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#### appianway

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##### Harder Physics Questions
« on: January 06, 2010, 11:06:21 am »
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Hi guys,

I know that a lot of you are only just beginning the course, but I thought I'd post a link to the National Qualifying Examination papers for the Physics Olympiad. Despite not being VCE papers (and being regarded as very difficult!), they contain a plethora of mechanics questions, ranging from multiple choice akin to slightly more difficult (or in some rare cases, standard) VCE questions to much harder extended response. In saying that, they're definitely doable... if you think about what they're asking! Have a look through the papers and attempt anything to do with forces if you feel ready, and if you want, have a go at the other questions too. They're quite enjoyable and they'll get you thinking in a physics-y way! This is also a good resource for students in year 11 (or under, but I think this year all of the students who made it into the summer school were in year 11 when they sat the paper for physics) who are considering sitting the National Qualifying Exam and want to practise the questions.

Anyway, have fun!

http://www.aso.edu.au/www/index.cfm?itemid=31&CFID=5544640&CFTOKEN=5035285bc2b0ffbd-00F13507-B30D-FED8-30006C2417BB550C

#### Akirus

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##### Re: Harder Physics Questions
« Reply #1 on: January 06, 2010, 11:57:06 pm »
0
I'm bored, it's late and physics is fun. Here we go...

http://www.aso.edu.au/www/docs/2007PhysicsNQEPaper-FINALWeb.pdf

Question 13:

Quote
In any collision the sum of the momenta of all the bodies involved should be the same immediately
before and after the collision. In an elastic collision the same is true for the sum of the kinetic
energies of the bodies.

a. If a small blob with a small mass m has a velocity v directly towards an extremely heavy
wall, and the ensuing collision is elastic, what is the approximate final velocity of the blob?
(1 mark)

As indicated in the opening paragraph, no kinetic energy will be lost. The ball has a small mass and the wall is extremely heavy so it is unlikely that the kinetic energy of either will change, thus: $E_{k - initial} = E_{k - final}$

However, the velocity of the ball is a vector quantity, and the elastic collision will result in a reversal of direction. We can therefore model this with the equation:

$v_{final} = -v$

Quote
There is a box with lots and lots and lots of these blobs with mass m, all moving with different
speeds in different directions so that the total momentum is initially (approximately) zero. The box
is held firmly in place. Assume that all collisions are elastic.
b. What happens to the total kinetic energy of all the blobs? Why? (2 marks)
c. What happens to the total momentum of the all the blobs? Why? (2 marks)

b. The total kinetic energy will remain constant, as per the law of conservation of energy.

c. The total momentum of the blobs will remain constant, as per the law of conservation of momentum.
Quote
If there were a hole in the side of the box that the blobs could escape through and the initial
distribution of velocities of the blobs were the same as before:
d. What happens to the total kinetic energy of the blobs left in the box? Why? (1 mark)
e. What happens to the total momentum of the blobs left in the box? Why? (2 marks)

d. The total kinetic energy of the blobs will decrease as blobs escape from the hole. This is because the box ceases to be a closed system; kinetic energy is transferred to other external bodies by the escaped balls.
e. The total momentum of the blobs will decrease as blobs escape; this can also be considered an extension of the loss of total kinetic energy, since they're both linked directly to the mass and velocity of the blobs. For the same reason, as the box is no longer a closed system, the total momentum will decrease as blobs escape.

Quote
When a blown up balloon is released before it is tied up it shoots off away from the hand of the
person holding it.
f. Why does the balloon shoot off in the direction that it does? (2 marks)

f. When the balloon is held in the hand, the force exerted by the hand balances the forces of the particles pushing against the balloon from within. However, when it is released, the equilibrium of the balloon is lost and it will "shoot off away from the hand" in a seemingly random direction (due to the various forces of the gas molecules within).

I know some of these are wrong and some are just overly brief.
« Last Edit: January 07, 2010, 05:10:53 am by Akirus »

#### appianway

• Guest
##### Re: Harder Physics Questions
« Reply #2 on: January 07, 2010, 02:53:31 pm »
0
I think 13 a's correct. b and c look fine, as does d. However, with a, be careful with your wording. The kinetic energy's DEFINITELY going to change, but it's going to be very small. With b and c, try and give a bit more of an elaboration rather than just quote the laws - that's already shown in the question. Instead, perhaps relate back to a and show why the energy's conserved once again.

e should be correct, but you need to give a proper justification and make it a bit detailed. Be more specific about how the momentum changes and in what direction. Maybe give a mathematical explanation as well.

f is a little bit correct, but relate it not just to equilibrium, but Newton's laws. Be more specific about how the balloon influences the air particles, and perhaps relate it back to how this affects the motion of the the balloon.

#### Akirus

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##### Re: Harder Physics Questions
« Reply #3 on: January 07, 2010, 03:15:56 pm »
0
Yeah, I'm well aware the written responses are really brief. It was late and I was hoping for more calculations, so I got lazy.

The one regarding momentum with the escaped blobs is wrong in that I didn't consider its vector properties, and for the balloon I didn't mention the effect of the change in momentum due to the escaped particles.
« Last Edit: January 07, 2010, 03:17:51 pm by Akirus »

#### appianway

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##### Re: Harder Physics Questions
« Reply #4 on: January 07, 2010, 03:19:16 pm »
0
Yup. In my opinion though, the most interesting part is actually elaborating on the explanation to cement why it happens and understand the interactions between different things. Fun!

I'm actually leaving for the Summer School for this in 45 minutes, but I'll try and formulate some of my own questions like these (perhaps I'll be so exposed to them that I'll be able to? Here's hoping in 2.5 weeks when I come back

#### Akirus

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##### Re: Harder Physics Questions
« Reply #5 on: January 07, 2010, 10:34:08 pm »
0
I actually prefer doing tricky mathematical proofs rather than written theory. Of course, that's not to say I don't enjoy that as well, just to a lesser extent.

#### appianway

• Guest
##### Re: Harder Physics Questions
« Reply #6 on: January 23, 2010, 11:19:16 am »
0
That's the opposite of me! I don't like the mathematical manipulation (although 3 dimension integration and 4-vectors are fun), but I like looking at a problem and considering the physical principles at play.

Although the maths IS useful for proving things.

#### Cthulhu

• Guest
##### Re: Harder Physics Questions
« Reply #7 on: January 31, 2010, 07:45:19 pm »
0
Suppose that the radius of the Sun were increased to $5.90 * 10^{12}m$ (the average radius of orbit for pluto), that the density of the expanded sun was uniform and that the planets revolved within it. Calculate earths orbital speed within the 'new' sun and what would be the earths new period of revolution.

Errata: When I say the sun increases it's radius I also meant to say it's volume and mass. Volume is implied however. Uniform density as well.
« Last Edit: February 01, 2010, 08:10:03 pm by Cthulhu »

#### appianway

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##### Re: Harder Physics Questions
« Reply #8 on: January 31, 2010, 07:51:53 pm »
0
Ahh, would you do it something like this? I'm not going to compute it, but yes...

- Use Gauss' law to calculate the gravitational field strength at the point of the earth by considering the mass enclosed within the orbit (it should be the original mass x Volume of earth's orbit as a sphere/new volume of the sun)
-v^2/r = g because circular motion occurs, and solve for v

However, the current radius of the orbit is unknown, as is the mass of the sun. At present (without the new expansion), g = v^2/r, and v can be found by considering the period of the orbit and the distance (2 pi r). g also = G(mass of sun)/r^2, so if the two expressions for g are equated, one of the values (either the mass or the radius) can be found in terms of the other.

#### Cthulhu

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##### Re: Harder Physics Questions
« Reply #9 on: January 31, 2010, 08:08:01 pm »
0
Note that I said the density was (should be is) uniform? $\rho = \frac{m}{V}$ so if the density is uniform you could find the original density of the sun $\rho_{orig}$ to help you find $\rho_{new}$ which can then help you find the new mass of the sun.
Hint:$\rho_{orig} = \rho_{new}$
You're on the right track with $g = \frac{v^{2}}{r}$ but think in terms of forces...

#### appianway

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##### Re: Harder Physics Questions
« Reply #10 on: January 31, 2010, 08:11:11 pm »
0
Ah, just wondering, why are both densities equal? Shouldn't one be much smaller (seeing as the same mass is distributed over a larger volume)?

Oh, and edit: Just making sure I've understood the forces acting. So, is there a gravitational force and a buoyancy force?
« Last Edit: January 31, 2010, 08:26:23 pm by appianway »

#### Cthulhu

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##### Re: Harder Physics Questions
« Reply #11 on: January 31, 2010, 08:25:37 pm »
0
But it isn't the same mass it is the same density. The mass and volumes can be different as long as the density is the same.

#### appianway

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##### Re: Harder Physics Questions
« Reply #12 on: January 31, 2010, 08:27:12 pm »
0
Ah, it's not as thought the sun itself was expanded to fill the same volume (rather, additional matter is added to expand the volume).

#### Cthulhu

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##### Re: Harder Physics Questions
« Reply #13 on: January 31, 2010, 08:31:49 pm »
0
yes! something like that.

#### appianway

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##### Re: Harder Physics Questions
« Reply #14 on: January 31, 2010, 08:37:48 pm »
0
If the only two forces acting are the buoyancy force and the weight force, I thinkkkk I've got an expression... but I can't be bothered typing it up (it could be wrong anyway!)