Computer Science Canada

OAPT Physics Contest 2011 - Practice
Author:  whoareyou [ Tue Apr 05, 2011 8:39 pm ]
Post subject:  OAPT Physics Contest 2011 - Practice
So, because a lot of people here are interested in physics and I need help practicing for the upcoming physics competition, I am going to post the 2010 contest, and any other questions that I may need help on.

This could give other members something to do in their spare time, and will also help me to study! Very Happy

*EDIT: So apparently, there isn't an option to upload anything here, so I will post a link to it.
*Link to the OAPT 2010 Contest - uploaded by me: http://www.mediafire.com/?7fxmpmka8qafcgh
Author:  whoareyou [ Tue Apr 05, 2011 8:52 pm ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
Question 10
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It would be E right, because Fg = mg, so 12.2kg x 9.81N/kg = 120N ?
Author:  RandomLetters [ Tue Apr 05, 2011 9:05 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Did you consider the verticle component of Fa?

Also, I just realized how much it sucks to have 2nd sem physics, because I am totally clueless regarding the electricity questions, Shocked
Author:  whoareyou [ Tue Apr 05, 2011 9:15 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Are you writing this contest too ?

I had physics first semester, so I (think I) know everything!

And thanks for the advice - I forgot to consider the vertical component. By the way, how would you implement that into the equation? We never did any kind of force problems with angles in them.
Author:  RandomLetters [ Tue Apr 05, 2011 9:21 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Forces with angles can be broken down into component vectors.

You can use trig to break down forces exactly like vectors.

Then, you would just use the verticle component separately, so Fn = Fg + Fay
(while the horizontal motion would use Fax and Ff)
Author:  whoareyou [ Wed Apr 06, 2011 2:55 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
That wouldn't work ... because Fg + FAy > 120N which is not an option.
Author:  RandomLetters [ Wed Apr 06, 2011 3:40 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Fg is downwards, FAy is upwards, so the magnitude is subtracted, giving 70N down. Sorry, I should have said they were vectors.
Author:  andrew. [ Wed Apr 06, 2011 5:16 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Basically:
F_Ay + F_N = F_g
F_Ax - F_r = ma

You only need the first equation. Solve for F_N to get F_N = 69.7 N.
Author:  whoareyou [ Wed Apr 06, 2011 6:12 pm ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
Thanks for the guidance!

The following questions I am having troubles with.

QUESTION 5
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QUESTION 6
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QUESTION 7
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I would think A or B since the passenger would exert a force downwards on the helicopter while the helicopter is flying up.

QUESTION 9
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I would think A because both the dog and the flatcar are accelerating, so the dog is moving with the flatcar, so there would be no force. Although, since there is acceleration, the net force of the dog shouldn't be zero.
Author:  A.J [ Wed Apr 06, 2011 8:25 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Don't merely state that you are having trouble with them. Instead, you should also include any headway you made with any of them. All the questions you posted seem to be ordinary Kinematics and Dynamics problems, not contest problems.
Author:  DemonWasp [ Wed Apr 06, 2011 8:28 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Question 5: You know v1 = 0, d1 = 20. Use the equation d = v1 * t + 1/2 * a * t ^ 2 . You know that since v1 = 0, you can cross out the first term. So you have d1 = 1/2 * a * t ^ 2. If you double the time period, you have to double the value of t, so we would write d2 = 1/2 * a * (2*t) ^ 2. Solve that for d2, which you will find to be in terms of a and t. Look at the similarities between the equation you have now and the first one we wrote down, then think about what you can do with them. Remember that d2 is the distance travelled in BOTH time periods.

Question 6: Use conservation of energy for each rock, solve for v2. Compare.

Question 7: When the helicopter flies up, do you feel a force pushing you into the chair, or a force lifting you out of the chair? The answer to that should let you choose whether it's A or B.

Question 9: Since the dog is accelerating at the same rate as the flatcar (and the question doesn't mention anything else), we can assume that friction is doing all the work. Calculate the force required to produce that acceleration in the dog and you will have your answer. This is partially a trick question too, since you don't actually need the coefficients of friction. They'd be useful if the question was whether the dog slipped or not, but it isn't.
Author:  whoareyou [ Thu Apr 07, 2011 6:06 pm ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
I have attempted all the questions.

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For question 7, I would think this: since Fnet = ma, the mass of the of the passenger are m, and the total acceleration of the passenger would be (g+a) as vectors, so the correct answer would be A.

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Author:  DemonWasp [ Thu Apr 07, 2011 10:59 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Good work.
Author:  whoareyou [ Fri Apr 08, 2011 6:33 pm ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
This question was taken from the 2009 OAPT Physics Contest, and my answer is wrong.

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Author:  RandomLetters [ Fri Apr 08, 2011 7:09 pm ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
Average velocity is:
v = Ad/At
We know that each velocity lasts for Ad/2, so we need to find At.
How do we find At, since a slower velocity takes longer to traverse Ad than a faster velocity?

(break it down to two movements for each velocity to find each At)
Author:  whoareyou [ Fri Apr 08, 2011 7:36 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
t=d/v

so t = 0.5d / (0.5d/v1)

and the same for v2 ... ?

is that what you meant?
Author:  RandomLetters [ Fri Apr 08, 2011 8:14 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
v = d/t

t = d/v

t1 = 0.5d/v1

same for v2

v average is total displacement over the total time,
so v = d/(t1+t2)
Author:  whoareyou [ Fri Apr 08, 2011 8:26 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Ok, so I need something to be clarified. So, under what circumstances is average speed = (v1 + v2)/2 because in this case, that isn't the right answer.
Author:  apython1992 [ Fri Apr 08, 2011 8:43 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
That would be the case when the speeds were traveled at for the same amount of time, because they have the same "weight". So for example traveling at 50km/h for an hour, and then 60km/h for an hour.
Author:  whoareyou [ Mon Apr 11, 2011 10:14 am ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
Posted Image, might have been reduced in size. Click Image to view fullscreen.

Is there a special equation that you need to use? I mean, I just looked at the picture, and got the right answer, 3. Could you do it by setting vt=1/2at^2 ?
Author:  A.J [ Mon Apr 11, 2011 11:45 am ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
The answer is 3 because the displacement of the ball during the 3rd interval (i.e. the distance between the 3rd and the 4th image of the ball) is the same on either plane, whereas this isn't the case for any of the other time intervals.
Author:  chrisbrown [ Mon Apr 11, 2011 1:55 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Are you sure about that A.J?

Speed will be equal at that point, yes, but the balls have different displacement functions.

Nitpicking, I know, I'm just saying...
Author:  apython1992 [ Mon Apr 11, 2011 2:11 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Speed is defined as the rate of change of distance. That being said, the speeds are the same because they traveled the same distance in the same amount of time. It doesn't matter that their displacement functions are different; within that interval, they are the same (or very close, because the ball rolling down the ramp picks up a tiny bit of speed on the way to point 4).
Author:  Zren [ Mon Apr 11, 2011 2:23 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
That question makes me want to hurt someone. You should never fully have to trust graphs for exactness.

The bottom plane is feasably impossible, as even a slight dip will show signs of acceleration, while here it appears to have constant velocity. Then again, it's probably like that to make the answer more apparent.

Edit: Nevermind. Stupid eyes stopped playing tricks on me to realize it's a flat surface in a world without other forces. >.> Released is a bad term for this as it doesn't imply it's moving other than from the picture.
Author:  apython1992 [ Mon Apr 11, 2011 2:28 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
What dip? And I agree, I don't usually like visual questions like that, but as long as they are apparent enough I can tolerate them.
Author:  chrisbrown [ Mon Apr 11, 2011 2:38 pm ]
Post subject:  Re: RE:OAPT Physics Contest 2011 - Practice
apython1992 @ Mon Apr 11, 2011 2:11 pm wrote:
(or very close, because the ball rolling down the ramp picks up a tiny bit of speed on the way to point 4).

That's not insignificant. If the slope of the plane was near vertical, you'd see a significant difference in the distances covered by the two balls. And yes sorry, I meant distance, not displacement.

All the information you need is given in the description:
Ball 1:
v(1) = 0
v(5) = ?

Ball 2:
v(t) = constant

Equal distances in equal time... sounds like an average velocity problem.
Author:  apython1992 [ Mon Apr 11, 2011 4:11 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Quote:
Equal distances in equal time... sounds like an average velocity problem.

Yes, that's exactly what this is. High school level physics doesn't assume calculus knowledge, and for a visual question such as this that will have to suffice.
Author:  A.J [ Mon Apr 11, 2011 5:18 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Yes, as mentioned above, it looks like you want to determine the time interval where the ball on both the planes had the same average velocity.
Author:  whoareyou [ Mon Apr 11, 2011 5:38 pm ]
Post subject:  Re: RE:OAPT Physics Contest 2011 - Practice
Is it supposed to be determined logically, visually, or mathematically?

I don't see any numbers given in the question to do it mathematically.

I kind of did what A.J did, by looking at the spaces to see the distance covered by each interval and I assumed it to be 3 (which is right btw).

apython1992 @ Mon Apr 11, 2011 2:11 pm wrote:
Speed is defined as the rate of change of distance. That being said, the speeds are the same because they traveled the same distance in the same amount of time. It doesn't matter that their displacement functions are different; within that interval, they are the same (or very close, because the ball rolling down the ramp picks up a tiny bit of speed on the way to point 4).


His/Her reasoning makes sense to me Very Happy!
Author:  apython1992 [ Mon Apr 11, 2011 6:59 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Well, logic is part of any deduction! So yes, logically and visually. You're right, no numbers were given, so you'd have to think of a way to determine the answer much like A.J did. And for the record, I'm a "he" Wink.
Author:  whoareyou [ Mon Apr 11, 2011 7:46 pm ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
For this question from the OAPT Physics Contest 2006, the correct answer is E because of some inertial frame of reference. However, I still don't quite understand it because if the duck swims with the current, it would swim at a speed of 3m/s and get to bread number 2 faster than if it would swim at 1m/s against the current to get bread number one. This is something I still don't understand.

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The next question (2007), I believe it's a projectile motion question. First I would split the x-component up to find the time it takes to travel x, then 1/2 it. Then I would created a new equation for the y-component, and sub in the time into that equation. However, I don't know if the equation I'm creating is correct : P-h = 1/2at^2 ?

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Author:  apython1992 [ Mon Apr 11, 2011 8:17 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
For the first question, you would be correct if the pieces of bread were on the shore (moving relative to the duck). However, because the pieces of bread are in the water, and thus have speeds equal to the current (like the duck), they remain stationary relative to the duck. That said, it doesn't matter which way the duck travels if they are not moving at all relative to each other...in that frame of reference, it's no different than having everything on "solid" ground, because nothing moves relative to one another.


For the next question, this isn't so much a plug-and-chug mathematical question as it is to test your general understanding of the independence of vector components, like in projectile motion. We can leave numbers out of this completely. You know that in the vertical direction, displacement is proportional to the square of time because of acceleration, and also that the acceleration ONLY occurs in the vertical direction. Since the horizontal velocity remains constant in projectile motion, cutting the horizontal distance traveled in half cuts the time taken to travel that trajectory in half as well. Make sense? That said, if the time is cut in half, that reduces the vertical displacement by 4 (making it a quarter as large as the original). Perhaps it would help to see the ratio:

code:

   New h / Old h
= [(1/2)t]^2 / t^2
= (1/4)t^2 / t^2
= 1/4


While I used equal signs, they really should be proportionality signs (though the last step actually is equal). Does this make sense?
Author:  Zren [ Mon Apr 11, 2011 8:21 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
The rivers flow affects both the bread and the duck.

If the duck does nothing, it will forever be stuck between them (ignoring other forces). Whenever bread1 gets closer, duck moves further away downriver, and vice versa for bread2. Thus the only vector actually affecting the distances between each other is the ducks effort.

It's easier to notice if you draw the vectors and add them together.
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Starting positions.
River's flow
If duck swims to bread1
If duck swims to bread2
Author:  whoareyou [ Wed Apr 13, 2011 6:46 pm ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
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(1/R) + (1/R) + (1/R) = 3/R <-- not an option ?
Author:  chrisbrown [ Wed Apr 13, 2011 7:01 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Check the formula for resistors in parallel. You're missing one crucial step.
Author:  whoareyou [ Wed Apr 13, 2011 7:18 pm ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
Embarassed omg, i can't believe i missed that.

1/R = 3/R --> R = R/3 Very Happy
Author:  whoareyou [ Thu Apr 21, 2011 7:54 pm ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
This question is from the OAPT 2003 contest.

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I would assume that the first step is to find the wavelength, given the resonance frequencies at 2 different lengths. For a closed air column, the resonance frequency occur at L1 = 1/4λ and L2 = 3/4λ. However, when you do this, the result is two different wavelengths.

Quote:

The standing waves formed in the air column have a node at the waterline and an antinode near the open end. The length of the air column must be one-quarter of a wavelength for the first possible loud sound, and three-quarters of a wavelength for the second loud sound. Subtracting these and multiplying by 2 gives you the wavelength. From the wavelength and frequency, you can use the Universal Wave Equation to calculate the speed of the sound. Subtract 332 m/s and divide by 0.6 to obtain the ambient temperature of D) 20?C.


Why are they subtracting and multiplying to get the wavelength? And why isn't it working when I do the L1, L2 stuff ?
Author:  DemonWasp [ Thu Apr 21, 2011 9:41 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Why are they subtracting, then multiplying?

Quote:

L1 = 1/4λ
L2 = 3/4λ
L2 - L1 = 1/2λ
2 * (L2 - L1) = λ


The reason it isn't working is because you don't know that L1 is necessarily the first antinode. It could be the second, or the third!

Since you know that antinodes exist at (2N + 1) * λ / 4, and you know two of them are at 60 and 100cm respectively, you have:

Quote:

L1 = 60 = ( 2 * N1 + 1 ) * λ / 4
L2 = 100 = ( 2 * N2 + 1 ) * λ / 4

Subtracting and simplifying, you get:
100 - 60 = [ ( 2 * N2 + 1 ) - ( 2 * N1 + 1 ) ] * λ / 4
100 - 60 = 2 * ( N2 - N1 ) * λ / 4
40 * 2 = ( N2 - N1 ) * λ
80 / ( N2 - N1 ) = λ


This is really just a generalization of an assumption made above: because of the wording of the problem, we're assuming that L2 and L1 are consecutive antinodes, and therefore N2 - N1 = 1. So this also gives you λ = 80cm.


Interesting side note: because you know that N1 and N2 are both whole numbers (not fractional / decimal / irrational, greater than zero), you can find a set of reasonable solutions. Given the information that N2 - N1 = 1, you can narrow it down even further. As it happens, N1 = 1 and N2 = 2 works out. The reasoning:

Quote:

L2 / L1 = 100 / 60 = ( 2 * N2 + 1 ) / ( 2 * N1 + 1 )

Cross-multiply...

100 * ( 2 * N1 + 1 ) = 60 * ( 2 * N2 + 1 )
5 * ( 2 * N1 + 1 ) = 3 * ( 2 * N2 + 1 )
10 * N1 + 5 = 6 * N2 + 3
10 * N1 = 6 * N2 - 2
5 * N1 = 3 * N2 - 1


There's ways to solve that last equation and get a real set of answers, but they're all from first year uni math so I've (of course) completely forgotten them. You can, however, just start plugging in numbers for N1 to see when whole numbers come out for N2. As it happens, with N1 = 1, you get:

Quote:

5 * (1) = 3 * N2 - 1
3 * N2 = 5 + 1 = 6
N2 = 2


Edit: Why oh why are things inside [ code ] blocks HTML-escaped, guys? My λ's were showing up as an λ escape-sequence.
Author:  whoareyou [ Tue May 10, 2011 8:45 pm ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
So I did the contest; I got a really embarrassing mark, however, I did place very high out of all of the 993 who wrote it. Anyways, even though the contest is finished, I am still very much interested in physics and wish to continue posting questions that I may need help on.

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First, the velocity of the car is 19.8 m/s [45]. So I split that up into x and y components, then I solved for the time, and finally used the time to find the distance. When I do that, I get an answer of 39.something and it doesn't easily round up to 40.0m which is a choice, so I was wondering if I had the right logic.

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Is this the kind of question where Qgained + Qlost = 0 --> mc(tf-ti) = -mc(tf-ti) ?
Author:  A.J [ Wed May 11, 2011 1:53 am ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
For #5, I get A (i.e. ~20.0 m). The vertical component of the velocity is v_x = 19.8*sin(45), and the horizontal component is 19.8*cos(45). Solving for the time it takes the car to get to its highest point (i.e. when the vertical component of the velocity gets to 0) we get:
v_final = v_initial + 2*a*t
0 = 19.8*sin(45) - 2*g*t => t ~= 0.714322 s
Therefore, it takes the car 2t ~= 1.42864 s to land on the other side. Thus the horizontal distance traveled is:
v_x*2t = 19.8*cos(45)*1.42864 ~= 20.0 m

For #13, the idea is:
heat gained by the calorimeter + heat gained by the cold water = heat lost by the hot water, or in terms of variables: Q_cal + Q_CW = Q_HW. Here Q_cal = 0 (since we are saying that the cup is well insulated).

Now, Q_CW = m*c*dT = (126.4 - x)*4.184*(26-16)
and Q_HW = m*c*dT = (41.3)*4.184*(50-26). Setting these two equations equal and solving for x yields: x = mass of calorimeter = 27.28 g. Therefore the mass of the water mixture was 167.7 - 27.28 = 140.42 g
Author:  Zren [ Wed May 11, 2011 2:39 am ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Quote:
v_final = v_initial + 2*a*t


Er, why is gravity acceleration twice as effective?
Author:  A.J [ Wed May 11, 2011 4:01 am ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Whoops, my bad. I guess then it would be 40.0.
Author:  whoareyou [ Mon May 30, 2011 8:51 pm ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
code:

A student rode her bicycle half the distance from school to her house at a leisurely 5.0 km/h. Noting that it was almost time for dinner, she increased her speed for the remaining half of the distance. What speed would she need to maintain for the remaining half in order to average 10 km/h for the entire trip?


I remember somewhere in this forum that you can only do the average speed (v1 + v2 divided 2) if they have the same "weight". But, that wouldn't apply here since v1 and v2 have to be different. so since vt = d/2 in this case, the speed would have to be the same to travel the same distance again. But since v1t = d/2, v2 /= v1, so therefore it's impossible to average 10km/h?

Is that the right answer? and is my rationale correct?
Author:  DemonWasp [ Mon May 30, 2011 10:24 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
This is the right answer, for a totally incorrect reason. It took me a while to spot it, so don't feel bad. The correct rationale goes something like this (fill in the gaps):

average speed = ( d1 + d2 ) / ( t1 + t2 )
but, average speed is also double v1: Vav = 2 * v1, and d1 = d2 = d
so:
2 * v1 = 2 * d / ( t1 + t2 )
t1 + t2 = d / v1

Now, why does d / v1 look familiar? When you figure out what that's equal to, what does that suggest about t2?
Author:  apython1992 [ Tue May 31, 2011 9:05 am ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
"Same weight" means that each speed must be travelled at for the same amount of time for you to take a simple arithmetic mean (would the average speed be 75 km/h if you travelled at 50 km/h for five hours and 100 km/h for one minute?). As DemonWasp has pointed out, the average speed should be measured as the total change in position over the total change in time.
Author:  whoareyou [ Tue May 31, 2011 9:40 am ]
Post subject:  Re: RE:OAPT Physics Contest 2011 - Practice
DemonWasp @ Mon May 30, 2011 10:24 pm wrote:
This is the right answer, for a totally incorrect reason. It took me a while to spot it, so don't feel bad. The correct rationale goes something like this (fill in the gaps):

average speed = ( d1 + d2 ) / ( t1 + t2 )
but, average speed is also double v1: Vav = 2 * v1, and d1 = d2 = d
so:
2 * v1 = 2 * d / ( t1 + t2 )
t1 + t2 = d / v1

Now, why does d / v1 look familiar? When you figure out what that's equal to, what does that suggest about t2?


Well, d/v1 = t1, so if t1 + t2 = t1, then t2 is impossible (you can't travel in 0 time) ?
Author:  DemonWasp [ Tue May 31, 2011 10:09 am ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Well, I wouldn't say t2 is impossible, just that it's zero. That means that, even if the student were travelling at the speed of light for the remainder of the journey, she would average below 10km/h for the entire journey. Crazy to think about, but true.
Author:  whoareyou [ Wed Sep 21, 2011 8:20 pm ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
Frames of Reference and Relative Velocity

A swimmer who achieves a speed of 0.75 m/s in still water swims directly across a river 72 m wide. The swimmer lands on the far shore at a position 54 m downstream from the starting point.
(a) Determine the speed of the river current.
(b) Determine the swimmer?s velocity relative to the shore.
(c) Determine the direction the swimmer would have to aim to land directly across from the starting position.

I am having trouble understanding the actual physics of this question. I mean I get the math: a) 72/0.75 = 96s --> 54/96 = 0.56m/s but I don't understand why. If the swimmer is swimming with the current, should it the speed be faster? And are my calculations even right because it says he swims across the river, but is then 54m downstream. So there are two different directions ...

Can anybody explain the physics of this question to me please?
Author:  DemonWasp [ Wed Sep 21, 2011 9:38 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Poorly-drawn ASCII-art diagram:
code:

    target  land
     x       x
-------------------------
>    |      /
>    |     /    ---->
>    |    /     water
>    |   /      flow
>    |  /       ---->
>    | /
>    |/
-------------------------
     x
    start


The swimmer is swimming directly across the river. His speed of 0.75m/s is in the "straight up" direction in my diagram. Because of the speed of the water, he won't follow the path straight across; instead, he will move on a straight diagonal (also shown in the diagram, using "/" ).

You have correctly calculated the speed of the river-water. This will also be the speed of the swimmer in the "downstream" direction. His speed, however, has to be considered in two "components". The first component is the "across river" direction. The second component is the "down river" direction. The actual speed of the swimmer will be the speed along the hypoteneuse (diagonal) formed by adding the two directed components together. You can find his actual velocity using Pythagorean Theorem (because this is a right-angled triangle, and you know the lengths of two of the sides -- in m/s, in this case). That's the answer for question B.

For question C, you will have to draw a different diagram; try it first and if you can't get it, ask back here.
Author:  whoareyou [ Sun Oct 09, 2011 8:37 pm ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
Atwood's Machine - One mass and one force

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I'm unable to scan my work onto the computer, but I've gotten to the part where I this equation:

ay = (FA - Fgblock)divided by (mA + mblock)

The problem is now, I have two unknowns ... FA and mA. This is where I'm stuck and I don't know if I missing something from before or ...

By the way, this is a question from the 2011 OAPT Physics Contest.
Author:  RandomLetters [ Sun Oct 09, 2011 9:36 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
What is mA?
Author:  whoareyou [ Mon Oct 10, 2011 12:12 pm ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
This is the work that I have.

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I'm stuck here because the m in the equation is for the total mass of the system, but I don't have it. I tried doing something like:

Posted Image, might have been reduced in size. Click Image to view fullscreen.

but you still need to know FA to get the mass, so I don't know what I'm doing wrong Sad
Author:  Tony [ Mon Oct 10, 2011 12:30 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
sure you know it. The block is the only thing with mass in this system.
Author:  Raknarg [ Mon Oct 10, 2011 5:40 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
I think I learned this in grade eight. When you're using pullys, the force required to equal the mass of the load is mass/# of pullies. This means the answer is e. though it might be something else, i didnt really pay attention in science when i was in junior high scool.
Author:  whoareyou [ Mon Oct 10, 2011 7:33 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
I used Tony's approach; I don't have the right answer though so I'm not able to check:

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Author:  whoareyou [ Mon Oct 10, 2011 8:10 pm ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
SIN 85-5

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To be honest, I don't know where to start. I've picked an approach by trying to find the acceleration (it says that it is constant), but every time I do, the acceleration won't give me the displacement at each given time interval. I've assumed that there was no initial velocity since the machine looks like it just let the rock fall.
Author:  Homer_simpson [ Mon Oct 10, 2011 8:35 pm ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
you can not assume 0 initial velocity since its not explicitly mentioned. So you can write two equations with two unknowns v0 based on the given information and solve for the two. then using the values you obtain calculate the displacement after three glargs
Author:  whoareyou [ Tue Oct 11, 2011 3:14 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
I got an equation: Δd = v1Δt + 0.5aΔt2 --> Δd = 20Δt+10Δt2. It gives me the correct answer to the question; the height is 150 NARGS. But I tried it a different way after.

v1 = v2 - aΔt, and v1 is Δd/Δt, but here I get that the initial velocity is 20, but the acceleration is 10. Is this a calculation error ( I don't think so). Why doesn't this method work?
Author:  Zren [ Tue Oct 11, 2011 4:39 pm ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
Where does it say v1 is 20?


v2 = v1 + a*t
v(t2->t3) = v(t1->t2) + ?*1

You had to use simple math to get v(t1->t2), and you needed to calculate the 'constant' acceleration. Why not just use a table of values, then fill in the velocity between the distances, then the acceleration between velocities. Then it's just simply filling out the table.

What I mean (contains answer):
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Deduce velocity.
Deduce acceleration between velocities.
Acceleration is constant! Fill rest of that column.
Determine velocity from 2-3 ___.
Determine (final) distance!

Bonus: discover init velocity and acceleration.

/me wishes compsci had toggle-able spoiler tags. v4 needs to use jQuery! :[
Author:  whoareyou [ Fri Oct 21, 2011 6:50 pm ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
This question isn't related to physics, but math. So, I have a function f(x) = 1/ (2x-1). By textbook says that to find the behaviors near the vertical asymptotes, you test values for 1/2 ^ + and 1/2 ^ -. Notice that the exponent is + and minus. Does anybody know what that means? If somebody does, I have a follow up question. Those signs don't mean positive of negative right? They mean approaching 1/2 from the left (-) and from the right (+). But when finding the horizontal asymptotes, you write as x --> + ∞, you write f(x) approaches 0, not 0^+ right? Because the + approaching from the right, not from the positive side? Also, I think that the +/- only applies to x-coordinates not f(x). Am I right?
Author:  apython1992 [ Fri Oct 21, 2011 7:39 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Yes, you are right. They mean approaching from the right or left (i.e. considering f(x) = 1/x, the limit as x approaches zero from the left is negative infinity, while approaching from the right it is positive infinity).

You're right again, you just check the limit as x approaches plus/minus infinity for horizontal asymptotes.
Author:  whoareyou [ Tue Oct 25, 2011 8:57 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Is there any source you can provide for this? Because I was searching Google, and came across http://en.wikipedia.org/wiki/One-sided_limit, and it said

"One should write either:

\lim_{x\to a^+}f(x)\ or \lim_{x\downarrow a}\,f(x) or \lim_{x \searrow a}\,f(x)

for the limit as x decreases in value approaching a (x approaches a "from the right" or "from above"), and similarly

\lim_{x\to a^-}f(x)\ or \lim_{x\uparrow a}\, f(x) or \lim_{x \nearrow a}\,f(x)

for the limit as x increases in value approaching a (x approaches a "from the left" or "from below")."
Author:  apython1992 [ Tue Oct 25, 2011 10:22 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
What Wikipedia says is correct, in the case of a being some finite number such that you can approach it from either side (for instance, x can approach 3 from the left, say by taking values 2.9, 2.99, 2.999, and so on). For infinite, it doesn't really make sense to approach it from both sides because infinite is taken as an arbitrarily large number. This goes for negative infinite as well. So we don't really consider taking one sided limits for finding horizontal asymptotes. But we do for vertical asymptotes, which always occur at some finite value.
Author:  whoareyou [ Tue Nov 15, 2011 9:30 pm ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
Trigonometric Identities

I was just wondering if it was possible to prove the sum to product identity --> cos - cosy = -2sin(x+y/2)sin(x-y/2) using compound angle formulas. Every time I try to do it, I start with the right side, and expand as I would sin(A+B)sin(A-B). I always end up with the line -2(cos^2B - cos^2A) and I don't think there is anything that I can do from there. I was able to do the other sum to product identities (cosC + cosD = 2cos(C+D/2)cos(C-D/2) and sinC + sinD = 2sin(C+D/2)cos(C-D/2)) so I don't really think I;m doing anything wrong ... Is there some kind of expansion or some special trick when I get to that last line?[/u]
Author:  bbi5291 [ Thu Nov 17, 2011 12:27 am ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
LS = cos x - cos y
= cos ((x+y)/2 + (x-y)/2) - cos ((x+y)/2 - (x-y)/2)
= [cos (x+y)/2 cos (x-y)/2 - sin (x+y)/2 sin (x-y)/2] - [cos (x+y)/2 cos (x-y)/2 + sin (x+y)/2 sin (x-y)/2]
= -2 sin(x+y)/2 sin (x-y)/2
= RS

or, read that in reverse if you wish
Author:  whoareyou [ Thu Nov 17, 2011 10:59 pm ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
Posted Image, might have been reduced in size. Click Image to view fullscreen.
mass of Fuzzy = 100.0kg
mass of cart = 200.0kg

I did my work by separating them into 2 different events, but it still never produced the right answer.

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The correct answer is 0.685m/s. Confused
Author:  apython1992 [ Fri Nov 18, 2011 5:10 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
I didn't go through your entire solution, but your first (and maybe only) mistake is in the third line of event 1, where you distributed m_F into the brackets. The mass is 100 kg, so it should be 500 + 100v_c, not 500 + 500v_c. Smile
Author:  whoareyou [ Fri Nov 18, 2011 6:11 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Oh yes that's right! But I redid it earlier and got the right answer. Thanks for the help anyways.
Author:  whoareyou [ Tue Feb 14, 2012 5:55 pm ]
Post subject:  Re: OAPT Physics Contest 2011 - Practice
MCV 4U0 - Instantaneous Rate of Change Using the Difference Quotient

In this example on Wikipedia, they evaluate the slope of the tangent for the function x^2 at x = 3 using the difference quotient. However, at the last step, the equation simplifies to lim [h-->0] f'(3) = 6+h, h cannot equal 0. So then, how come the limit can be evaluated if it is equal to zero? I mean, if h cannot equal 0, then why do we substitute it in after?

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Author:  DemonWasp [ Tue Feb 14, 2012 6:18 pm ]
Post subject:  RE:OAPT Physics Contest 2011 - Practice
Presumably you've already gotten some proof (from "first principles") that lim(h) as h->0 = 0. The corollary proof that lim(C+h) as h->0 = C [for some C which is constant with respect to h] is then pretty straightforward.

You never substitute in h = 0 per-se. It's more like you know the limit of that general form (with C = 6), so you can conclude about the limit of this specific example.

That said, I nearly failed most of my calculus courses, so don't take that answer as authoritative.

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