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: