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# What happened to the http://www.rdoh.com/exams/ watch

1. anyone knows? just wanted to get answers for M2 specimen paper but the site is not working.
2. dont know about that site but you can always try

they should have it
3. (Original post by lexazver203)
anyone knows? just wanted to get answers for M2 specimen paper but the site is not working.
pete2004 was indirectly responsible.
4. ignore my post above,just had a look and they dont have the one you want
5. (Original post by Jonny W)
pete2004 was indirectly responsible.
Really? How come? Did edexcel find they were breaching copyright laws?
6. (Original post by chats)
ignore my post above,just had a look and they dont have the one you want
yeah just looked there are no solutions for any of the M2 papers
7. (Original post by Bezza)
Did edexcel find they were breaching copyright laws?
Possibly. Or the owner of the site took it down as a precaution.
8. Well maybe you can help

partivle A mass m moving with speed 3u
particle B mass 2m moves in the opposite direction to A at speed of u.
After collision the direction of A is reversed.
(a) Show that speed of B after the collision = 1/3(1+4e)u - done that
(b) Show that e>1/8 cannot do that one.

9. Let p and q be the velocities of A and B after the collision, both measured in the direction of A's initial movement.

(i)
Conservation of momentum:
3mu - 2mu = mp + 2mq,
u = p + 2q.

Newton's experimental law:
(q - p) = 4eu.

So:
q - (u - 2q) = 4eu,
3q = u(1 + 4e),
q = u(1 + 4e)/3.

(ii)
p
= u - 2q
= u(1 - (2/3)(1 + 4e))
= u(1 - 8e)/3

We know that the direction of A is reversed after the collision. So p < 0. So e > 1/8.
11. (Original post by Jonny W)
Let p and q be the velocities of A and B after the collision, both measured in the direction of A's initial movement.

(i)
Conservation of momentum:
3mu - 2mu = mp + 2mq,
u = p + 2q.

Newton's experimental law:
(q - p) = 4eu.

So:
q - (u - 2q) = 4eu,
3q = u(1 + 4e),
q = u(1 + 4e)/3.

(ii)
p
= u - 2q
= u(1 - (2/3)(1 + 4e))
= u(1 - 8e)/3

We know that the direction of A is reversed after the collision. So p < 0. So e > 1/8.
you see the thing is we havent done newtowns experimentaal law and i have no idea whatit is.
12. Newton's experimental law
= Newton’s law of restitution
= What the coefficient of restitution e means
13. (Original post by Jonny W)
Newton's experimental law
= Newton’s law of restitution
= What the coefficient of restitution e means
ok we've done that one just never heard it was called an experimental law.

Are q and p the speeds of 2 particles after the collision then?
14. if you're looking for the answers for the specimen, go the the edexcel website (or whichever board you're doing) and download the specimen materials.

It includes the markscheme.

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