10*h
675j/1500=1500h/1500
h= 0.45 M.

If a spring with a constant K of 115 has 280 j of energy stored, how far was it compressed?
  Eel=1/2kx^2
280j= 0.5*115*x^2
280j=57.5x^2
280j/57.5=(57.5x^2)/57.5
4.87=x^2
x=2.2 m

Unknown weight.

The goal of this challenge was to solve for the additional weight on one can, and being allowed to gather the velocity of the cart's and the mass of both cart's without the addition of the can.

In order to solve for the unknown additional weight on one of our two carts, we used the stuck to unstuck formula as our carts were sprung from a joined position. This is 

                stuck=unstuck

m(a+b)*v(ab)=mava+mbvb.

Our initial given's were the mass of each cart on its own (0.5Kg), and we solved for speed using pasco capstone software.

In order to do this we took the formula m(a+b)*v(ab)=mava+mbvb and added the pertinent info resulting in:

m(500+ (500+x)*v(ab)=mava+mbvb. The initial velocity was 0, and cart a's final velocity was -55, while cart b's final velocity was 0.24. This resulted in the further change to the formula (1000+x)*(0)=(500)(-55) + (500 + x)(0.24). This further resulted in 0=-278+220+0.24x. Continuing on this series of equations 0=-158+0.27x, which simplifies to 158=0.24x, and when we divide both sides by 0.24, it results in .658Kg =x. 

We predicted it would be .658 kg, and it turned out to be .609 Kg. We were within 10 percent error rate as (.658-.609)/.609=8.05 % error.

Rocket prompt

Hello dear readers,
In this rocket challenge, our goal was to solve for where our rocket would land within a 10% error rating by using calculated the average velocity and a new angle degree that we had not previously tested. We chose the 40 degree wooden block as our final, but tested with 30 and fifty degrees.
The following formula's were used throughout our challenge:
V=change in x/change in time
change in x=1/2at^2+vit
SOHCAHTOA

In order to gather data, we used a stopwatch, a rolling meter stick, and two wooden blocks which caused a 30 and fifty degree angle.

The above velocities are only the velocities in the horizontal position, not the vertical.

Then when we continue, to solve time we use the Xy = 1/2(a)t^2 + Viy(t) = 0= (0.5*-10)t^2+15.4t = 0=-5t^2+15.4t, where b is 15.4, and a is -5to solve for time, inputting it into the quadratic formula: 
x = [-b ± √(b² - 4ac)]/2a
 t = [-15.4± √(-15.4² - 0)]/2(-5)
 t = -30.6/-10
 t = 3.082 s

So to find our final distance traveled, we can use vx=change in xx/change in time.
This becomes 12.25=delta (change in)x/3.6
 to continue this concept, 3.082*12.25= 37.75 meters. 

Our final data was that we were at 40.4 meters, and potentially attributed to either strong wind or miscalculations, resulting in a percentage error of (40.4-37.75)/40.4=6.5% error.

This is basically the same as previous, same equations, but instead we divide it into viy and vix vectors.

Rocket post

In this challenge, we were attempting to solve for how far our ball would go, after having solved for velocity at the 30& and 50 degree angles in order to solve for the 40 degree,
In order to do this, we collected the following data:

This leads to our estimates of



This leads to the math of

Video Analysis of moving objects

Have you ever wondered what the video analysis of a ball thrown through the air would be?
Prepare to be amazed, because here it is.

Here is our ball's track, the same video done twice:

These captions show the path of our ball over the course of its flight. One of the things that needs attention is that we have the meter stick, which gives a frame of reference for the flight of the ball of 1m.

This is the graph that shows both our change in position in the x direction (green), and in the y-direction (red). We can see how much distance it covers, how long it takes, and why.

When we ran our velocities, we found the following data:

This graph shows the velocities, and thus the accelerations of our graphs. In the x direction, we can see that the displacement is very low, and our graph peaks then seems to come back to almost the same as when it started. However in the y direction, our ball seems to increase its speed for the first few seconds, but then it comes to reaches a crescendo, the peak, and slows down until it hits zero, which is the point where it starts to descend.

A and B. 

Our acceleration in the y direction is 9.81 m/s^2 due to the force of gravity, and our acceleration in the x direction is 0m/s^2 as x-direction does not have an acceleration capacity, rather an ability to exist in that of constant velocity as the change in v evens out, making it constant velocity.

C & D. 

Our initial velocity in the x-direction is 5 m/s which remains constant the whole way through, but our initial velocity in the y-direction 3.09 m/s as that is the initial slope on the position vs time graph for the y direction.

E&F. 

The velocity at the top of our path in the x direction is 5 m/s, while the velocity at the top of our path in the y direction is 0 m/s as that is the point where it is about to turn around, thus reaching standstill for a moment.

G&H

The final velocity in the x direction remained the same, 5 m/s as it was travelling at a constant velocity, but had reached -9.65 on the last part of the graph.

                                                   I&J.

In order to discover the height of the ball, we can see where it was on our position graph, and it was 1.8m. The distance our ball traveled can found using x_final-x_initial=change in x. In this case, our change in x was 2-0.4=1.6 m.

K&L

It took 2 seconds for the ball to reach the top of its path (27.5s-25.5s), and it

Remained in the air (29.15-25.5) 3.65s. 

Conclusion:

1. Vertical accelerations are constant at 9.81 m/s^2 as a result of the force of gravity, but horizontal acceleration is 0 m/s^2, as it is travelling at constant velocity, not acceleration. I know this both through equations, and also the visible aspect of our data. 

2. For Vertical formula's, I can use the Δx = 1/2 aΔt² + ViΔt formula, but for the horizontal, the only formula we can use is v = Δxv/Δt, and let that solve it. In order to do so, I would solve for time through the change in x formula, and that input that into the velocity formula. These two formula's work together, not against each other. 

3. In order to solve for the displacement, I would use the Δx=xf-xi, and find it from there, but the vertical formula is a longer verison that has 0's in it to cancel it out. 

4. In order to solve for height, we would use the Δx = 1/2 aΔt² + ViΔt, input time, and then divide our x by two, as it would take equal amounts of time to fall to its peak, and to fall to the ground. 

5. At the top of its peak, there is only the horizontal velocity, as the vertical velocity is nonexistent. 

Cart and Ball post

UFPM Challenge

A modified Atwood machine:

[Image upload of diagram]

In this lab, we were attempting to calculate two variables. The constant velocity of our moving cart, and the rate at which our cart on the track was accelerating due to the hanging weight which was attached.

Our cart’s track had no angle, so all we needed to find was the rate at which our weight (50g, 0.5n) was causing it to accelerate.

Thus, the following force diagram was created.

Please note the lack of friction, thus causing our diagram to be Unbalanced.

Thus, our diagram becomes an unbalanced force particle model. In order to solve for such a situation, we use the a(acceleration)=f_net/m(kg)

Thus, our following formula became a=0.5/.6394

As a result, a=.78 m/s²

We measured how far our change in x was, we discovered that it was 90 cm which equals .9m. When we input our known variables into the Change in X=vit+ at².

The final formula to solve for time is .9=.78t². As a result, our time was 1.15 s.

Part 2 was testing the velocity of our cart on the ground, and thus we solved for the velocity of our cart, finding that it was 0.3 m/s. Thus we multiplied 0.3*1.15s ending up with 34.5 centimeters away.

In order to test our conclusion, we let our ball drop when our cart had reached 34.5 cm’s away from the meeting point.

Once we tested our predicament, it was found to be correct.

UFPM Blog

Unit Summary UFPM

This unit covered a vast amount of information in regards to unbalanced forces but was also really exciting. Examples of such forces are all balls in the air, as only gravity acts upon them, all accelerating objects, and anything really that is not maintaining a constant velocity.

The formula’s used were:

1.       vf = at + vi

2.       vf² = vi² + 2aΔx

3.       a = F_net/mass (in kg)

4.       Δx = 1/2 aΔt² + ViΔt

5.       a = Δv/Δt

We started this unit exploring what happens when there is more force going one way, than holding it back, by looking at what happened if friction, or air resistance were not factored in, or if they were, that they were not equal in force.

Examples of such scenarios are cars that are speeding up or slowing down. Part of this was learning how to solve for the frictional force, as shown below.

If a car that weighs 240 kg with an F_push  of 300n, and and F_F of 80, the following diagram looks like:

In order to solve for acceleration, we must use the a=fnet/m formula as shown below. 

We have solved for acceleration, it is .91m/s^2

If we know that a parachutist covered 1500 meters before expanding his parachute, from rest, how long did it take him?

                     It took him 17.3 seconds to fall 1500 meters.

Now, what is his final velocity right before he opens his parachute?

Thus it is 173m/s before he pulls his parachute cord open. 

This is the summary of our blog, if you have questions, please feel free to comment below. 

Collision? Where?

In this challenge, we were seeking to understand where two carts, going at different speeds would collide starting from 3 meters apart.

In order to solve the challenge, we measured the rate of acceleration (5 trials averaged) for both carts and then inserted the following data into a formula we had learnt, X = v_ot + (1/2)at^2, and set the x's equal to each other.

Our first cart had an acceleration of .0909 m/s^2, and the second cart had an acceleration of .137 m/s^2.

Thus,when you set the formula's equal to each other:

1/2(0.909)t^2 = 1/2 (-0.137)t^2=3

Which ends up being:
0.4545t^2+ 0.0685t^2=3

Which ends up with 3 meters ( our displacement) equaling 0.4545t^2 + 0.685t^2, which ends up being 0.11395^2. 

Thus, 3=0.11395t^2, divided by 0.11395 on both sides, we find t^2=25.9, wherein the square root of both sides is t=5.08s. Thus, when we let both of them run for 5.08 seconds, our prediction results in an expected collision of 1/2(0.909)(5.08^2)= collision.
Thus our collision should occur 1.2 meters from the top, 80 cm from the bottom.
When we tested it, our cart ran 1.23 meters, coming in at 77 centimeters from the bottom.
This difference, of 3.75 degrees of error, is much less than 10 percent error.