HSC Physics Question Thread

[RuiAce]

hi Jake

yeah i did do it just need confirmation for the time.
seems the answers may be wrong again.
my teacher is giving us the old version of surfing which has quite a few mistakes

For a virtually identical question, if you want checking you should post up your own solutions to be looked at

[katnisschung]

For a virtually identical question, if you want checking you should post up your own solutions to be looked at

hi could somebody confirm my answers for ux, uy and t

A cannon ball is fired at 40 degrees to the horizontal fro mthe top of a 218.7 m cliff and hits
 a taret 300m from the base of the cliff

surfing says the answers are
ux=30.64 ms-1
uy=25.7ms-1
t=9.8 secs

ignore the massive cross through my answer 

[and1_98]

hi could somebody confirm my answers for ux, uy and t

A cannon ball is fired at 40 degrees to the horizontal fro mthe top of a 218.7 m cliff and hits
 a taret 300m from the base of the cliff

surfing says the answers are
ux=30.64 ms-1
uy=25.7ms-1
t=9.8 secs

ignore the massive cross through my answer 

Think the answers from surfing are correct.

delta(x) = vtcos(40) = 300 - so, by rearranging, t = 300/(vcos(40)) when the range is achieved (when the ball hits the ground).

Then, sub this value of t into the the quadratic for delta(y) of form - delta(y) = -1/2at^2 + vsin(40) + 218.7

Allowing this quadratic to be = to 0 (as, when the ball has achieved the range of 300m, the vertical displacement of the projectile will be 0 (or -218.7m; depending on how you set up the quadratic initially)) proceed to solve for 'v'.

I had v^2 = 1597.481175... so v = 39.96850228
so uy = vsin(40) = 39.97 x sin(40) = 25.69 m/sec
and ux = vcos(40) = 39.97 x cos(40) = 30.62 m/sec

Then, for the time of flight, I just subbed the aforementioned 'v' back into the initial quadratic for delta(y), letting it = 0 (i.e. delta(y) = 0 since when the time of flight is achieved, the projectile reaches the ground); solving it with the quadratic formula.

You should get two values for t (the positive value being the one you want as t>0), one of them being 9.798270249... secs.

Hope this helps.

[jakesilove]

Think the answers from surfing are correct.

delta(x) = vtcos(40) = 300 - so, by rearranging, t = 300/(vcos(40)) when the range is achieved (when the ball hits the ground).

Then, sub this value of t into the the quadratic for delta(y) of form - delta(y) = -1/2at^2 + vsin(40) + 218.7

Allowing this quadratic to be = to 0 (as, when the ball has achieved the range of 300m, the vertical displacement of the projectile will be 0 (or -218.7m; depending on how you set up the quadratic initially)) proceed to solve for 'v'.

I had v^2 = 1597.481175... so v = 39.96850228
so uy = vsin(40) = 39.97 x sin(40) = 25.69 m/sec
and ux = vcos(40) = 39.97 x cos(40) = 30.62 m/sec

Then, for the time of flight, I just subbed the aforementioned 'v' back into the initial quadratic for delta(y), letting it = 0 (i.e. delta(y) = 0 since when the time of flight is achieved, the projectile reaches the ground); solving it with the quadratic formula.

You should get two values for t (the positive value being the one you want as t>0), one of them being 9.798270249... secs.

Hope this helps.

Looks great to me! Thanks for the answer!

[bholenath125]

Guys i have a lot of trouble with Projectile motion is there someone who could help me out

[RuiAce]

Guys i have a lot of trouble with Projectile motion is there someone who could help me out

Well what's the primary concern? What exactly troubles you with projectiles?

Some of Jamon's guides in the meantime

[thataveragevegan]

So I was wondering if i could get a typical exam response style for these questions? My teachers have verbally explained it in class but i was looking for a specific exam response, so i can compare it to my response. 

9. Explain the reason for the selection of infinity
as the place of zero gravitational potential
energy.

10. Explain how this selection of zero level results
in any point within a gravitational field having
a negative gravitational potential energy.

[jamonwindeyer]

So I was wondering if i could get a typical exam response style for these questions? My teachers have verbally explained it in class but i was looking for a specific exam response, so i can compare it to my response. 

Hey there! I'll give you how I'd answer these (very rough, by no means Gospel). These are also a little different to what you'd normally get in an exam style scenario! Especially the second one

Explain the reason for the selection of infinity as the place of zero gravitational potential energy.
Gravitational fields are infinite in size, and all objects inside a gravitational field, by definition, have GPE. Since they are infinitely large, to have a point with zero GPE, we need to be infinitely far away. Thus, we choose infinity as the place of zero GPE. This is also more convenient on an cosmological scale than using the surface of a planet as the zero point.

Explain how this selection of zero level results in any point within a gravitational field having a negative gravitational potential energy.

It is convention for GPE to increase as our distance from the centre of the gravitational field increases. However, since we have chosen zero as our infinity point, we would need to be increasing towards zero. Therefore, every point within a gravitational field must have a negative GPE, which approaches zero with increasing distance from the centre of the field.

I hope these help! Again, definitely not Gospel, there are many ways to answer questions like this. You could use dot points, or structure it differently, but if I had those questions in front of me that is how I'd respond

[thataveragevegan]

Hey there! I'll give you how I'd answer these (very rough, by no means Gospel). These are also a little different to what you'd normally get in an exam style scenario! Especially the second one

Explain the reason for the selection of infinity as the place of zero gravitational potential energy.
Gravitational fields are infinite in size, and all objects inside a gravitational field, by definition, have GPE. Since they are infinitely large, to have a point with zero GPE, we need to be infinitely far away. Thus, we choose infinity as the place of zero GPE. This is also more convenient on an cosmological scale than using the surface of a planet as the zero point.

Explain how this selection of zero level results in any point within a gravitational field having a negative gravitational potential energy.

It is convention for GPE to increase as our distance from the centre of the gravitational field increases. However, since we have chosen zero as our infinity point, we would need to be increasing towards zero. Therefore, every point within a gravitational field must have a negative GPE, which approaches zero with increasing distance from the centre of the field.

I hope these help! Again, definitely not Gospel, there are many ways to answer questions like this. You could use dot points, or structure it differently, but if I had those questions in front of me that is how I'd respond

Thanks heaps!

[Yasminpotts1105]

Which of the following is a true statement about scientific theories, such as Einstein's special theory of relativity?
- They are useful in making predictions.
- They are concepts that lack an experimental basis.
- They are ideas that can't be accepted until they have been tested.

[FallonXay]

Which of the following is a true statement about scientific theories, such as Einstein's special theory of relativity?
- They are useful in making predictions.
- They are concepts that lack an experimental basis.
- They are ideas that can't be accepted until they have been tested.

Hello!

"They are concepts that lack and experimental basis" is clearly wrong as Einstein used numerous forms of thought experiments to explain his special theory of relativity i.e looking at a mirror on a train travelling at the speed of light. Furthermore, Einstein also used experiments such as the Michelson-Morley experiment part of the evidence to support his theory. (Remember: The null result of the Michelson-Morley experiment didn't disprove the aether but was useful as partial evidence in supporting Einstein's theory)

"They are ideas that can't be accepted until they have been tested" This is clearly false as Einstein's Special Theory of Relativity was an accepted model before physical tests such as the Hafele-Keating experiment or muon decay in particle accelerators (which exhibit relativistic effects) were applied.

"They are useful in making predictions" is the correct answer as Einstein's theory of special relativity revolves around the constancy of light predicted the relativistic effects (time dilation, mass increase, length contraction). This, although could not be tested at the time, through numerous thought experiments predicted the effects of the constancy of light.

Hope this helped 

[jamonwindeyer]

Hello!

"They are concepts that lack and experimental basis" is clearly wrong as Einstein used numerous forms of thought experiments to explain his special theory of relativity i.e looking at a mirror on a train travelling at the speed of light. Furthermore, Einstein also used experiments such as the Michelson-Morley experiment part of the evidence to support his theory. (Remember: The null result of the Michelson-Morley experiment didn't disprove the aether but was useful as partial evidence in supporting Einstein's theory)

"They are ideas that can't be accepted until they have been tested" This is clearly false as Einstein's Special Theory of Relativity was an accepted model before physical tests such as the Hafele-Keating experiment or muon decay in particle accelerators (which exhibit relativistic effects) were applied.

"They are useful in making predictions" is the correct answer as Einstein's theory of special relativity revolves around the constancy of light predicted the relativistic effects (time dilation, mass increase, length contraction). This, although could not be tested at the time, through numerous thought experiments predicted the effects of the constancy of light.

Hope this helped 

Legend, thanks for the awesome answer

[thataveragevegan]

hey so i was wondering if i could get some clarification from hopefully Jake but any ATAR notes lecturer will do
In my physics notes that i bought from ATAR Notes  Newtown's law of Universal Gravitation is defined as
" Newtown's Law of Universal Gravitation determines the gravitational force acing on two objects due to their gravitational attraction.
G is the gravitational constant, m1 and m2 are the two masses and d is the distance between them. Remember this force acts on both objects . The less the mss of the object, the more it is affected. (F=ma).

What exactly do you guys mean by "the more it is affected", affected by what exactly ?

Thanks  

[jamonwindeyer]

hey so i was wondering if i could get some clarification from hopefully Jake but any ATAR notes lecturer will do
In my physics notes that i bought from ATAR Notes  Newtown's law of Universal Gravitation is defined as
" Newtown's Law of Universal Gravitation determines the gravitational force acing on two objects due to their gravitational attraction.
G is the gravitational constant, m1 and m2 are the two masses and d is the distance between them. Remember this force acts on both objects . The less the mss of the object, the more it is affected. (F=ma).

What exactly do you guys mean by "the more it is affected", affected by what exactly ?

Thanks

Hi! I wrote those notes, so I'd say I'm pretty qualified

So that statement refers to the fact that it is affected more by the gravitational force. In this case, I mean that it will accelerate more.

According to \(F=ma\), because the force is the same, larger masses will accelerate less when experiencing the same force

Reading those sentences in isolation like that, I agree it's a little ambiguous! The notes will be getting a little tidy up soon, so I'll actually make a note of adding a little extra to that bit hope this helps!

[Yasminpotts1105]

Could you please show worked solutions so we know how to actually get the answer, after trying for ages?