Trinity Admissions Test Solutions

Original post by -Gifted-

Thanks a ton, it makes sense to me. However, I used conservation of mechanical energy, but that yielded a different result where the velocity doesn't cancel out. Let me scan it.

Unfortunately energy isn't conserved.

Reply 22

Here it is,

Spoiler

Reply 23

Original post by Renzhi10122

Unfortunately energy isn't conserved.

Why is energy not conserved? I feel like I am missing a main concept here :/

Reply 24

Zacken

22

Original post by StrangeBanana

ah, some competition for me, then :P

What made you choose Pembroke? I hear the food's fantastic.

We may see each other at the interviews. ^^ Why King's, if you don't mind me asking? :tongue:

Ah - I swear I've seen you around quite a bit. :tongue:

King's just looked so impressive and spacious, it just gave off this certain feeling of grandeur that I appreciated. :tongue:

Reply 25

Renzhi10122

Original post by -Gifted-

Why is energy not conserved? I feel like I am missing a main concept here :/

Because the dust sticks, it's like a coliision between two particles where they stick together.

Reply 26

username1763791

OP

4

Original post by StrangeBanana

What made you choose Pembroke? I hear the food's fantastic.

That is pretty much the reason I chose it, I'm not gonna lie :lol:

Reply 27

Zacken

22

Original post by Renzhi10122

Because the dust sticks, it's like a coliision between two particles where they stick together.

Non-elastic collision so coefficient of restitution

e \neq 1

hence energy isn't conserved as kinetic energy is lost.

Reply 28

username1258398

19

Original post by -Gifted-

Thanks a ton, it makes sense to me. However, I used conservation of mechanical energy, but that yielded a different result where the velocity doesn't cancel out. Let me scan it.

You're welcome! :smile:

Like Renzhi and Zacken have said, energy isn't conserved; that's only when the collision is perfectly elastic, i.e. the coefficient of restitution is 1.

Original post by Zacken

Ah - I swear I've seen you around quite a bit. :tongue:

King's just looked so impressive and spacious, it just gave off this certain feeling of grandeur that I appreciated. :tongue:

Yeah, same here. That chapel is amazing.

Original post by Jordan\

That is pretty much the reason I chose it, I'm not gonna lie :lol:

It's a good enough a reason as any, to be honest. :tongue:

(edited 8 years ago)

Reply 29

Original post by Renzhi10122

Because the dust sticks, it's like a coliision between two particles where they stick together.

I see, so regardless of whether I take into account of the combined mass when the particles stick together energy is never conserved whenever they combine?

Reply 30

Renzhi10122

Original post by -Gifted-

I see, so regardless of whether I take into account of the combined mass when the particles stick together energy is never conserved whenever they combine?

Correct, but momentum always is, thus the approach by strangebanana..

Reply 31

Original post by Renzhi10122

Correct, but momentum always is, thus the approach by strangebanana..

It makes sense now, thank you guys :smile:

Reply 32

DFranklin

18

Original post by atsruser

You're right. Damn. sin^2 breaks this. Haven't got time to think about it in detail now. I'll delete that post forthwith.

It's not sin^2x that breaks it - it would work fine for

\int x \sin^2x \,dx

. It's the x^2 term that fails.

It works for multiplying by x because when you pair x with pi - x and add you get a constant (pi).

But when you pair x^2 with (pi-x)^2 it doesn't work so nicely.

Reply 33

Original post by DFranklin

It's not sin^2x that breaks it - it would work fine for

\int x \sin^2x \,dx

. It's the x^2 term that fails.

I meant to write xsin^2x i.e you can't write the integral as

\int x \cdot x\sin^2 x dx = \int xf(\sin x) dx

which is where my application of the result fails, I think. Not sure I followed your earlier comment re: arcsin BTW.

BTW I stole the (incorrect) approach from this post: http://www.thestudentroom.co.uk/showpost.php?p=60163271&postcount=19

I just did the second part of this, and IMHO, it's beautiful - I think you'll like it.

I can't see any easy way to get it though, apart from deriving the identity for x^3 from scratch though, which seems pretty challenging for a STEP candidate.

Reply 34

Original post by Zacken

Specimen Test 1, Question 9

Warning: this is going to be fairly rigorous, but it's the best I can do, if anybody can suggest how I can formalise/improve my solution, do let me know! :smile:

Spoiler

Reply 35

Zacken

22

Original post by atsruser

I meant to write xsin^2x i.e you can't write the integral as

\int x \cdot x\sin^2 x dx = \int xf(\sin x) dx

which is where my application of the result fails, I think. Not sure I followed your earlier comment re: arcsin BTW.

BTW I stole the (incorrect) approach from this post: http://www.thestudentroom.co.uk/showpost.php?p=60163271&postcount=19

I just did the second part of this, and IMHO, it's beautiful - I think you'll like it.

I can't see any easy way to get it though, apart from deriving the identity for x^3 from scratch though, which seems pretty challenging for a STEP candidate.

The tried exploiting the fact that it was an even function and the symmetry between sine and cosine, but to no avail - brute forcing it using IBP turned out to be the easiest solution. Unless, there's some kind of reduction formulae by considering

I_n = \int_0^{\pi} (x \sin x)^n \, \mathrm{d}x

although I can't see that being any simpler.

Reply 36

Original post by DFranklin

From memory, the x->pi-x trick works (the numerator of the integrand has been chosen carefully by the examiners).

It does - that's how I did it. But is that really expected in STEP without a hint?

I must admit, the form of the numerator didn't tell me much until I was almost finished with

\int x^3 f(\sin x) dx

so if they're expecting people to recognise the numerator, *and* see the substitution, I feel that they're angling for the budding von Neumanns only.

Reply 37