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A sphere of mass m1 m_1 moving with speed u1 u_1 collides directly with a similar sphere of mass m2 m_2 moving with speed u2 u_2 in the same direction, with u1>u2 u_1>u_2 . The coefficient of restitution between the two spheres is e e . Show that the loss of kinetic energy E E due to the collision satisfies the equation
2(m1+m2)E=m1m2(1e2)(u1u2)2 2(m_{1}+m_{2})E=m_{1}m_{2}(1-e^{2})(u_{1}-u_{2})^2

And here's the link for the paper with the projectile q. (q11) :
http://www.mathshelper.co.uk/STEP%20II%202006.pdf
Original post by EnglishMuon
A sphere of mass m1 m_1 moving with speed u1 u_1 collides directly with a similar sphere of mass m2 m_2 moving with speed u2 u_2 in the same direction, with u1>u2 u_1>u_2 . The coefficient of restitution between the two spheres is e e . Show that the loss of kinetic energy E E due to the collision satisfies the equation
2(m1+m2)E=m1m2(1e2)(u1u2)2 2(m_{1}+m_{2})E=m_{1}m_{2}(1-e^{2})(u_{1}-u_{2})^2

And here's the link for the paper with the projectile q. (q11) :
http://www.mathshelper.co.uk/STEP%20II%202006.pdf


Thank you! :smile:
Reply 22
Original post by EnglishMuon


Solving 'compound variable' Simultaneous Equations in STEP


Aye - this is always a good technique, what I tend to do is use the explicit sub u=xyu=xy and v=x+yv = x + y (or more generally, u=f(x,y)u = f(x,y) and v=g(x,y)v = g(x,y)), it is usually the intended solution to spot that the pattern in both equations is the same, so to speak and to exploit this fact. Anywho, this gives us a linear system in u,vu,v which is easier to solve.

I like making the explicit sub because it's easier to just power through the algebra, it saves some time re:writing xy and x+y over and over again, although admittedly not much here but it certainly would in other problems where either of f(x,y)f(x,y) or g(x,y)g(x,y) is nasty and I find that it helps avoiding silly mistakes as well.
Original post by Zacken
Aye - this is always a good technique, what I tend to do is use the explicit sub u=xyu=xy and v=x+yv = x + y (or more generally, u=f(x,y)u = f(x,y) and v=g(x,y)v = g(x,y)), it is usually the intended solution to spot that the pattern in both equations is the same, so to speak and to exploit this fact. Anywho, this gives us a linear system in u,vu,v which is easier to solve.

I like making the explicit sub because it's easier to just power through the algebra, it saves some time re:writing xy and x+y over and over again, although admittedly not much here but it certainly would in other problems where either of f(x,y)f(x,y) or g(x,y)g(x,y) is nasty and I find that it helps avoiding silly mistakes as well.

Yep very true. I'll definitely show the explicit substitution in future !
Original post by Zacken
Aye - this is always a good technique, what I tend to do is use the explicit sub u=xyu=xy and v=x+yv = x + y (or more generally, u=f(x,y)u = f(x,y) and v=g(x,y)v = g(x,y)), it is usually the intended solution to spot that the pattern in both equations is the same, so to speak and to exploit this fact. Anywho, this gives us a linear system in u,vu,v which is easier to solve.

I like making the explicit sub because it's easier to just power through the algebra, it saves some time re:writing xy and x+y over and over again, although admittedly not much here but it certainly would in other problems where either of f(x,y)f(x,y) or g(x,y)g(x,y) is nasty and I find that it helps avoiding silly mistakes as well.


Yep i do this aswell. Those subs are very useful.


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Day 3 Summary
Saving time on Fiddly Questions!
So the one thing I dislike most is spending valuable time rearranging fractions and expanding brackets when I could be thinking about an interesting problem, so here are a few things I look out for in manipulation, however obvious some of them may seem:

Not repeating your working:
A good example is with complex numbers and geometry- Consider the points P,Q,R,S represented by complex numbers p,q,r,s p,q,r,s such that PQRS (clockwise) is a parallelogram. In order to prove this fact is true, we need to show PQ=SR,SP=RQ |\vec{PQ}|=|\vec{SR}|, |\vec{SP}|=|\vec{RQ}| and the opposite sides are parallel. By the vector nature of complex numbers we can prove two opposite sides are equal in length and parallel just by showing PQ=SR,SP=RQ \vec{PQ}=\vec{SR}, \vec{SP}=\vec{RQ} . But note that we have not made any assumptions regarding which points P,Q,R,S may be, so we only need to show true for two of the sides and the other result follows.

Subtracting expressions during Induction:
Consider the recurrence Hn+1(x)+2xHn(x)+1=0 H_{n+1}(x)+2xH_{n}(x)+1=0 where Hn(x) H_{n}(x) is some predefined function. If we wanted to prove Hn+2(x)+(Hn(x))2=f(x) H_{n+2}(x)+(H_{n}(x))^{2}=f(x) for a given f(x) for example, it is often easier to consider Hn+2(x)+(Hn(x))2(Hn+1(x)+(Hn1(x))2 H_{n+2}(x)+(H_{n}(x))^{2}-(H_{n+1}(x)+(H_{n-1}(x))^{2} for example in the inductive step (as here we just need to show the resulting terms=0). (This is just a random example so may not give anything sensible in this case!)
Reply 26
Original post by EnglishMuon

Subtracting expressions during Induction:
Consider the recurrence Hn+1(x)+2xHn(x)+1=0 H_{n+1}(x)+2xH_{n}(x)+1=0 where Hn(x) H_{n}(x) is some predefined function. If we wanted to prove Hn+2(x)+(Hn(x))2=f(x) H_{n+2}(x)+(H_{n}(x))^{2}=f(x) for a given f(x) for example, it is often easier to consider Hn+2(x)+(Hn(x))2(Hn+1(x)+(Hn1(x))2 H_{n+2}(x)+(H_{n}(x))^{2}-(H_{n+1}(x)+(H_{n-1}(x))^{2} for example in the inductive step (as here we just need to show the resulting terms=0). (This is just a random example so may not give anything sensible in this case!)


Do you mean doing a concrete example of this? I'm not sure where the multiple squares are coming from here. :redface:
Original post by Zacken
Do you mean doing a concrete example of this? I'm not sure where the multiple squares are coming from here. :redface:


Yeah sure! Let me refer to my good friends the chebyshev polynomials defined by T0(x)=1,T1(x)=x,Tn+1(x)=2xTnTn1 T_{0}(x )=1, T_{1}(x)=x , T_{n+1}(x)=2xT_{n}-T_{n-1}.Then we can prove (Tn(x))2Tn+1(x)Tn1(x)=(T1(x))2T2(x)T0(x) (T_{n}(x))^2-T_{n+1}(x)T_{n-1}(x)=(T_{1}(x))^2-T_{2}(x)T_{0}(x) if you try the method above it comes out a little neater than other methods :smile:
Reply 28
Original post by EnglishMuon
Yeah sure! Let me refer to my good friends the chebyshev polynomials defined by T0(x)=1,T1(x)=x,Tn+1(x)=2xTnTn1 T_{0}(x )=1, T_{1}(x)=x , T_{n+1}(x)=2xT_{n}-T_{n-1}.Then we can prove (Tn(x))2Tn+1(x)Tn1(x)=(T1(x))2T2(x)T0(x) (T_{n}(x))^2-T_{n+1}(x)T_{n-1}(x)=(T_{1}(x))^2-T_{2}(x)T_{0}(x) if you try the method above it comes out a little neater than other methods :smile:


You're given the expression and you want to prove it, right? You're not meant to conjure it out of thin air? :redface:
Original post by Zacken
You're given the expression and you want to prove it, right? You're not meant to conjure it out of thin air? :redface:


Oh yea, I just meant on a basis of any induction proof that you may need to do.
So apparently my phone doesn't like actually posting my posts, but here are the some things I've come across the last couple of days :smile: :

Day 4 Summary
Using clues from the question to help find a method:

So almost EVERY STEP Q contains some 'funny wording' or an expression given in a certain form or something that may strike you as not just being a coincidence- mainly because its not!

Consider Q10 STEP III 2003. It's pretty easy to construct the differential equation
x¨=kxx˙ \ddot{x}=kx\dot{x} . So before actually seeing how to solve it, it can be useful to look through what the question is asking. Here it says "...and show that x x tends to infinity as t tends to π2dk \frac{\pi}{2dk} . I then asked myself "What function could this apply to" and it's pretty clear the answer is likely to be of the form Atan(dkt)+c Atan(dkt)+c . Thinking how this result can be achieved can then help find the correct method easily!

Also, using the form of the question is useful in checking answers.
In Q6 STEP III 2008, the question is some straight forwards solving of first order differential equations. But a silly detail I missed is eliminating the constant of integration after- There is not enough info provided by the initial conditions to get rid of the "+c"s, but the question directly states " Given that y=p2+2xp y=p^2+2xp ..." which does not include the constant terms, so you can just deduce they are 0. This could cost a couple of easy marks!
Hey Insight, did I remember correctly that you are doing m4/m5?
Original post by EnglishMuon
Hey Insight, did I remember correctly that you are doing m4/m5?


True that.
Original post by Insight314
True that.


Please would you be able to look over a strange solution for a question in a markscheme for me? I really can't see what they did, but it only took them about 1 line of working and drawing a couple of lines. Its q2 on the june 05 paper (https://013960f3b9a5a7ae938f512b8c85e3fe8578e56e.googledrive.com/host/0B1ZiqBksUHNYVF9pRzgzbXd2YkE/June%202005%20QP%20-%20M4%20Edexcel.pdf) if you wouldnt mind looking some time itd be appreciated :smile:
Original post by EnglishMuon
Please would you be able to look over a strange solution for a question in a markscheme for me? I really can't see what they did, but it only took them about 1 line of working and drawing a couple of lines. Its q2 on the june 05 paper (https://013960f3b9a5a7ae938f512b8c85e3fe8578e56e.googledrive.com/host/0B1ZiqBksUHNYVF9pRzgzbXd2YkE/June%202005%20QP%20-%20M4%20Edexcel.pdf) if you wouldnt mind looking some time itd be appreciated :smile:


ImageUploadedByStudent Room1459357616.052226.jpg

This was quite an easy question. What happened?

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Also, have you finished Mechanics 5? Or are you currently practicing Mechanics 4 before starting M5?

I am kind of fine with Relative Motion right now, but need to practice questions on Stability and maybe Damped Harmonic Motion, I still struggle to understand how extensions work in M3 SHM sometimes but there is enough time to grind papers before those exams.
Original post by Insight314
ImageUploadedByStudent Room1459357616.052226.jpg

This was quite an easy question. What happened?

Posted from TSR Mobile


I got the correct answer but was unsure of why we could construct that vector triangle like we did. So all I needed is that qVp is perpendicular to vq, and using similar triangles we can form that triangle drawn inside the circle? (lol sorry for my ignorance to these questions, ive never seen any of this written down anywhere so have to try and derive it first time when doing papers). But yeah Ive covered m5 and Im happy with that, seems easier than m4 in a way as its more pure maths inclined. The only questions I seem to mess up are the 'find the direction of velocity to have distance between these two objects as small as possible type questions". Thanks for the help!
Original post by EnglishMuon
I got the correct answer but was unsure of why we could construct that vector triangle like we did. So all I needed is that qVp is perpendicular to vq, and using similar triangles we can form that triangle drawn inside the circle? (lol sorry for my ignorance to these questions, ive never seen any of this written down anywhere so have to try and derive it first time when doing papers). But yeah Ive covered m5 and Im happy with that, seems easier than m4 in a way as its more pure maths inclined. The only questions I seem to mess up are the 'find the direction of velocity to have distance between these two objects as small as possible type questions". Thanks for the help!


Don't you have the M4 textbook?

ImageUploadedByStudent Room1459359140.606489.jpg
ImageUploadedByStudent Room1459359170.584614.jpg




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Original post by Insight314
Also, have you finished Mechanics 5? Or are you currently practicing Mechanics 4 before starting M5?

I am kind of fine with Relative Motion right now, but need to practice questions on Stability and maybe Damped Harmonic Motion, I still struggle to understand how extensions work in M3 SHM sometimes but there is enough time to grind papers before those exams.


Lol damped harmonic motion always ****s me up. One of my constants in the 2ODE is always opposite sign since i always write it the wrong way. Although Im on the review exercises so I shud be able to 95+ with some practice.
Relative morion is a treat for geomters :wink:
M5 is kinda hard since i dont know what In doing with forces sometimes lol!


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Original post by Insight314
Don't you have the M4 textbook?

ImageUploadedByStudent Room1459359140.606489.jpg
ImageUploadedByStudent Room1459359170.584614.jpg




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Wait whaaat? I asked my teacher and he said there are only old ones which dont cover the new spec. I just went straight into past papers and try to work it out as I go along. If you feel that books useful Ill definitely get one.

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