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course Phy 241
October 9 around 11:45am.
Questions 101006Deflection 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?
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16 cm
Horizontal velocity: 40cm/s (16cm/0.4sec=40cm/s)
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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?
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15 cm away from original fall mark (magnet was 1.5cm away)
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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?
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37.5 cm/s (15cm/0.4 sec=37.5cm/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?
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0.133 seconds (5cm/37.5cm/s=0.133 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?
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2250 g*cm/s
Momentum=v*mass
(37.5cm/s^2)*(60g)=2250g* cm/s
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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?
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16917g*cm/s
((2250g*cm/s^2)/(0.133sec)=16917g*cm/s)
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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?
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I used 2 dominos to prop up my ramp (approx. 1.8cm) and my ramp was 30cm long.
Slope will be rise (1.8cm) divided by run (30cm)=0.06cm (assuming the table top was the (0,0))
1.8 cm / 18 cm = .06, not .06 cm
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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?
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Each time was about one 8-count.
Since one of my counts is 0.17 seconds, the time interval would be approx. 1.4 seconds.
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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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Same question as above.
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`qx010. What was the distance the object fell (1 block = 8 inches or about 20 centimeters)?
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To floor: 27 blocks (216 in, 540 cm, 18 ft)
To first stair: 25 blocks (200 in, 500cm, 16.6ft)
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`qx011. Using the distance of fall and the time interval in seconds, find the acceleration of the falling object.
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‘ds= ˝ a’dt^2
540cm= ˝ a (1.4 sec)^2
540cm= (0.98 sec^2) a
A= 551 cm/s^2 (to floor)
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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).
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83.05cm/s
I attempted this. Here’s the measurements I used:
G=-980cm/s^2
X=15cm (mark across from original position)
Y=16cm (mark down from paper)
X0=0
Y0=0
Slope=0.06cm
I then plugged everything into the equation that we solved for “v0” and got v0 to be 83.05cm/s. I’m not too sure on x0 and y0 and if they’re 0 or not. I also didn’t know if I was supposed to just use “sin (theta)” or take the inverse of that. I used just “sin (theta)” when I solved for “v0.” Since g=-9.81m/s^2, I converted that to cm and used -980cm/s^2.
x is horizontal, y is vertical
your 15 cm and 16 cm observations were horizontal, but perpendicular to one another
let x - x0 be the displacement of the undeflected ball; I believe this was your 16 cm observation
y - y0 is displacement from edge of ramp to floor, about -95 cm
I believe youll get a result in the neighborhood of 40 cm/s
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`qx013. Find the time of fall of the ball.
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0.18sec
I used the equation: (x-x0)/(v0cos(theta))=t
=(15cm)/(83.05cm/s*cos(0.06))
=0.18sec
based on 83 cm/s that's about right, but the horizontal velocity is only about half as great as your result
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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?
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I wasn’t sure if I was supposed to use my slope (0.06) and just do sin(0.06) OR sin^-1(0.06). In my calculations above, I just used sin(0.06)=0.001. If I was supposed to use the inverse, it would give me 3.44. This would definitely make a big affect on the determination of v0.
the sine of that angle is rise / hypotenuse = 1.8 cm / (30 cm) = .06
the cosine is sqrt(1 - sin^2(theta)) = sqrt(1 - (.06)^2) = .9985, I believe, and that's very close to 1
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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.
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Comparing it to the trapezoid of v vs. t, here’s what I can come up with:
Slope: (change in force)/(change in position)
Area: (Force)*(change in position), will give us work done
I'd be a little more specific and say 'average force * change in position', but your answer is good.
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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.
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Slope: (change in PE)/(change in position) which will give us FORCE (average roc of PE wrt s) or you could say (force is the derivative of PE wrt position)
Area: Integrate (Force)*(‘ds) which will give you change in PE or the average force
Derivatives always have comprehensible meanings as rates of change; integrals don't necessarily have significant meaning
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&#Your work looks good. Let me know if you have any questions. &#
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