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course PHY 242
22:50; 10/7
Deflection of ball down ramp from path`qx001. How far did the ball travel after leaving the end of the ramp, in the trial without the magnet, and what therefore was its horizontal velocity during the fall, assuming a fall time of .4 second?
****14cm is the diff in distance for the ball falling without the effect of the ramp vs. with the ramp propped up with 2 dominoes.
B.c. the change in distance outward is 14cm and the time falling is .4sec then ‘ds[14cm]/`dt[.4s] = 35cm/s = v over the .4s interval.
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`qx002. What is the maximum deflection of the ball, due to the presence of the magnet, from its original straight-line path?
We [Jonathan and I] adjusted the length of distance away from the path of the ball at the same height. The max affect was the nearest we could get the magnet w/o annihilating its path of travel. The closest we came was 1.75cm from the path. The effect on the path of travel was 10cm change in the perpendicular direction [I will call this the z axis] of our preset x and y axis.
That's a very significant change. Good work.
****
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`qx003. What velocity does the ball therefore attain, in the direction perpendicular to its original straight-line path, as a result of the magnet?
****the `dz = the change in position. The `dt is still consistent [this is not validated] with the above time. `dz[10cm]/`dt[.4s]=25 cm/s
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`qx004. Assume the ball was close enough to the magnet to have its motion significantly influenced for a distance of 5 centimeters. How long did it take for the ball to travel this distance?
****if we examined the non—v0 in the x direction we can easily find the change in time over the 5cm interval. i.e. the time it took the ball to travel 5cm w/o the magnet in the x direction [we had the magnet positioned s.t. it effected the ball for the entire 5 cm]. The `dt is equal to `ds / V constant => cm/(cm/s) = the distances cancel in unit and we are left w `dt => `ds [5cm]/v[35 cm/s] = 1/7 sec = 0.143 sec.
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`qx005. Assume that the ball has a mass of 60 grams. How much momentum did it gain, in the direction perpendicular to its original line of motion, as a result of the magnet?
****the velocity in the z direction is `ds/`dt=>5cm/.4s=12cm/s. the mass is given to be 60g so momentum = v Ave * mass = 12 [cm/s]* 60 g = 720g*cm/s.
you previous calculated a velocity of 25 cm/s in this direction; 5 cm is the length of the magnet and isn't relevant to the current calculation
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`qx006. What therefore was the average rate of change of its momentum with respect to clock time, for the 5-cm interval during which the magnet significantly affected its motion?
**** the change in time is still .4 sec. and we found the momentum to be 720g*cm/s. so the average rate of change of momentum where p momentum with respect to clock time is p [720 g*cm/s] / `dt [.4s] = 1800 s*cm / s^2
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`qx007. What were the measurements by which you can calculate the slope of your ramp? What was the slope of the ramp?
****the length of the ramp was 30 cm the height is 2cm so we have the hypo and the opp of the directional angle we are seeking so sin^-1(2/30) = about 3.82 degrees.
that is the angle with horizontal; the slope is very very close to 2/30 = 1/15
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Acceleration of gravity
`qx008. What was your count for the object dropped in the stairwell? To what time interval in seconds does your count correspond?
****my counts varied overall by about 15percent. My count was 1.0, 8counts in the time the bead dropped. The distance my bead dropped was 26 cinderblocks or 524 cm @ 8`` per block*2.54cm *#blocks. the time for the ball in the air was about .7 sec with 8 numbers 1,2,3,…,8 @ 1-- 8 count per .7 seconds. So the time of travel is .7sec.
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`qx009. What was your count for the object dropped in the stairwell? To what time interval in seconds does your count correspond?
****
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`qx010. What was the distance the object fell (1 block = 8 inches or about 20 centimeters)?
****ans in above question.
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`qx011. Using the distance of fall and the time interval in seconds, find the acceleration of the falling object.
****the time interval is .7s and the `ds with respect to y is 524cm so the `ds[524cm]/`dt^2[.7s]^2 = 1069 cm/s^2.
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University Physics Students Only
`qx012. Find the initial velocity of the falling ball, based on your measurements of x, y, x0, y0 and your determination of sin(theta) and cos(theta).
****v0=sqrt[{2/a}{y-y0-sin(theta) *x /cos(theta) + sin(theta)*x0/cos(theta)}]* [{x-x0}/cos(theta)]
=after plugging everything I get 1092.1cm/s = v0. If this is true then 1092.1[cm/s]*cos(3.82)[direction]*t[s]=`dx[14cm]=>14[cm]/1092.1[cm/s](cos3.82)=0.0128sec
Ps. I do not believe this is correct. I will type up a complete solution and send it later.
Something is clearly wrong and you did well to notice and document the inconsistency.
You didn't show the details but I suspect you just neglected to take the square root (which means you also neglected to acutally do the unit calculations, which if done could well have shown you the nature of your error). If you do that you get a velocity very near the one you calculated previously, based on the .4 second fall.
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`qx013. Find the time of fall of the ball.
****1092.1[cm/s]*cos(3.82)[direction]*t[s]=`dx[14cm]=>14[cm]/1092.1[cm/s](cos3.82)=0.0128sec
that's going to come out very close to .4 second, once your velocity is corrected
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`qx014. Questions about uncertainties:
* What is your estimated uncertainty in the measurements you used to determine sin(theta)?
* What therefore is your uncertainty in sin(theta)?
* What uncertainty does this introduce into your determination of v0?
****1).3cm is the uncertainty of the height though the uncertainty in length of the ramp is about .5cm so the total uncertainty is +-.3 and +-.5cm => T uncertainty / T (l [ramp] + h) = +-.8/32 = +-2.5 percent
certainly you measured the 2 cm height more accurately than that
however that's not a big deal
The important thing is that you don't divide the uncertainties. If you have +-.3 cm on a 2 cm measurement, that's 15% right there, and an additional uncertainty doesn't reduce that.
Because of the way the differential of a division of two variable quantities works (per previous class notes), the uncertainties add.
2) Because we carry the same values for theta the percent error will remain the same.
Note we could also solve this with differentiation but it seems trivial with an apparent ans available.
good observation, which leads to the following comment:
note that the quotient rule is (f ' g - f g ' ) / f^2, not f ' / g '.
3) +-2.5 percent plus the additive percent error in any one or multiple variables.
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`qx015. Based on a single trapezoid of the graph of F vs. x for magnet forces, explain the meaning of the slope and the area of this graph.
****the area is the distance * F = W so integrating any definite force function yields the work done within the limits of integration [distance]. The slope of the curve is the rise over run so if F = kg*m/s and the run is the distance `ds = m the the slope = (kg*m/s^2)/m=kg/s^2 which is the joule or unit of potential energy.
My GUT feeling about this is negative! The rise in force over `dt Is resulting joule
positive and negative require careful thinking but with a little practice it isn't difficult
however it's always very easy to get it backward
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`qx016. Based on a single trapezoid of the graph of PE vs. x for magnet forces, explain the meaning of the slope and the area of this graph.
****the slope I would guess to be power providing the function of PE is with respect to time is power of course we can define the slope as rise over run or the derivative of a function between two points in time. Naturally when we integrate PE we get the area under the “curve” which is the force on that time interval.
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time isn't involved in these graphs; the operant variable here is position
because of this the graphs relate force and energy
if the graphs involved time then we would be in the context of quantities like momentum and power as opposed to work and energy
good instincts
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