Physics marathon (hsc) (1 Viewer)

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Assume g remains as 9.8ms^-2?
The original gravity is 10 or 9.8 (whatever you want, 10 creates less messy numbers lol).

But gravity does change throughout flight, it requires use of a specific formula
 
Last edited:

LlamaBoi

Member
Joined
Oct 23, 2012
Messages
31
Gender
Male
HSC
2012
Equations of motion are:
x = 0
y = Vt - 4.9t^2
y = 10t - 4.9t^2
vertical velocity is given by: Vy = 10 - 9.8t
Ball reaches maximum height when Vy = 0
i.e. 0 = 10 - 9.8t
t = 1.0204 seconds
then silly Mysterious S decides to be a kent and reduces radius of earth to 6400000/4
now to find our new g, we use g = GM/r^2
g = 6.67*10^-11*6*10^24/(1600000^2)
g = 156.33 ms^-1 (we feel realllyy fat)
so new equation of vertical motion (which is all we need) is y = 10t - 156.33/2 *t^2
when y = 0,
0 = 10t - 78.165*t^2
0 = t[10 - 78.165t]
t = 0 is when we launched so we use t = 0.1279
but t = 0.1279 is the total time of flight, and we only need half, so t = 0.06395
Then for total time we add our time with normal gravity with our time for crazy gravity and we get
t = 1.0204 + 0.06395
t = 1.08 seconds :D
hopefully thats right cause this was a pain to type up
 

Nooblet94

Premium Member
Joined
Feb 5, 2011
Messages
1,044
Gender
Male
HSC
2012
Equations of motion are:
x = 0
y = Vt - 4.9t^2
y = 10t - 4.9t^2
vertical velocity is given by: Vy = 10 - 9.8t
Ball reaches maximum height when Vy = 0
i.e. 0 = 10 - 9.8t
t = 1.0204 seconds
then silly Mysterious S decides to be a kent and reduces radius of earth to 6400000/4
now to find our new g, we use g = GM/r^2
g = 6.67*10^-11*6*10^24/(1600000^2)
g = 156.33 ms^-1 (we feel realllyy fat)
so new equation of vertical motion (which is all we need) is y = 10t - 156.33/2 *t^2
when y = 0,
0 = 10t - 78.165*t^2
0 = t[10 - 78.165t]
t = 0 is when we launched so we use t = 0.1279
but t = 0.1279 is the total time of flight, and we only need half, so t = 0.06395
Then for total time we add our time with normal gravity with our time for crazy gravity and we get
t = 1.0204 + 0.06395
t = 1.08 seconds :D
hopefully thats right cause this was a pain to type up
Nope. It's beyond the scope of the course though. You need 4u maffs to do it.
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Equations of motion are:
x = 0
y = Vt - 4.9t^2
y = 10t - 4.9t^2
vertical velocity is given by: Vy = 10 - 9.8t
Ball reaches maximum height when Vy = 0
i.e. 0 = 10 - 9.8t
t = 1.0204 seconds
then silly Mysterious S decides to be a kent and reduces radius of earth to 6400000/4
now to find our new g, we use g = GM/r^2
g = 6.67*10^-11*6*10^24/(1600000^2)
g = 156.33 ms^-1 (we feel realllyy fat)
so new equation of vertical motion (which is all we need) is y = 10t - 156.33/2 *t^2
when y = 0,
0 = 10t - 78.165*t^2
0 = t[10 - 78.165t]
t = 0 is when we launched so we use t = 0.1279
but t = 0.1279 is the total time of flight, and we only need half, so t = 0.06395
Then for total time we add our time with normal gravity with our time for crazy gravity and we get
t = 1.0204 + 0.06395
t = 1.08 seconds :D
hopefully thats right cause this was a pain to type up
That looks about right, nice work :)

Ill post another question soon (part 2 of Mysterious S's adventure!)
 

Sy123

This too shall pass
Joined
Nov 6, 2011
Messages
3,730
Gender
Male
HSC
2013
Nope. It's beyond the scope of the course though. You need 4u maffs to do it.
Well the way I saw it, first find the time taken for the projectile to reach max height, under gravity 10, it takes 1 second. Now lets find the new gravity of the earth, using our formula. We find it to be 160. Then substitute into our displacement equation for y again, and done, finding 0.25 seconds, add together 1.25 seconds?
 

Nooblet94

Premium Member
Joined
Feb 5, 2011
Messages
1,044
Gender
Male
HSC
2012
That looks about right, nice work :)

Ill post another question soon (part 2 of Mysterious S's adventure!)
The force due to gravity doesn't change simply because you've shrunk the earth. It's proportional to the distance from the centre of mass, which wouldn't change since you said it maintains its shape.

Well the way I saw it, first find the time taken for the projectile to reach max height, under gravity 10, it takes 1 second. Now lets find the new gravity of the earth, using our formula. We find it to be 160. Then substitute into our displacement equation for y again, and done, finding 0.25 seconds, add together 1.25 seconds?
To do it properly you'd need to take into account that acceleration is proportional to the inverse square of the distance from the centre of mass and then do entirely new equations of motion. Any other way is going get a totally wrong answer because of the large distances involved. It's still a very good question though, just more suited for 4U.
 
Last edited:

bleakarcher

Active Member
Joined
Jul 8, 2011
Messages
1,509
Gender
Male
HSC
2013
Problem with that question though is that you can no longer assume g is constant as it begins falling after it reaches max. height really. It should increase significantly as it falls closer to the surface of the planet. It works obviously if you assume that g does stay constant though.
 

joeyz_da_ak2plz

New Member
Joined
Sep 30, 2012
Messages
14
Gender
Male
HSC
2012
the work done to move an object from a very large distance away(infinity) to a point within a gravitational field
GPE=-GMeMo/r
 

traiwit

Member
Joined
Nov 11, 2011
Messages
243
Gender
Male
HSC
2012
Uni Grad
2016
1. describe how did hertz determine the speed of EMR
2. How does quantm teory satisfactorily explain blakbody radiation ?

assume both of them are 4marks question
 

deswa1

Well-Known Member
Joined
Jul 12, 2011
Messages
2,256
Gender
Male
HSC
2012
1. describe how did hertz determine the speed of EMR
2. How does quantm teory satisfactorily explain blakbody radiation ?

assume both of them are 4marks question
1. Maxwell predicted that since a changing electric current induced a magnetic field and a changing magnetic field induced an electric current, then theoretically electromagnetic waves should be able to self- propogate throughout space. Hertz used a high voltage induction coil in order to create an oscillating electric field-> this changing electric field induced an oscillating electric spark in a receiving coil (draw diagram). He then reflected the created EMR (radio waves) off a metal plate into the receiving coil in order to create intereference effects between the original EMR and the reflected EMR. Through considering this EMR, he was able to determine the wavelength of the radio waves. The frequencey of the waves was the same as the frequency of the induction coil so that was already known. Through using the equation v=f(lambda), he was thus able to generate a value for the speed of EMR- roughly 3x10^8 m/s.

2. Black body radiation refers to radiation emitted by a perfect absorber in thermal equilibrium. Classical theory suggested that as the wavelength of the radiation emitted approached zero, its intensity would approach infinity. This is impossible as the total energy emitted (area under the curve) would be infinite, contravening the law of conservation of energy. In addition, it didn't agree with experimental results and this was thus known as the ultraviolet catastrophe (draw curve). Planck realised this problem and attempted to resolve it by quantising the radiation emitted. Whilst classical theory stated that the radiation emitted could occupy all values, Planck said that the energy of black body radiation had to be in the form of E=hf (where h is Planck's constant and f is the frequency). This 'mathematical trick' allowed the theoretical black body radiation curve to match the empirical data, thus satisfactorily explaining black body radiation. Einstein later took the concept of quantisation and applied it to the photoelectric effect, further solidifying its strength as the idea of quantam theory was applied to solve a different problem.

Please criticise ANYTHING that is even a tiny bit wrong in the above or misworded or I missed anything etc. Don't worry about my feelings haha- smash it please. Thanks

New question: Discuss Einstein and Planck's differing views about whether scientific research is removed from social and political forces
 

manscux

Member
Joined
Oct 29, 2011
Messages
289
Gender
Male
HSC
2012
1. Maxwell predicted that since a changing electric current induced a magnetic field and a changing magnetic field induced an electric current, then theoretically electromagnetic waves should be able to self- propogate throughout space. Hertz used a high voltage induction coil in order to create an oscillating electric field-> this changing electric field induced an oscillating electric spark in a receiving coil (draw diagram). He then reflected the created EMR (radio waves) off a metal plate into the receiving coil in order to create intereference effects between the original EMR and the reflected EMR. Through considering this EMR, he was able to determine the wavelength of the radio waves. The frequencey of the waves was the same as the frequency of the induction coil so that was already known. Through using the equation v=f(lambda), he was thus able to generate a value for the speed of EMR- roughly 3x10^8 m/s.

2. Black body radiation refers to radiation emitted by a perfect absorber in thermal equilibrium. Classical theory suggested that as the wavelength of the radiation emitted approached zero, its intensity would approach infinity. This is impossible as the total energy emitted (area under the curve) would be infinite, contravening the law of conservation of energy. In addition, it didn't agree with experimental results and this was thus known as the ultraviolet catastrophe (draw curve). Planck realised this problem and attempted to resolve it by quantising the radiation emitted. Whilst classical theory stated that the radiation emitted could occupy all values, Planck said that the energy of black body radiation had to be in the form of E=hf (where h is Planck's constant and f is the frequency). This 'mathematical trick' allowed the theoretical black body radiation curve to match the empirical data, thus satisfactorily explaining black body radiation. Einstein later took the concept of quantisation and applied it to the photoelectric effect, further solidifying its strength as the idea of quantam theory was applied to solve a different problem.

Please criticise ANYTHING that is even a tiny bit wrong in the above or misworded or I missed anything etc. Don't worry about my feelings haha- smash it please. Thanks

New question: Discuss Einstein and Planck's differing views about whether scientific research is removed from social and political forces
For question 2 the question is saying how..... would you need to go into modes and how energy can only be absorbed in equal multiples of lambda. Than explain how excitation occurs at different band levels causing the energy absorbed to be dependant on the position of the electron on the shell. i.e since the electrons cannot be in the forbidden gap, the energy absorbed can only be in discrete quantities....
 

BoredAzn

New Member
Joined
May 24, 2011
Messages
12
Gender
Undisclosed
HSC
2012
What is the p-njuction? in terms of structure, electric field, postive holes. With diagram?
thanks
 

deswa1

Well-Known Member
Joined
Jul 12, 2011
Messages
2,256
Gender
Male
HSC
2012
For question 2 the question is saying how..... would you need to go into modes and how energy can only be absorbed in equal multiples of lambda. Than explain how excitation occurs at different band levels causing the energy absorbed to be dependant on the position of the electron on the shell. i.e since the electrons cannot be in the forbidden gap, the energy absorbed can only be in discrete quantities....
Ah yes- cheers :) How would you write the bolded bit as an answer? Thanks bro

What is the p-njuction? in terms of structure, electric field, postive holes. With diagram?
thanks
Ok here's a diagram:



Basically at the start, you have a P type semiconductor and an N-type semiconductor. The P type has an excess of positive holes and the N-type electrons. Now the some of the electrons from the N- type diffuse into the P-type and some of the holes from the P-type diffuse into the N-type. This makes the N-type side positively charged and the P-type side negatively charged. It also establishes an electric field in between them known as the depletion zone.
 

RishBonjour

Well-Known Member
Joined
Aug 14, 2011
Messages
1,261
Gender
Male
HSC
2012
What is the p-njuction? in terms of structure, electric field, postive holes. With diagram?
thanks
p-n junction is just a p type and n type joined together. This essentially forms solar cells.
we know that n type has surplus of electrons while p type has holes (but they are both neutral since electrons = protons)
When they are joined, electrons from n type migrate to p type and consequently the n type becomes positively charged while the p type becomes negatively charged (only close to where they are joined). This is known as the depletion zone - an electric field between - can act as a diode now.




EDIT: deswa1 beat me to it with pretty diagrams



How does the law of conservation of momentum apply to the motion of a rocket?
 
Last edited:

kiinto

Member
Joined
Sep 10, 2011
Messages
40
Gender
Male
HSC
2012
Ok here's a diagram:



Basically at the start, you have a P type semiconductor and an N-type semiconductor. The P type has an excess of positive holes and the N-type electrons. Now the some of the electrons from the N- type diffuse into the P-type and some of the holes from the P-type diffuse into the N-type. This makes the N-type side positively charged and the P-type side negatively charged. It also establishes an electric field in between them known as the depletion zone.
Careful. The p/n type semiconductors only gain a charge very close to the junction, in the depletion zone. The rest of them remains neutral, as far as I know.



How does the law of conservation of momentum apply to the motion of a rocket?
(Diagram) The Law of Conservation of Momentum states that total momentum must remain constant (i.e. P-rocket + P-exhaust = 0). As the rocket flies, it burns fuel, causing it to lose mass. This means that, as the momentum of the exhaust remains constant, the velocity of the rocket must increase (mv-rocket + mv-exhaust = 0). (If worth more marks: engage F=ma/g-force choip).

"Sometimes scientists are ahead of their time"
Analyse this statement with reference to the work of a scientist you have studied in your HSC Physics course
 
Last edited:

nirukk

Member
Joined
Jun 26, 2011
Messages
62
Gender
Male
HSC
2012
Explain how J.J Thompson calculated the charge to mass ratio of the cathode rays (3/4 marks)
 

kiinto

Member
Joined
Sep 10, 2011
Messages
40
Gender
Male
HSC
2012
Explain how J.J Thompson calculated the charge to mass ratio of the cathode rays (3/4 marks)
Thompson used a modified CRT, fitted with a collimator, magnetic (hemmaholtz) coils and electric plates, to define the charge to mass ratio. First he equated the force on the cathode ray due to the magnetic and electric fields respectively, such that the beam passed through unaffected. This allowed him to mathematically derive the velocity of the particles by equating the two formulas (qvB=qE). He then switched off the electric plates and measured the radius of the path of the cathode ray, due to the magnetic field alone. By combining the two aforementioned results (i.e. velocity and radius) he was able to define the charge to mass ratio. (q/m = v/Br)
 

Ichiii

New Member
Joined
Feb 23, 2011
Messages
7
Gender
Male
HSC
2012
He first used the magnetic field to find the deflection of the cathode ray
F(b) = F(c)
qvB = mv^2/r
q/m = v/Br

Then he aligned the beam until it was straight by using an electric field
F(b)=F(E)
qvB = qE
v=E/B

Then he subbed the velocity back into the charge to mass ratio (q/m) to get
q/m = E/B^2r
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top