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Author Topic: VCE Physics Question Thread!  (Read 603290 times)  Share 

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Camo15

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Re: VCE Physics Question Thread!
« Reply #765 on: November 11, 2014, 05:44:12 pm »
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Well, if you've got a rope or something at an angle holding up the cantilever, the force from the wall will just ensure the whole system is equilibrium. So find the value of tension in a cable/rope using torque calculations (to oppose the vertical weight force of the cantilever), then look at translational equilibrium - if the system isn't in translational equilibrium then the wall is likely applying a force to the cantilever in some direction.

(Image removed from quote.)
So in this example, we'd find the tension in the cable such that it opposes the weight of the beam. But the tension in the cable would be made up of a vertical component which opposes the weight of the beam, and a horizontal component (because the tension is at an angle). So the wall (point A) will provide a horizontal force to work against the tension so the whole thing stays in equilibrium.

Ok, so the rebound force is working against the horizontal component of the tension?

Say that there's a cable with a tension of 5N that's got an upward component of 4N and a horizontal component of 3N, what would I use to figure out the rebound force?

If anything I've said is completely wrong let me know haha

Marrogi12

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Re: VCE Physics Question Thread!
« Reply #766 on: November 11, 2014, 07:58:42 pm »
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Hey guys,  I've got like 1 quarter of a page empty on my summary sheet , and I was wonder if I should put examples or find other theory to put in there , I don't really find examples helpful as I only use my summary sheet when I forget a formula or constant , but you never know in an exam ahaha 
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S61778

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Re: VCE Physics Question Thread!
« Reply #767 on: November 11, 2014, 08:16:14 pm »
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Hi,

Could someone please explain how standing waves support the existence of discrete energy levels in the atom?

Camo15

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Re: VCE Physics Question Thread!
« Reply #768 on: November 11, 2014, 09:15:44 pm »
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Hi,

Could someone please explain how standing waves support the existence of discrete energy levels in the atom?

Standing waves mean that if orbit is equal to a whole number of wavelengths, then a standing wave is formed. This means that electrons of those particular wavelengths can exist in the atom. Since different wavelengths correspond to different energies, only those energy states are possible for electron orbit, meaning that discrete energy levels exist.

allstar

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Re: VCE Physics Question Thread!
« Reply #769 on: November 11, 2014, 09:36:49 pm »
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hello,
can someone please explain this question?
why can't i use"
mgh = (1/2)kx^2
1 x 10 x 0.3 = (1/2) k (0.3)^2

thank you

Camo15

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Re: VCE Physics Question Thread!
« Reply #770 on: November 11, 2014, 09:46:14 pm »
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hello,
can someone please explain this question?
why can't i use"
mgh = (1/2)kx^2
1 x 10 x 0.3 = (1/2) k (0.3)^2

thank you

What's the answer to that?

I think it has to do with the fact that you can't take ugh from a certain point.

allstar

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Re: VCE Physics Question Thread!
« Reply #771 on: November 11, 2014, 09:47:47 pm »
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ans says k =33?

how would i use their formula f=kx?
« Last Edit: November 11, 2014, 09:51:17 pm by allstar »

theshunpo

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Re: VCE Physics Question Thread!
« Reply #772 on: November 11, 2014, 09:55:29 pm »
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ans says k =33?

how would i use their formula f=kx?

When it is stationary, the change in x is 30cm (0.3m). At this point The netforce =0, which means the force kx-mg=0. using x=0.3 you should be able to work out k=33
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speedy

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Re: VCE Physics Question Thread!
« Reply #773 on: November 11, 2014, 09:55:36 pm »
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hello,
can someone please explain this question?
why can't i use"
mgh = (1/2)kx^2
1 x 10 x 0.3 = (1/2) k (0.3)^2

thank you

Just do -> F = kx -> mg = kx -> 10 = k(.03) -> k = 33.33
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lzxnl

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Re: VCE Physics Question Thread!
« Reply #774 on: November 11, 2014, 09:59:28 pm »
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Standing waves mean that if orbit is equal to a whole number of wavelengths, then a standing wave is formed. This means that electrons of those particular wavelengths can exist in the atom. Since different wavelengths correspond to different energies, only those energy states are possible for electron orbit, meaning that discrete energy levels exist.

Your answer is fine for VCE purposes, but I'm going to say again: this model is heavily oversimplified. The Bohr model of the atom only works for hydrogen atoms, where there is only one electron. Even for helium, the fact that you have two electrons and repulsions between electrons means you'll have problems. Indeed, Bohr was utterly unable to do anything about atoms with more than one electron.

hello,
can someone please explain this question?
why can't i use"
mgh = (1/2)kx^2
1 x 10 x 0.3 = (1/2) k (0.3)^2

thank you

Let me ask you this: what scenario does the question ask for? It's NOT an energy question because the masses are stationary at both times. This means the net force = 0. An extra extension of 30 cm counters the addition of 1 kg of mass. So k*0.3m = 1kg * g = 9.8 N
k = 9.8/0.3 N/m or around 33 N/m

I'll explain why you can't use energy. When you use mgh = 1/2 kx^2, you are saying the spring energy change equals the gravitational energy change. As mechanical energy is conserved and as the total potential energy seems to have remained constant, the kinetic energy must also have remained constant. Initially, before adding a mass, the spring wasn't moving. Therefore the kinetic energy initially is zero. Solving mgh = 1/2 kx^2 thus solves for the next position where the spring isn't moving either. This is all assuming NO EXTERNAL INFLUENCE! Conservation of energy only applies to a closed system.

Let me ask you this. If you got a spring, attached a mass to it and let go, what would happen? It'd oscillate up and down like a slinky. At its lowest point where the kinetic energy is momentarily zero, the net force isn't zero. It was moving down before, it's now not moving. Clearly the net force is going straight up. But in the diagram your mass is remaining stationary, so the net force is zero. So evidently, the stationary point you solve for with mgh = 1/2 kx^2 is a different point to the point where you can just hang the mass indefinitely.

The solution to this dilemma is: once you put the new mass on, you yourself have to then gradually pull the spring down until it stretches a bit more and the extra stretch of the spring balances the extra weight force. This is when you let go of the spring and nothing happens. See how you had to actively hold the spring? The system is no longer closed. Conservation of energy no longer applies.

In summary, use net force because the spring isn't moving and is staying still. That's not an energy question.
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Marrogi12

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Re: VCE Physics Question Thread!
« Reply #775 on: November 11, 2014, 10:02:18 pm »
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Your answer is fine for VCE purposes, but I'm going to say again: this model is heavily oversimplified. The Bohr model of the atom only works for hydrogen atoms, where there is only one electron. Even for helium, the fact that you have two electrons and repulsions between electrons means you'll have problems. Indeed, Bohr was utterly unable to do anything about atoms with more than one electron.

Let me ask you this: what scenario does the question ask for? It's NOT an energy question because the masses are stationary at both times. This means the net force = 0. An extra extension of 30 cm counters the addition of 1 kg of mass. So k*0.3m = 1kg * g = 9.8 N
k = 9.8/0.3 N/m or around 33 N/m

I'll explain why you can't use energy. When you use mgh = 1/2 kx^2, you are saying the spring energy change equals the gravitational energy change. As mechanical energy is conserved and as the total potential energy seems to have remained constant, the kinetic energy must also have remained constant. Initially, before adding a mass, the spring wasn't moving. Therefore the kinetic energy initially is zero. Solving mgh = 1/2 kx^2 thus solves for the next position where the spring isn't moving either. This is all assuming NO EXTERNAL INFLUENCE! Conservation of energy only applies to a closed system.

Let me ask you this. If you got a spring, attached a mass to it and let go, what would happen? It'd oscillate up and down like a slinky. At its lowest point where the kinetic energy is momentarily zero, the net force isn't zero. It was moving down before, it's now not moving. Clearly the net force is going straight up. But in the diagram your mass is remaining stationary, so the net force is zero. So evidently, the stationary point you solve for with mgh = 1/2 kx^2 is a different point to the point where you can just hang the mass indefinitely.

The solution to this dilemma is: once you put the new mass on, you yourself have to then gradually pull the spring down until it stretches a bit more and the extra stretch of the spring balances the extra weight force. This is when you let go of the spring and nothing happens. See how you had to actively hold the spring? The system is no longer closed. Conservation of energy no longer applies.

In summary, use net force because the spring isn't moving and is staying still. That's not an energy question.
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Yacoubb

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Re: VCE Physics Question Thread!
« Reply #776 on: November 12, 2014, 08:16:21 pm »
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Did SO bad in Physics lol. Easily lost ~30 marks. Oh well, it's my bottom subject!

lolaishappy

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Re: VCE Physics Question Thread!
« Reply #777 on: December 19, 2014, 08:00:25 pm »
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Hellooo, I need clarification on tension in these types of problems (like below) and how to solve this one... part (ii)

(a) A car of mass 1400kg tows a trailer of mass 600kg due north along a level road at constant speed. The forces resisting the motion of the car and trailer are 400N and 100N respectively.

(b)If the car and trailer in part (a) with the same resistance forces, have a northerly acceleration of 2.0ms^2, what is:
(ii) the magnitude of the tension in the bar between the car and trailer?
Newb coming through

Zealous

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Re: VCE Physics Question Thread!
« Reply #778 on: December 20, 2014, 12:27:26 am »
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Hellooo, I need clarification on tension in these types of problems (like below) and how to solve this one... part (ii)

(a) A car of mass 1400kg tows a trailer of mass 600kg due north along a level road at constant speed. The forces resisting the motion of the car and trailer are 400N and 100N respectively.

(b)If the car and trailer in part (a) with the same resistance forces, have a northerly acceleration of 2.0ms^2, what is:
(ii) the magnitude of the tension in the bar between the car and trailer?
Let's resolve the forces acting on both the mass and trailer - so let's treat them as one object and ignore the trailer for now.

The car will need to apply a force to accelerate the object at 2ms-2 as well as overcoming friction so the required force is:


Now to find the tension in the bar and the trailer, we can resolve forces on the individual items - the car and the trailer so lets look at just the car. Have a look at the diagram below.



There's 3 forces acting on the car: the applied force of 4500N, the friction force of 400N and some tension force.  We also know the car by itself will also accelerate at 2ms-2. Therefore the NET force acting on just the car is F=ma=1400x2=2800.

Now we can setup an equation of forces:

F(NET)=F(Applied)-F(Friction)-F(Tension) which means F(Tension)=F(Applied)-F(Friction)-F(NET)=4500-400-2800=1300. Therefore the tension force is 1300N.

You can also check this result using the trailer - the forces acting on the trailer would be the tension force and the friction on the trailer. So F(NET)=1300-100=1200, a=1200/600=2m/s2. So we can confirm that the trailer will accelerate correctly if the tension force is 1300N.

So in summary, when solving tension questions:
1) Resolve the forces acting on the system as a whole (without considering the tension forces).
2) Resolve the forces of an individual item in the system to find the tension.

Hope this helps!
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lolaishappy

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Re: VCE Physics Question Thread!
« Reply #779 on: December 20, 2014, 09:02:20 am »
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Let's resolve the forces acting on both the mass and trailer - so let's treat them as one object and ignore the trailer for now.

The car will need to apply a force to accelerate the object at 2ms-2 as well as overcoming friction so the required force is:


Now to find the tension in the bar and the trailer, we can resolve forces on the individual items - the car and the trailer so lets look at just the car. Have a look at the diagram below.

(Image removed from quote.)

There's 3 forces acting on the car: the applied force of 4500N, the friction force of 400N and some tension force.  We also know the car by itself will also accelerate at 2ms-2. Therefore the NET force acting on just the car is F=ma=1400x2=2800.

Now we can setup an equation of forces:

F(NET)=F(Applied)-F(Friction)-F(Tension) which means F(Tension)=F(Applied)-F(Friction)-F(NET)=4500-400-2800=1300. Therefore the tension force is 1300N.

You can also check this result using the trailer - the forces acting on the trailer would be the tension force and the friction on the trailer. So F(NET)=1300-100=1200, a=1200/600=2m/s2. So we can confirm that the trailer will accelerate correctly if the tension force is 1300N.

So in summary, when solving tension questions:
1) Resolve the forces acting on the system as a whole (without considering the tension forces).
2) Resolve the forces of an individual item in the system to find the tension.

Hope this helps!

Thanks Zealous! Btw are applied force and Thrust the same thing? ???
« Last Edit: December 20, 2014, 09:16:17 am by lolaishappy »
Newb coming through