HSC Physics Question Thread

[david.wang28]

Howdy! So first, just to set it up how you did in the image, the impulse of \(I=3\text{kgms}^{-1}\) you found is also equal to the change in momentum, so:



Obviously mass is constant, so it's just the velocity change:



However, this is applied in the opposite direction to the initial velocity! The final direction of travel is opposite to the initial. So the final velocity is actually given by:



But because it is speed, we ignore direction, so the negative doesn't matter that's all you missed, you had a bit of a sign error with how the final speed was related to the initial speed

Yeah, the process was there, but the understanding wasn't. Thanks Jamon!

[david.wang28]

Hello,
I have trouble with question 3 and 4 in the link below. I have posted my working out, but somehow the answer for a) was 2.1 m/s and b) was 78%. Can anyone please help me with this question? Thanks

[fun_jirachi]

Hey there!

First part is perfect, got that right! The major flaw in the next part is that the question asks for what fraction of Laurence's initial kinetic energy is converted into other forms of energy. What you've done there is include Marcus's mass in the final calculation of kinetic energy, causing your value to be heavily skewed. Make sure you just use 39 x 1/2 x 2.1^2, and not 84 x 1/2 x 2.1^2 for the final.

Hope this helps

[david.wang28]

Hey there!

First part is perfect, got that right! The major flaw in the next part is that the question asks for what fraction of Laurence's initial kinetic energy is converted into other forms of energy. What you've done there is include Marcus's mass in the final calculation of kinetic energy, causing your value to be heavily skewed. Make sure you just use 39 x 1/2 x 2.1^2, and not 84 x 1/2 x 2.1^2 for the final.

Hope this helps

Yes this really helps. Thanks for the help!

[r1ckworthy]

Hey all,

How would you respond to the following question: Explain, in terms of the principles of physics involved, why gravitational potential energy is a negative quantity?

I kind of know the answer already, but I just don't know how to format and 'write' the answer itself. My answer right now is all over the place and isn't quite formal or logical.

[jamonwindeyer]

Hey all,

How would you respond to the following question: Explain, in terms of the principles of physics involved, why gravitational potential energy is a negative quantity?

I kind of know the answer already, but I just don't know how to format and 'write' the answer itself. My answer right now is all over the place and isn't quite formal or logical.

I reckon let's start with your answer! Can you share it? Chances are even if it isn't quite logical yet we can get it there with just a bit of moving things around

[r1ckworthy]

I reckon let's start with your answer! Can you share it? Chances are even if it isn't quite logical yet we can get it there with just a bit of moving things around

Alright, here is my answer:

At a distance of infinity away from the gravitational field of a central mass, the gravitational potential energy of an object will be zero. However, when work is done to raise the object a distance against the gravitational field, gravitational potential energy must increase. Therefore, gravitational potential energy must be negative so it can match both descriptions.

I'm fine with most of my answer, except the last part.

[jamonwindeyer]

Alright, here is my answer:

At a distance of infinity away from the gravitational field of a central mass, the gravitational potential energy of an object will be zero. However, when work is done to raise the object a distance against the gravitational field, gravitational potential energy must increase. Therefore, gravitational potential energy must be negative so it can match both descriptions.

I'm fine with most of my answer, except the last part.

Cool! Yeah, I like your answer. It pretty much lines up with the one in my head! I like your last statement, you could even say it as, "GPE must therefore be negative for both of these principles to be true." or something like that. Logically though, your answer works fine, nice work

[Jefferson]

For a transformer

Primary Coil:

• 2000W
• 1000V
• 2A

Secondary Coil

• 2000W
• 200V
• 10A

Question:
If this transformer is 80% efficient, calculate the output current.
Is it as simple as 0.8 × 10A?

If so, then why does efficiency effect current and not voltage? Shouldn't it affect power overall, (P=IV).
i.e. 0.8P = 0.8(IV).

[jamonwindeyer]

If so, then why does efficiency effect current and not voltage? Shouldn't it affect power overall, (P=IV).
i.e. 0.8P = 0.8(IV).

Hey! Really good question - The efficiency of transformers is actually a fairly complicated topic, the HSC glosses over it just a little bit. Essentially, the exact effect of efficiency on output voltage and current depends a bit on where the power loss occurs. Is it due to resistance in the windings? Eddy currents in the core? There are different models for transformer efficiency and different ways of accounting for this. The HSC tends to not worry about this and give you extra values to help you hone in on the answer.

Problem here is that you aren't given that additional direction - Like your induction question, I don't think this is worded in the best way. So you've got to make some assumptions to answer this. I think your assumption is fair - It says, "Okay, my output voltage is locked, so all of my power loss occurs due to some of the current being used elsewhere in the transformer." Incidentally, this would occur due to the current needed to magnetise the core, and you can also account for eddy current loss in this way. You've assumed there is no resistance in either primary or secondary coil, and that there is no leakage flux.

Are you expected to know all of that in your assumption? Of course not. You just assumed the output current was what was affected by efficiency, which given the lack of information suggesting otherwise, is a fair assumption to make

[Jefferson]

Hey! Really good question - The efficiency of transformers is actually a fairly complicated topic, the HSC glosses over it just a little bit. Essentially, the exact effect of efficiency on output voltage and current depends a bit on where the power loss occurs. Is it due to resistance in the windings? Eddy currents in the core? There are different models for transformer efficiency and different ways of accounting for this. The HSC tends to not worry about this and give you extra values to help you hone in on the answer.

Problem here is that you aren't given that additional direction - Like your induction question, I don't think this is worded in the best way. So you've got to make some assumptions to answer this. I think your assumption is fair - It says, "Okay, my output voltage is locked, so all of my power loss occurs due to some of the current being used elsewhere in the transformer." Incidentally, this would occur due to the current needed to magnetise the core, and you can also account for eddy current loss in this way. You've assumed there is no resistance in either primary or secondary coil, and that there is no leakage flux.

Are you expected to know all of that in your assumption? Of course not. You just assumed the output current was what was affected by efficiency, which given the lack of information suggesting otherwise, is a fair assumption to make

Wow!
Thank you so much for a detailed explanation yet again.
That really helped. <3.

[david.wang28]

Hello,
I have trouble with the question 18 in the link below. Can anyone please help me with this question (I have no idea how to do it)? Thanks

[fun_jirachi]

Hey there!

In an elastic collision, both KE and momentum are preserved. In an inelastic collision, the masses 'smoosh together' and move off as one mass.

Because m₁ = 3m₂, you can just sub in that whenever you would otherwise sub in m₁.


Because of the nature of an inelastic collision, this is a much easier calculation.


Hope this made sense!

[david.wang28]

Hey there!

In an elastic collision, both KE and momentum are preserved. In an inelastic collision, the masses 'smoosh together' and move off as one mass.

Because m₁ = 3m₂, you can just sub in that whenever you would otherwise sub in m₁.


Because of the nature of an inelastic collision, this is a much easier calculation.


Hope this made sense!

Yes this made sense. Thanks a lot!

[david.wang28]

Hello,
I'm having trouble with Q 2.5 in the link below (I have no idea how to do it). Can anyone please help me with these questions? Thanks!