HSC Physics Question Thread

[Bubbly_bluey]

Hey guys! I've never see a question likethis before. How do I even start? Does the ball have both and initial horizontal and vertical velocity because the lift? 
Thanks

[jamonwindeyer]

Hey guys! I've never see a question likethis before. How do I even start? Does the ball have both and initial horizontal and vertical velocity because the lift? 
Thanks

Hey hey! This was a doozy last year. Think of it this way - When you are in a lift coming to the top, or a rollercoaster slowing down near the top about to drop, do you know how you feel like you are lifting off the ground? Or out of the chair? That's a slight upwards acceleration due to the motion of your reference frame, and that is what the ball experiences!

This slight upwards acceleration acts against gravity, and increases the flight time (the ball reduces height more slowly). The answer is therefore C

This is a tough one to explain - Does this make sense?

[teapancakes08]

"Two massless parallel current carrying wires are supported by strings as shown in the diagram. If each of the wires is 1 m long and carries a current of 50 mA, and the supporting strings are 85 cm long, calculate the force acting on each of the wires when the angle between the strings is 60º."

....

I'm not really sure how to start this off honestly...

[jamonwindeyer]

"Two massless parallel current carrying wires are supported by strings as shown in the diagram. If each of the wires is 1 m long and carries a current of 50 mA, and the supporting strings are 85 cm long, calculate the force acting on each of the wires when the angle between the strings is 60º."

....

I'm not really sure how to start this off honestly...

Hey! So this is the formula we need:



So, we know the wires are 1 metre long (\(L=1\)), we know the currents are 50mA (\(I_1=I_2=0.05\), and \(k\) is a constant. Only thing missing is the distance between them! We find that by drawing an imaginary line between the two wires and using the cosine rule:



Note that we could also get this result by noticing that the triangle is equilateral (since it is isosceles and the two base angles need to add to 120, they must be 60 degrees each), but this is the way you'd do it in general. So, with \(d=85\text{cm}=0.85\text{m}\) we can plug all our values into that formula above to get our answer

Hopefully that makes sense!

[teapancakes08]

Hey! So this is the formula we need:



So, we know the wires are 1 metre long (\(L=1\)), we know the currents are 50mA (\(I_1=I_2=0.05\), and \(k\) is a constant. Only thing missing is the distance between them! We find that by drawing an imaginary line between the two wires and using the cosine rule:



Note that we could also get this result by noticing that the triangle is equilateral (since it is isosceles and the two base angles need to add to 120, they must be 60 degrees each), but this is the way you'd do it in general. So, with \(d=85\text{cm}=0.85\text{m}\) we can plug all our values into that formula above to get our answer

Hopefully that makes sense!

I see ^^ (guess this is where have a maths background comes in handy )

I was really confused because the answers in the textbook threw me off...

[jamonwindeyer]

I see ^^ (guess this is where have a maths background comes in handy )

I was really confused because the answers in the textbook threw me off...

I feel like those answers are possibly incorrect! Because their method would make sense, but that angle at the top isn't 12.5 degrees. They've basically cut that top angle in half - So it should be 30 degrees!

[teapancakes08]

I feel like those answers are possibly incorrect! Because their method would make sense, but that angle at the top isn't 12.5 degrees. They've basically cut that top angle in half - So it should be 30 degrees!

That's what I thought too! But then since it was the textbook and my physics skills aren't very impressive I just stared at it for god knows how long until I decided to put it on ATARNotes...

[jamonwindeyer]

That's what I thought too! But then since it was the textbook and my physics skills aren't very impressive I just stared at it for god knows how long until I decided to put it on ATARNotes...

Ahaha don't worry, I was always afraid to contradict my textbook too!! Always good to check with us if you just don't think it seems quite right - No textbook is immune from errors

[Bubbly_bluey]

Hey! There was a question that goes along the lines of: An astronaut lands and stands on an astoroid that is orbiting the Sun between (I forgot the planets) Mars and Jupitur. Why does the astronaut not fall off?? or why does the astronaut feel weightless? -Something along those lines. It was a multiple choice:
A and B)was about the astoriod having a stable orbit around the sun
c) The force of the astoriod is negliable
D) Theres and equal opposite reaction
Thanks heaps! So sorry if the question is very vague (brain fart)

[jamonwindeyer]

Hey! There was a question that goes along the lines of: An astronaut lands and stands on an astoroid that is orbiting the Sun between (I forgot the planets) Mars and Jupitur. Why does the astronaut not fall off?? or why does the astronaut feel weightless? -Something along those lines. It was a multiple choice:
A and B)was about the astoriod having a stable orbit around the sun
c) The force of the astoriod is negliable
D) Theres and equal opposite reaction
Thanks heaps! So sorry if the question is very vague (brain fart)

Hey Bubbly! Sorry mate, but I can't really decipher it with the info you've given - Might just have to wait until you get the paper back. How do you think you went?

[Bubbly_bluey]

Hey Bubbly! Sorry mate, but I can't really decipher it with the info you've given - Might just have to wait until you get the paper back. How do you think you went?

haha its ok ill try get back to you when I get the question. As for the exam, a couple of questions from the exam were past HSC and I happened to have studied briefly the exact same questions! How crazy was that XD. But seriously wished I had paid more attention to the solutions to word my responses better

[jamonwindeyer]

haha its ok ill try get back to you when I get the question. As for the exam, a couple of questions from the exam were past HSC and I happened to have studied briefly the exact same questions! How crazy was that XD. But seriously wished I had paid more attention to the solutions to word my responses better

Ahaha gotta love it! A few people have had that - Definitely lucky! I had an exam in university last year that was literally one half the Exam for 2014, one half the Exam for 2015...  Crazy well done!! Enjoy your break friend, you've earned it!

[pikachu975]

a little clueless as to how they got v..

i got r to be 1.00x10^17 and then i can't exactly sub
this into the escape velocity formula cos it isn't the radius ..

Pretty sure it's asking for escape velocity FROM 1x10^17 m away from Earth, so just sub that as r and you're done!


Question:
If they tell you the torque acting on the WHOLE coil, would the torque acting on ONE side be half of the original torque?

[jamonwindeyer]

Pretty sure it's asking for escape velocity FROM 1x10^17 m away from Earth, so just sub that as r and you're done!


Question:
If they tell you the torque acting on the WHOLE coil, would the torque acting on ONE side be half of the original torque?

Totally missed that question, sorry Katniss! Think you are almost definitely right pikachu too, cheers heaps

Hmm, that's an interesting question. Torque is a rotational force, I don't think it would quite make sense to define it on one side of the coil only. Because if you separate one half, it's more just a linear force on one side if you catch my drift? Have never really thought about it before!

If you loosen the definition a bit though, I think it would be! The torque on a coil is generated by the forces acting on either side, and those forces are equal in magnitude. So yeah, I think it stands to reason that the torque would be half

[jakesilove]

Totally missed that question, sorry Katniss! Think you are almost definitely right pikachu too, cheers heaps

Hmm, that's an interesting question. Torque is a rotational force, I don't think it would quite make sense to define it on one side of the coil only. Because if you separate one half, it's more just a linear force on one side if you catch my drift? Have never really thought about it before!

If you loosen the definition a bit though, I think it would be! The torque on a coil is generated by the forces acting on either side, and those forces are equal in magnitude. So yeah, I think it stands to reason that the torque would be half

Torque doesn't act on an 'area' of the coil; it acts on the whole coil (or rather, the plane of the coil). That's why, when you try to figure out the angle relevant for the calculation of torque, you use the PLANE instead of a particular wire. If a question EVER asked you for the torque, even if it's like 'on wire XY', you still need to take the entire coil into account (as the magnitude is the same everywhere, you just do the normal calculation). So, I wouldn't focus too much on the physical representation of Torque, and just sub the hell out of the formula they've given you