Physics Lab Help

  • Thread starter ji707
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  • #1
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1.hi, im new to this forum and i have a question regarding my final physics lab:
A solid spherical ball, R= 1.4 cm and m=95 grams, is released from point A(1m above the). After travelling through the loop, the ball exits the set up as a projectile and land at point B with the horizontal distance of X= 37 cm.

the picture of the setup is here:
lab#13.jpg

2. i have to find the theoretical distance x, but im not sure how to do this...

The Attempt at a Solution


I tried finding the velocity at the end of the track by saying mgh = K.E.r+K.E+mgh(loop)+mgh(end of track) and i get the velocity once the ball leaves the track to be .843m/s. i think this seems to easy though, is it?

Im stuck on how to even start this and this is our final lab so we have to use all of our physics we've learned so far.
Can someone give me a hint on how to start?
 
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Answers and Replies

  • #2
tiny-tim
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welcome to pf!

hi ji707! welcome to pf! :smile:
I tried finding the velocity at the end of the track by saying mgh = K.E.r+K.E+mgh(loop)+mgh(end of track) and i get the velocity once the ball leaves the track to be .843m/s.
yes, that's right :smile:

(except that the loop makes no difference)

now resolve the velocity into x and y components, and use the standard constant acceleration equations :wink:
 
  • #3
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ahh, thank you i was stuck on this for a while but knowing the loop cancels out helps a lot, thanks : D
 
  • #4
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How would i go about finding the exit velocity given an x value?
ive tried plugging into the kinematic equations but i cant solve for both v and t...
 
  • #5
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I don't understand your question. Use the conservation of energy to solve for the exit velocity, decompose the vector, then use the kinematic equations to solve for what you want. You know the Vi, Vf and a for the x component and Vi, a and displacement for the y component.
 
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  • #6
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Ill try to clarify, after all of this, they gave us an x value and told us to use that given value to find a new exit velocity corresponding to the new x value of the projectile,
so when i tried to plug in the x displacement value they gave us i figured that since the exit velocity had changed so would the time, now i have two unknown variables, V and t and I dont know how to go about finding them. hope that clears up my question.
 
  • #7
tiny-tim
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hi ji707! :smile:
… after all of this, they gave us an x value and told us to use that given value to find a new exit velocity corresponding to the new x value of the projectile …
now i have two unknown variables, V and t …
yes, it means, ignore everything except the angle at the exit, and find v as a function of x

two equations with two unknowns should be enough …

use one to find t, then put that value of t into the other to find v :wink:
 
  • #8
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thank you tiny-tim.
so i did the problem and found my x to be 1.43m does that seem right, i feel like im way off...
 
  • #9
tiny-tim
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so i did the problem and found my x to be 1.43m does that seem right, i feel like im way off...
(which question is this now?)

if you want us to check your answer, you need to show us your calculations :wink:
 
  • #10
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ok, im finding the theoretical x value so i did


Mgh(point A)=1/2 Mv^2+1/5 Mv^2+Mgh(end of track)

.913J= .0475v^2+ .019vt^2+ .484J

v_exit=2.539 m/s

Then i found the Y component

2.539*sin(3pi/4) = 1.79 m/s

then into kinematic for a free falling object

0 = 1.43+1.795t-4.9t^2

t = .8

then again into the kinematics

x = -1.79*.8

x = 1.43m

I think I messed up somewhere, because the actual value when we did the lab was .37m...
 
  • #11
tiny-tim
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looks ok :confused:

(but would have been a lot easier if you'd left M out of the equation completely :wink:)

perhaps you lost a lot of energy from air resistance in the tube …

how much was the diameter of the tube compared with the diameter of the ball?
 

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