Oxbridge Physics Prelims Revision

HELP!

I'm getting thoroughly frustrated at a question here, please please someone help I really cant see how to keep this simple:

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By considering the zero momentum frame for a collision between a mass m1, at speed v (in the lab frame) and a stationary (in the lab frame) mass m2, calculate the fraction of initial kinetic energy transferred to the mass m2.
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It's driving me crazy cos I just get stuck in a mess of algebra. Someone please help!

Reply 582

Willa

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By considering the zero momentum frame for a collision between a mass m1, at speed v (in the lab frame) and a stationary (in the lab frame) mass m2, calculate the fraction of initial kinetic energy transferred to the mass m2.
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In the ZMF (zero-momentum...), particle 1 has speed

u_1=v-\frac{m_1v}{m_1+m_2}=\frac{m_2v}{m_1+m_2}

, and particle 2 has speed

u_2=-\frac{m_1 u_1}{m_2}=-\frac{m_1v}{m_1+m_2}

.

Now use the energy and linear momentum conservation equations:

m_1u_1+m_2u_2=m_1v_1+m_2v_2

m_1u_1^2+m_2u_2^2=m_1v_1^2+m_2v_2^2

You know u_2 in terms of u_1, so you can eliminate v_1 and v_2 from the 2 equations. The algebra is generally a bit messy, but just plough through the working, making sure that your main aim is to eliminate variables, convert back into lab frame, and calculate kinetic energies, as required.

What I say may be obvious, but the key thing is not to mess around with variables, just try and eliminate them! Keep track of which variables are known, and which ones you are trying to determine in terms of these known variables.

Reply 583

Bezza

2

LennonMcCartney

In the ZMF (zero-momentum...), particle 1 has speed

u_1=v-\frac{m_1v}{m_1+m_2}=\frac{m_2v}{m_1+m_2}

, and particle 2 has speed

u_2=-u_1

.

but they're different masses so u2 doesn't equal -u1 - you need no momentum so doesn't

m_1 u_1 = -m_2 u_2\ so\ u_2 = -\frac{m_1 u_1}{m_2}

Reply 584

Bezza

but they're different masses so u2 doesn't equal -u1 - you need no momentum so doesn't

m_1 u_1 = -m_2 u_2\ so\ u_2 = -\frac{m_1 u_1}{m_2}

Sorry, yes, me being silly. He is correct of course, but the point is that u_2 can be written as a known function of u_1.

Reply 585

grrr I have tried doing it but i always end up with a nasty square root part which just is really pissing me off! Cos I have to find v2, which is the square root of something horrid, and add something to it...and it just isnt working! ARGH this is driving me crazy!

Reply 586

yay i finally did it. I hope to god a question like that doesnt come up in the exam!!

Reply 587

That is the problem with mechanics, you can never really tell if you're doing it right or not. But if the velocities fit the equations, then it must be correct.

Reply 588

Quick relativity question, just to check i've got things the right way round:

If an event occurs at a distance x' (as seen by the S' observer) at t'=0, then this means that a stationary observer determines that the event happened at t = gamma * vx'/c^2

that right?

Reply 589

Yes. I tend to remember that

x'=\gamma(x-\beta ct)

and

ct'=\gamma(ct-\beta x)

for the symmetry of it and the easier application to energy and momentum. And to switch the dashes, just replace

\beta

with

- \beta

.

Reply 590

OK I'm trying to analyse a situation in two different ways which should give the same answers but arent!

I have a rocket moving at speed 2c/3 and it emits a photon from x' which travels to the origin and then back to x' again. This obviously means t' = 2x'/c
So I want to see what time the stationary observer records, and using the standrd equation: t = gamma(t' + vx'/c^2) I obtain 4t'/root(5)

Now I wanted to check that this corresponded to length contraction, so I imagined the stationary observer seeing the rocket to be length x. That means that the first leg of the photon journey is t=x/(c+v) because it the photons relative speed to the rocket is c+v. And then adding on the return journey, t=x/(c+v) + x/(c-v)
substitute in the value for v above, to get: t=3x/5c + 3x/c = 18x/5c
But that should equal the previous calculated t, so 18x/5c = 4t'/root(5) = 8x'/croot(5)
so x = x' * (40/18root(5))

but using the standard transformation, that factor should be gamma = 3/root(5)

so where's my mistake!?

Reply 591

Anyone know of a good website that explains phasor diagrams well, I cant find any which help me understand how to use them. We seemed to cover them in lectures really quickly but they pop up everywhere and I dont think I can do them v well. Can anyone help?

Reply 592

Try hyperphysics. Look that up in google, and it should give you some ideas.

Reply 593

LennonMcCartney

Try hyperphysics. Look that up in google, and it should give you some ideas.

No that hasnt really helped me, I'm still confused. I can't do this subject damn it, why is nothing going right for me.....I couldnt do chemistry and computing, physics was my one strength, and now I can't even do that :frown:

And my rowing coach this week seems to be picking on me which makes me feel like I can't row either....and these new blades (shiny as they are) are causing a whole set of new blisters. Hmmmmpth nothing is going right for me :frown:

And with my first exam on monday, I am so underprepared that I'm gonna barely be able to answer any of the questions. This sucks, and it's all my fault!

Reply 594

Aww! Will, chill out! OK so this is a fairly tense time of the year. Worry a little bit, but not too much. Make sure to take plenty of breaks, just for yourself, where you don't worry, don't revise, don't do anything stressful.

Reply 595

I dont have time to take breaks, I've taken too many breaks already! I dont have the time to revise enough to get a good mark in these exams - well at least a mark which represents what I think I can achieve. I'm only just getting the hang of some modules, and I reckon I dont have time to learn entire chunks of some of them!

Reply 596

Hoofbeat

OP

14

GOOD LUCK ALL THE OXFORD PHYSICISTS IN THEIR EXAMS WHICH START TOMORROW (sorry Will, don't know when yours start but Good Luck to you too!!!).

No doubt I'll have lots of questions in the morning.

Reply 597