HSC Physics Question Thread

[jamonwindeyer]

Jamon, i'm interested in the solution of the question that mq123 posted, can you please have a look at Jake's solution and comment on it?
Thanks

Sure thing!! Jake and I's Physics background is virtually identical, so I don't think I'll add much 

The approach to the first question is (I'm quite sure) definitely correct, I saw something similar in Physics last year. I do think there would be other methods, perhaps examining the components of force on the beam, but I like Jake's approach!

Less sure about the next bit, but I like Jake's approach and the reasoning is sound. However, I think the fact that the sign is not at the centre of the beam will change the answer, because the distance from the pivot point makes a massive difference to the force on the beam.

In general, that sign is what makes this question tricky. I'm not quite sure how to compensate for it. One possibility would be to move the sign to the end of the beam, so that it can be compensated for more easily, then find the lesser, equivalent mass that makes this valid. For example, if we find that moving it to the end of the beam doubles the torque, we could then adjust the mass to return it to its initial value. This would allow us to compensate for the vertical force of the sign in our vector diagrams on the right of the beam, which not changing the overall effect.

Sorry if that last thing was not explained quite nicely, but on the whole, I like Jake's approach! Just that one finer point I'd disagree with, but this is a bit beyond my capabilities 

[jakesilove]

Sure thing!! Jake and I's Physics background is virtually identical, so I don't think I'll add much 

The approach to the first question is (I'm quite sure) definitely correct, I saw something similar in Physics last year. I do think there would be other methods, perhaps examining the components of force on the beam, but I like Jake's approach!

Less sure about the next bit, but I like Jake's approach and the reasoning is sound. However, I think the fact that the sign is not at the centre of the beam will change the answer, because the distance from the pivot point makes a massive difference to the force on the beam.

In general, that sign is what makes this question tricky. I'm not quite sure how to compensate for it. One possibility would be to move the sign to the end of the beam, so that it can be compensated for more easily, then find the lesser, equivalent mass that makes this valid. For example, if we find that moving it to the end of the beam doubles the torque, we could then adjust the mass to return it to its initial value. This would allow us to compensate for the vertical force of the sign in our vector diagrams on the right of the beam, which not changing the overall effect.

Sorry if that last thing was not explained quite nicely, but on the whole, I like Jake's approach! Just that one finer point I'd disagree with, but this is a bit beyond my capabilities 

Having known really very little about classical mechanics, my reasoning for the final parts goes something like this. The example questions I found (linked above) bears a striking resemblance to our current predicament, except it doesn't include a sign. I think that the horizontal force approach is likely correct (it does not require taking into account any mass, just the Tension force and an angle). However, like Jamon said, the vertical force takes into account the mass of the beam, therefore perhaps the sign changes the formula (ie. it won't be just mass one + mass two * g).

However, that formula also doesn't require use of the length of the bar. Length appears to be independent of the force: all that is required is a Tension force, an angle and a 'mass'.

I'm not sure whether this is true, but this was my line of reasoning. If you add the mass of the sign at a point, this will change the center of mass. The new center (which will likely be somewhere *on* the point of contact between the sign and the rod, a % way along the infinitesimal connection) *should* have half of the mass on its right, and half of its mass on the left. Therefore, you could create a new rod (with a new length) that is the same mass as the original rod (including the sign), with a center of mass at the same point.

This new rod will have a mass of (mass one + mass two), probably be longer, and probably be denser. (Note: If you're worried about the length change affecting tension, just increase the density of the metal. No harm done) However, the formula doesn't care about the length of the rod (or density); only its mass. In all respects, this new rod is the same as the old, so I figure that the formula will still apply.

This could be completely off, I'm just explaining my reasoning from a currently-purely-quantum-mechanical-and-definitely-not-classical perspective.

I don't know if I explained myself well: hopefully it made sense!

Jake

[jamonwindeyer]

Having known really very little about classical mechanics, my reasoning for the final parts goes something like this. The example questions I found (linked above) bears a striking resemblance to our current predicament, except it doesn't include a sign. I think that the horizontal force approach is likely correct (it does not require taking into account any mass, just the Tension force and an angle). However, like Jamon said, the vertical force takes into account the mass of the beam, therefore perhaps the sign changes the formula (ie. it won't be just mass one + mass two * g).

However, that formula also doesn't require use of the length of the bar. Length appears to be independent of the force: all that is required is a Tension force, an angle and a 'mass'.

I'm not sure whether this is true, but this was my line of reasoning. If you add the mass of the sign at a point, this will change the center of mass. The new center (which will likely be somewhere *on* the point of contact between the sign and the rod, a % way along the infinitesimal connection) *should* have half of the mass on its right, and half of its mass on the left. Therefore, you could create a new rod (with a new length) that is the same mass as the original rod (including the sign), with a center of mass at the same point.

This new rod will have a mass of (mass one + mass two), probably be longer, and probably be denser. (Note: If you're worried about the length change affecting tension, just increase the density of the metal. No harm done) However, the formula doesn't care about the length of the rod (or density); only its mass. In all respects, this new rod is the same as the old, so I figure that the formula will still apply.

This could be completely off, I'm just explaining my reasoning from a currently-purely-quantum-mechanical-and-definitely-not-classical perspective.

I don't know if I explained myself well: hopefully it made sense!

Jake

Ohhh okay, I'm with you, that sounds reasonable!! If it worked in other examples then yep I reckon you are right!! I'm drastically out of my element regardless 

[jakesilove]

Ohhh okay, I'm with you, that sounds reasonable!! If it worked in other examples then yep I reckon you are right!! I'm drastically out of my element regardless 

I know exactly what you mean, I'll get back to you with a proper answer in about a year.

Jake

[jamonwindeyer]

I know exactly what you mean, I'll get back to you with a proper answer in about a year.

Jake

I will hold you to this 

[Maz]

Hey guys
firstly, thank you for having a go at the question
i had another go at it and i did it a different way and then i showed my teacher and he said it's right...i've put what i did below:
the sum of the clockwise torques has to equal the sun of the anticlockwise torques: C.W.T=A.C.W.T
doing that you get
(12kg x 9.8ms^-1 x 0.8 (half the beam's length)) + (45kg x 9.8 x 1.3 (1.6-0.3=1.3)) = Tsin20 x 1.6
evaluating that you get; 667.38= Tsin20 x 1.6
that gives the Tension (T) to equal 1.22x 10³ N

so thats what u do for the first bit

b) The vertical component is
45x9.8= 441 (down)               12x 9.8= 117.6
so then to get vertical component: 117.6+ 441 = 1.22x 10³ N + up
solving that gives a vertical component of 141.34N

Horizontal component
.22x 10³ x cos20 = 1146.42

then for part c you just need to find the resultant which can be done with pythagorus. i don't know how to get a triangle on here, but if u visualise a right angled triangle the 141.34N will be the vertical side and the 1146.42 will be the horizontal (adjacent to theta) side. this will leave the hypotenuse:
a² + b² =c²
(1.15x10³)² + 141.34² = c²
solving this gives c as 1.16 x 10³ which is the magnitude of the force
now as it is a vector we need an angle which is given simply by tan^-1(141.5/(1.15x10³)
which equals 7.015 degrees
so final answer for part c is 1.16 x 10³N at 7.015 degrees

hope this helps 

[jamonwindeyer]

Hey guys
firstly, thank you for having a go at the question
i had another go at it and i did it a different way and then i showed my teacher and he said it's right...i've put what i did below:
the sum of the clockwise torques has to equal the sun of the anticlockwise torques: C.W.T=A.C.W.T
doing that you get
(12kg x 9.8ms^-1 x 0.8 (half the beam's length)) + (45kg x 9.8 x 1.3 (1.6-0.3=1.3)) = Tsin20 x 1.6
evaluating that you get; 667.38= Tsin20 x 1.6
that gives the Tension (T) to equal 1.22x 10³ N

so thats what u do for the first bit

b) The vertical component is
45x9.8= 441 (down)               12x 9.8= 117.6
so then to get vertical component: 117.6+ 441 = 1.22x 10³ N + up
solving that gives a vertical component of 141.34N

Horizontal component
.22x 10³ x cos20 = 1146.42

then for part c you just need to find the resultant which can be done with pythagorus. i don't know how to get a triangle on here, but if u visualise a right angled triangle the 141.34N will be the vertical side and the 1146.42 will be the horizontal (adjacent to theta) side. this will leave the hypotenuse:
a² + b² =c²
(1.15x10³)² + 141.34² = c²
solving this gives c as 1.16 x 10³ which is the magnitude of the force
now as it is a vector we need an angle which is given simply by tan^-1(141.5/(1.15x10³)
which equals 7.015 degrees
so final answer for part c is 1.16 x 10³N at 7.015 degrees

hope this helps 

Ahhhh I see, nicely done! Jake and I were close but no cigar it seems  great job!!

[cajama]

Hi I've attached a multiple choice question from the 2008 paper (Q4.)
For graph questions like these, should I be memorising what certain graphs look like? I always find it difficult to determine what shape the graph should look like e.g. whether or not the curve should be concave down or up.

[RuiAce]

Hi I've attached a multiple choice question from the 2008 paper (Q4.)
For graph questions like these, should I be memorising what certain graphs look like? I always find it difficult to determine what shape the graph should look like e.g. whether or not the curve should be concave down or up.

No. You should be able to quote the relationship between time and vertical displacement in projectile motion.

The equation is given
Δy=u_yt + 1/2 a_y t²

It is clear that the relationship between height and time is a quadratic. This means that graphically:
If height was on the y-axis, and time on the x-axis, then you should have a parabola

Whereas if you INVERT IT:
If height is on the x-axis, and time is on the y-axis, then you should have a curve that looks like y=sqrt(x)

If you can't recognise the shape of y=sqrt(x) then that's a problem in itself. That's more of a maths problem.

[Alalamc]

Hi
How do you demonstrate the production of alternating current? (FIRST HAND INVESTIGATION)
Thankyou

[RuiAce]

Hi
How do you demonstrate the production of alternating current? (FIRST HAND INVESTIGATION)
Thankyou

There are heaps of ways to do this. One way is just to move a magnet back and forth in a solenoid.

[jamonwindeyer]

Hi
How do you demonstrate the production of alternating current? (FIRST HAND INVESTIGATION)
Thankyou

Rui is totally correct! Another (more interesting) way you can do it is get a really long bit of wire (a few metres) and get two people to swing it around like a skipping rope!! The wire is then moving through the earths magnetic field, and it is long enough that you actually do get a (very small) alternating current! You can even determine the direction of the earth's magnetic field by rotating the wire, the most current will be generated when the wire is perpendicular to the magnetic field, so when you have a maximum current, you know the wire is pointing East/West! At least with respect to magnetic north, which isn't perfectly north, but pretty close 

[jamonwindeyer]

Hi I've attached a multiple choice question from the 2008 paper (Q4.)
For graph questions like these, should I be memorising what certain graphs look like? I always find it difficult to determine what shape the graph should look like e.g. whether or not the curve should be concave down or up.

To flow on from Rui's answer, you definitely don't need to memorise the graph shapes. But don't stress, this is actually a really common thing people struggle with grasping!

There are two ways to think of it. If you are math-oriented, use Rui's approach above which requires the formula on your formula sheet!

Or, if you prefer, consider the graphs with some clever thinking. Note that the question says that the independent variable is graphed on the horizontal axis. The independent variable is what we are changing, that is, the height we are dropping from. Thus, graphs C and D are automatically excluded!! The variables aren't in the right spot. Additionally, graphs C and D are just rotated versions of Graphs A and B, they are perfectly valid graphs! Just not for this question 

With this done, we just need to choose between a straight line and a parabola. You can just know it is a parabola if you want. Alternatively, think of it this way. The ball will cover the first 10 metres of drop at some speed, then the next 10 metres faster, since it has accelerated. Then the next 10 metres will be faster again. So on. Every time we add 10 metres to the drop, it increases the time taken, but it increases it by a smaller and smaller increment each time. This way of thinking will give you the answer, Graph A 

Hope this helps! I had a bit of trouble with the reason the graphs were parabolic in questions like this when I first studied it, so I hope this helps!! 

[Happy Physics Land]

Hi
How do you demonstrate the production of alternating current? (FIRST HAND INVESTIGATION)
Thankyou

Another way to produce alternating current is just to simply turn the switch on and off quickly.

[RuiAce]

Another way to produce alternating current is just to simply turn the switch on and off quickly.

This is true if and only if we're speaking in the context of electromagnetic induction and basically a second coil is beside the first. Otherwise we just have a DC circuit that's being damaged because flicking the switch on and off constantly isn't even a good idea to begin with.