The Student Room Group

Scroll to see replies

Reply 1
just work out speeds in terms of U and E after the collision then sub into Kinectic energy befor eand after, subtract and you have your asnwer.
Oh i did this question today at school!!!!
my teacher had to photocopy his soolution for all of us - noone could understand it.
...

... 32 \frac{3}{2} A4 sheets for this question. .........
Here's my teachers solution.

And even he said it was definitely in his top 10 Maths/Further Maths questions he'd ever done.

Pretty nasty tbh.
Reply 5
Dude that question was so horrible, did it last year and it took like 3 sides of A4!!!
Reply 6
thx ppl
Reply 7
I asked the same question 4 weeks ago, and was told not to bother
http://www.thestudentroom.co.uk/showthread.php?p=16464595#post16464595
Reply 8
well now u have the answer!!
thanks to me - well my very smart maths teacher - but I'll take the credit!! :tongue:
Sorry, couldn't resist!
I wouldn't normally do full workings, but in this case....

Here is the initial set up, and the rest of the working is in the spoiler:

Conservation of momentum gives:

m1u1+m2u2=m1v1+m2v2m_1u_1+m_2u_2=m_1v_1+m_2v_2

Rearranging we get:

m1(u1v1)=m2(v2u2)m_1(u_1-v_1)=m_2(v_2-u_2)

Thus: m1m2=v2u2v1u1\displaystyle \frac{m_1}{m_2}=\frac{v_2-u_2}{v_1-u_1} And call this equation (1)

Consideration of restitution gives:

v2v1=e(u2u1)v_2-v_1=-e(u_2-u_1) And call this equation (2)

Now loss in KE is E. So:

E=12m1(u12v12)+12m2(u22v22)E=\frac{1}{2}m_1(u_1^2-v_1^2)+\frac{1}{2}m_2(u_2^2-v_2^2)

Spoiler

(edited 12 years ago)
Reply 11
Medicine Man
Here's my teachers solution.

And even he said it was definitely in his top 10 Maths/Further Maths questions he'd ever done.

Pretty nasty tbh.


Dear Medicine man, if your teacher has solved question 8, what about question 7. Please tell me that you have a similar working out for question 7. I am stuck on it for the past 3 days. I urgently need help now.

Thanks
:eek3:
ShafeeqSRS
Dear Medicine man, if your teacher has solved question 8, what about question 7. Please tell me that you have a similar working out for question 7. I am stuck on it for the past 3 days. I urgently need help now.

Thanks
:eek3:


sent you a PM

but you can always ask on the Maths forum, someone should be able to help you. :smile:
Reply 13
Medicine Man
sent you a PM

but you can always ask on the Maths forum, someone should be able to help you. :smile:


What is a PM?
How do I enter the maths forum?

(I am a new member)

Thanx
ShafeeqSRS
What is a PM?
How do I enter the maths forum?

(I am a new member)

Thanx


PM is a private message =]

To post a question in the maths forum, just click the link below and select "new thread" near the top of the page.

http://www.thestudentroom.co.uk/forumdisplay.php?f=38
Reply 15
If you use the fact that the kinetic energy loss does not change if you change your referential (you can think of it as the dissipation of energy only depends on the relative velocity, or you can just prove it), you can simplify the problem by making (before the colision) the first ball move at V and the second one not moving. The algebra gets way more simple.
Original post by Tiag0
If you use the fact that the kinetic energy loss does not change if you change your referential (you can think of it as the dissipation of energy only depends on the relative velocity, or you can just prove it), you can simplify the problem by making (before the colision) the first ball move at V and the second one not moving. The algebra gets way more simple.


6 years ago bro. These people probably have kids by now lolz


Posted from TSR Mobile
Reply 17
Original post by physicsmaths
6 years ago bro. These people probably have kids by now lolz


Posted from TSR Mobile


bit harsh ...
(are you getting notifications?)
Original post by TeeEm
bit harsh ...
(are you getting notifications?)


About what? This forum?


Posted from TSR Mobile
Yeah, gonna close this thread due to age :tongue:

Latest