This was a conservation of energy lab using an air track and blower/ air track c
ID: 1583988 • Letter: T
Question
This was a conservation of energy lab using an air track and blower/ air track cart/ 2 timing gates and ringstands.
Setup:
1. place the air track on the lab table so that each end of the track projects the edge of the table.
2. insert a post (<1cm diameter) into the top of the cart. measure and record the diameter of the post; 11.7mm.
3. position a timing gate at approximately 50cm (as marked on the track) and the second timing gate 1m away. the gates should be situated so that theu will note the times as the post in the track cart passes through them.
Using the information gathered it asks:
(1) using the measurments of the height at the ends of the track from the floor, calculate the slope of the track to determine the difference in height (?h) of the air track between the timing gates. (Hint: if the air track is 2m long and the timing gates are 1m apart, the difference in height between the gates will be 1/2 the difference in height between the ends)
(2) from the timing data for each run, determine the velocityof the cart as it passes through the fist gate (vi) and the velocity as it passes throught the second gate (vf). make a table showing your velocities and height for each run
given equations to use:
This is are recorded time for the gates, but how do you use the above equations to solve for ?h and ?i/?f ?
our initial height measured: 101mm and final height: 100mm
Trial #
(t1) at gate 1
(t2) at gate 2
(t3) between gates
?h
(vi) velocity passing through gate 1
(vf)
velocity passing through gate 2
(a) acceleration
1
13.337 and 13.231
13.417 and 13.422
2
11.838 and 11.443
11.925 and 12.971
3
14.295 and 13.978
14.380 and 13.994
4
11.783 and 12.031
11.862 and 11.889
5
10.595 and 11.057
10.674 and 10.597
6
10.351 and 10.366
10.426 and 10.627
Trial #
(t1) at gate 1
(t2) at gate 2
(t3) between gates
?h
(vi) velocity passing through gate 1
(vf)
velocity passing through gate 2
(a) acceleration
1
13.337 and 13.231
13.417 and 13.422
2
11.838 and 11.443
11.925 and 12.971
3
14.295 and 13.978
14.380 and 13.994
4
11.783 and 12.031
11.862 and 11.889
5
10.595 and 11.057
10.674 and 10.597
6
10.351 and 10.366
10.426 and 10.627
+ mgh, = +mgh ,Explanation / Answer
given post diameter, d = 11.7 mm
then t3 between the gates = (t2 - t1)
so tghere are two values of t at t1 and t2, so we take their average and find difference to calcualte t3
dh = hi - hf = 101 mm - 100 mm = 1 mm
vi = d/difference in the timeings at t1
vf = d/difference in timings at t2
acceleration a= (vf^2 - Vi^2)/2*0.5
hence the following is the completed table
# t1 -1 t1 -2 t2 -1 t2 -2 t3 dh(mm) vi(mm/s) vf (mm/s) a(mm/s^2)
1 13.337 13.231 13.417 13.422 0.1355 1 110.37735 2340 1365854.21
2 11.838 11.443 11.925 12.971 0.8075 1 29.620253 11.18546 -188.0611733
3 14.295 13.978 14.38 13.994 0.0505 1 36.908517 -30.31088 -110.8722891
4 11.783 12.031 11.962 11.889 0.0185 1 -47.17741 -160.277 5865.509349
5 10.595 11.059 10.674 10.597 -0.1915 1 -25.21551 -151.948 5613.097045
6 10.351 10.366 10.426 10.627 0.168 1 -780 58.2089 -151252.9294
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