University of Warwick

Coventry

This discussion is now closed.

Check out other Related discussions

Relativity Revision Thread

Scroll to see replies

Reply 20

.matt

benwellsday

Yeah I knew about eV in terms of J, I'm trying to work out how to get J in terms of kg, although I guess you just divide by c^2.

Ah yeah, that's all it is I think.

Slightly confused now though:
2007, 6e:
According to observations on the Earth, the nearest star to the solar system is
4.25 light-years away. A space ship which leaves the Earth and travels at a uniform
velocity takes 4.25 years, according to the ship-borne clock to reach this star. What
is the speed of the space ship, expressed as a fraction of the speed of light?

This is now making me think that my answer above for that question on the 2006 paper was wrong (about a 900 year round trip) - gamma obviously isn't 1 here, but using the same working as the 2006 question would make it 1...

EDIT: wait, think I've got it: taking proper time as 4.25 and time as 4.25c/v gives gamma = c/v, leading to v = c/sqrt(2)

University of Warwick

Coventry

Reply 21

benwellsday

How do you figure? It takes 900 years to the stationary observer on Earth, and in this question has the observer on the ship recording a time of 4.25 years.
Anyway for this one I got

\frac{1}{\sqrt{2}}c

which seems quite low.

Reply 22

.matt

benwellsday

How do you figure? It takes 900 years to the stationary observer on Earth, and in this question has the observer on the ship recording a time of 4.25 years.
Anyway for this one I got

\frac{1}{\sqrt{2}}c

which seems quite low.

Yup, just me being an idiot :o:

Reply 23

benwellsday

Did you get the same answer, because I don't trust my working looking at it now.
I did

\Delta T_0 = 4.25y

and

\Delta L_0 = 4.25 ly

4.25v = \frac{1}{\gamma} 4.25

by speed = distance / time. Then I did some messy stuff with units that looks dodgy.

Reply 24

01perryd

.matt

EDIT: wait, think I've got it: taking proper time as 4.25 and time as 4.25c/v gives gamma = c/v, leading to v = c/sqrt(2)

how do you get time as 4.25c/v sorry if its obvious. is it just distance/time

Reply 25

benwellsday

01perryd

how do you get time as 4.25c/v sorry if its obvious. is it just distance/time

I think it is. 4.25c is because it's light years I think, then v is velocity in Earth frame.

Reply 26

.matt

benwellsday

Did you get the same answer, because I don't trust my working looking at it now.
I did

\Delta T_0 = 4.25y

and

\Delta L_0 = 4.25 ly

4.25v = \frac{1}{\gamma} 4.25

by speed = distance / time. Then I did some messy stuff with units that looks dodgy.

t_0 = 4.25

t = {4.25c/v}

Using time dilation formula,

\gamma = \frac{t}{t_0} = \frac{c}{v} \Rightarrow v = c\sqrt{1 - \frac{v^2}{c^2}} \Rightarrow v = \frac{c}{\sqrt{2}}

01perryd

how do you get time as 4.25c/v sorry if its obvious. is it just distance/time

Yeah - 4.25 light years is the distance 4.25c, and the speed is v, so t = 4.25c/v

Reply 27

benwellsday

So how'd you all find it? Was anyone insane enough to try question 6?
I think it was a pretty average paper.
One question, 4) c) can you let the energies of the particles be the same because they have the same rest mass (plus conservation laws..)? It seemed a bit iffy, but since it was only 2 marks per part I assumed there was some sort of easy relation.

My top mistake in this exam :- Differentiating 1 to 1...that caused a 15 minute panic. Stupid.

Reply 28

rowzee

benwellsday

One question, 4) c) can you let the energies of the particles be the same because they have the same rest mass (plus conservation laws..)? It seemed a bit iffy, but since it was only 2 marks per part I assumed there was some sort of easy relation.

I didn't see at first if I should assume that they have the same velocity, but apparently they were supposed to have for the question to make sense.

But I agree that question 5. was easy marks compared to 6., the last part wasn't even strictly relativity, just differentiating stuff.

Reply 29

Essex_B

benwellsday

I did, fully regretting it now...

Reply 30

.matt

Yeah, I did question 6 too. Got halfway through question 5 and just thought, nah, and turned the page! It wasn't bad though, the only thing I wasn't sure about was the muons one - how were you supposed to find the 1% with the longest lifetimes without knowing the distribution of the lifetimes? I just worked out the average distance travelled and guessed 10km for the height :P

Reply 31

benwellsday

Looking at it now, the first 6 marks are substitution into an equation and verifying it (just the lorentz transforms right?), the next 7 are very similar to a past paper, the last 12 are the more annoying ones, although I reckon I could derive time dilation with a diagram and a light clock. Still prefer question 5), that muons thing looks weird, the 1% threw me off and the last one is one of those fiddly things I'd mess up more than likely.

Reply 32

.matt

A diagram probably would have been a better place to start, I started with the Lorentz transformation for time which made the question fairly short :o:

Bit worried that the last one might have been more complicated than I thought now: I worked out the velocity of one ship relative to the other and just stuck that in the length contraction formula.

Oh well, Relativity is done now, roll on Mechanics! :p:

Reply 33

Whiternoise

Physics Foundations was alright, relativity started off great - nice easy question 4, then i made a right numpty of myself and chose to do 5.

I really should have just given up and switched to 6 having around 20 minutes left :frown:

Reply 34

TheTallOne

Bump

Modules are now merged :colonhash:

but I'm still struggling with which inertial frame is which when considering Lorentz Velocity Transformations.

In the lecture he used two objects traveling at 0.75c apart in opposite directions relative to a planet. It got me in November and I'm still uncertain now.

What's the answer to Q12 in the problem sheet Wk 1-5? [Klingon ship moves at 0.2c away from a planet. Enterprise ship follows at 0.25c relative to Klingon ship, the Enterprise catches up with the Klingon at what speed when observed on the planet.]

I either get 3/7 or 9/19. Can't remember which one was correct :colonhash:

EDIT2: NM I got it now, at least for the case when you add two positive values. It has to get smaller (ie I think it is 3/7). Logic works. Explanation doesn't but I can work with this in the exam.

EDIT: All other top universities' physics departments are so uptight with their lecture material - they all require logins whereas Warwick doesn't.

Reply 35

.matt

TheTallOne

EDIT: All other top universities' physics departments are so uptight with their lecture material - they all require logins whereas Warwick doesn't.

Cambridge notes here seem decent enough. IIRC the University Physics textbook covers pretty much all of the module though, and explains it pretty well (don't know if it'll knocking about in the library or not)

(There's also a summary of special relativity at the start of Warwick's general relativity module here , but it doesn't really look like any of the first year stuff...)

Reply 36

TheTallOne

Ok, if anyone's here :yy:

A particle with rest mass

M_0

can decay at rest into a pair of particles each with rest
mass m0 . Calculate the following in the rest inertial frame of the original particle
using

M_0c^2=600

MeV and

m_0c^2=150

MeV .
(i) The total energy of each particle produced in the decay.
(ii) The magnitude of the relativistic linear momentum of each particle produced in
the decay.
(iii) The speed of the particles produced in the decay.

This is Relativity 2009, Q4 c)

i) As benwellsday above, I assumed that the total energy of each particle will be the same, so by conservation of energy each particle has total energy 300MeV.
ii) This is the part I'm then stuck on, I think I'm meant to use

E^2 = p^2c^2 + E_0^2

, where

E=300, E_0=150

. This gives me p=259MeV/c=0.866. Firstly is this correct and if so what are the units??
iii) I then used

E=\gamma m_0c^2

\gamma=2

and

v=\frac{\sqrt{3}}{2}c

Massive

for any help :h:

Reply 37

turgon

TheTallOne

Ok, if anyone's here :yy:

E^2 = p^2c^2 + E_0^2

, where

E=300, E_0=150

. This gives me p=259MeV/c=0.866. Firstly is this correct and if so what are the units??
iii) I then used

E=\gamma m_0c^2

\gamma=2

and

v=\frac{\sqrt{3}}{2}c

Massive

for any help :h:

Yeah that looks fine to me. Beware of using

E^2 = p^2c^2 + E_0^2

in all situations though (look at a similar question on the 2008 paper). You can keep the units as just MeV/c.

The mathmaticians on my floor have given up on revising for this for the time being. A lot of the physicists don't really know whats going on half the time either, but they've got other exams this week to stress over.

PS: If Minkowski space time diagrams come up, I'm gonna cry. I'll also cry if rolling friction, calculating moments of inertia or rockets come up.

Reply 38

TheTallOne

turgon

Yeah that looks fine to me. Beware of using

E^2 = p^2c^2 + E_0^2

If they ask you to show that

E^2 = p^2c^2 + E_0^2

holds then I think that's the time to use it :h:

.

Minkowski I'm OK with, with Mechanics I'm a bit iffy on questions similar to the asteroid one in 2008 and on pulleys in general, but I could just about blag it if I had a bit more time.

Reply 39

turgon

TheTallOne

If they ask you to show that

E^2 = p^2c^2 + E_0^2

holds then I think that's the time to use it :h:

.

Minkowski I'm OK with, with Mechanics I'm a bit iffy on questions similar to the asteroid one in 2008 and on pulleys in general, but I could just about blag it if I had a bit more time.