MTM Challenge blog post

MTM Challenge Blog post

For this challenge, nick and I had to calculate the mass of a can of soup. We were given 2 carts with a mass of 500 grams, the can, and two motion sensors.

To solve this experiment, we decided to keep the force constant and to measure the velocity of each cart,

Data and Calculations

Our plan was to calculate the total momentum before which was 0 because there was no movement,

the velocity of the car with no mass was .56 m/s and the velocity of the one with the can was 0.28m/s

Ptotal Before= Ptotal after

mava+mbvb=mava mbvb

.500(0)+ mb(0) = .500(56)+mb(-.28)

                        0= .28-mb(-.28)

                       mb= 1kg

the mass of cart 2 was 1 kg but since the mass of the cart was 500g we substracted that from 1 kg and got 500g for the mass of the can,

The can's actual weight is 478 grams so our prediction was really close, around 4% close. I think we are extremely close, and where we might have had an error was in the collection of the velocity, we took around 6 trials which might have not been enough, and next time we should take more data.

Video Analysis

Video Analysis

This picture is taken after the ball had already completed its course and the marks represent the position of the ball every 1/2 second

This shows our Position vs time, the yellow line shows the position on the X axis and the blue line shows the ball's possition on the Y axis

This is our Velocity vs time graph the blue line shows x axis and yellow shows the y axis

the direction of the acceleration on the y axis is negative because the only force that is being exerted on the ball is Fg so its pushing down (negative direction)

The acceleration of the ball in the X axis is negative becasuse of the way we oriented our axis, if the ball is being thrown to the left (like in the video) the acceleration would be negative

the initial velocity in the y axis is about 3 m/s because that is the initial point of our data and the Vi in the x axis is about -4 m/s

final velocity in both is 0 m/s because the ball finishes at rest

the ball was in the air for around 1.3 secs

Conclusions

the horizontal acceleration is 0 because the forces are balanced on the x axis but on the y axis the acceleration should be around 10 m/s^2

formulas displacement y: x=1/2at^2+vit

formulas displacement x axis: x=vit

formulas velocity y: vf^2=vi^2+2ax

formulas velocity x axis:  vf^2=vi^2+2ax

UBFPM challenge

UFPM Challenge

In this challenge, we were supposed to calculate the time that it would take a 50g hanger to drop from a table when it was being held back by a cart with a weight on it, and with this time calculate how far, cart with constant velocity would travel. and after calculating it we were supposed to make the mass land on top of the cart.

during this challenge we were not allowed to drom the mass until the moment of testing our prediction, but we were allowed to do trials with the constant velocity cart.

Data Collected

Mass of the system=0.6kg

Mass of weight=0.5kg

a=Fnet/M

a=0.5/0.6

a=0.833m/s^2

x=1/2at^2+vit

x=0.78

0.78=1/2(0.833)t^2

t=1.37sec

V=x/t

0.27=x/1.37

x=0.37m

Cart has to be 0.37m away from place of drop

Our prediction was made by calculating the acceleration of the hanger and then calculating how lon it would take it using the equation x=1/2at^2+vit and after that calculating how long the car would travel in that time.

we were exact on our prediction making the hanger land on the cart on our 2nd attempt.

Fan Cart Post

CAPM: Final Challenge

Challenge: Predict the meeting place of the two carts, within 10% of error

The data wer recorded, helped us calculate the average acceleration of the cart using the motion sensor determined the velocity at a certain time (1/2 second intervals) and gave us a V vs T graph and as we know, the slope of the graph is equal to the acceleration of the object, and this helped us determine where would the two cars meet.

a=(xf-xi)

         t

a=(0.61-0)

         4

a=0.1525m/s^2

 

we recorded the acceleration of cart A and compared it to car E, and using the acceleration (equation of the line) we found the intersection and find where they collide.

Δx = (1/2) * a * (Δt)^2 + V_it

Cart A:

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

^{Cart B:}

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

 

We plugged in the equations in the calculator and it found the intersection and gave us the distance away from the origin where the cars would collide.

the intersection was 29 centimeters away from the origin. and after running the test we found out that we were 3 cm off.

conclussion:

Unit 3:  Constant Acceleration

CAPM.1

On a position versus time graph, velocity is not shown explicitly but, the velocity of an object in a graph with a constant acceleration can be determined by doing two things.

Slope of the tangent

Velocity is usually the slope of a x vs t graph, but when there is an acceleration the slope is not calculable. So, to find the instantaneous velocity can be find by calculating the slope of the tangent of the point.

Mathematic Expression

The second method is from a velocity vs time graph, when you find the acceleration of the object (Δv/time) you can use a mathematic formula to find the velocity at any given time. This formula is v=at+vi

CAPM.2.

Displacement can be found from two ways in a velocity vs time graph.

Area Under the Line

To find the displacement in a constant acceleration, velocity vs time graph you can calculate the area of the shapes under the line

Mathematical Expression

The other way is using the matemathical expression of the line to find the displacement. x=1/2at^2+vi

CAPM.3

There are four ways to calculate acceleration

Slope of a v vs t graph.

Acceleration can be found as the slope in a v vs t graph as Δv/time

Rearrange formulas

Solving for acceleration in equations is another way to find it. Acceleration can be found by rearranging: x=1/2at^2+vi and v=at+vi.

Connection to Real World.

This unit, helps us understand acceleration of objects in the real world, objects like a skateboard going down the hill or a baseball being thrown.

Ramp Challeng

Ramp Challenge

In this challenge we were asked to predict the acceleration of a ball rolling down a table that was inclined, we were also asked to predict the velocity of this ball at 4 seconds. After many trials we found the average of the position of the ball, at 0.5, 1, 1.5, 2, 2.5, and 3 seconds to find as many points as posible without revealing the answer to the challenge.

 our data was the following:

Time              Position

 0.5s                  6cm

   1s                  20.5cm

1.5s                   35cm

2s                     56cm

2.5s                   81cm

3s                     105cm.

After doing this we found that the acceleration of the ball was constant, this created a top-opening parabola in excel.

To find the slope of this line we needed to square the value of T and since most velocity units are meters over seconds we divided our position by 100 to find the values in meters, giving us this chart.

As you can see in the chart, the slope of this graph, is 0.1157, and as we learnt in the classroom, the slope of an x vs t graph, is 1/2a, meaning that the acceleration of an object is double of the slope of it x vs t graph, giving us a constant acceleration of 0.2314m/s^2.

Because we have the graph that shows us position vs time, we were able to find the acceleration, giving us a velocity vs time formula that is v=a(t) giving us a velocity at 4 seconds of v=0.2314(4) v=0.9256m/s

Unknown mass

Unknown Mass Challenge

In the picture above we can see an object with an unknown mass hanging from 2 ropes with different tensions and at different angles.

In this Free body diagram we see the forces and the components that affect this object, in this challenge we were asked to find the weight of the object. To do this we need to find FT1y and FT2y since all the forces in the Y axis need to be equal to 0.

To find these two forces you need Cos, and the strength of the original force (FT1, and FT2).

FT1y= Cos(20)= Adj

                            2.2

 Adj= 2.2 Cos(20)

FT1y=2.067

FT2y= Cos(55) = Adj

                               1

Adj = Cos(55)

FT2y = 0.573N

FT2y + FT1y + Fg = 0N

Fg = 2.067 + 0.573

 Fg = 2.64 N