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What happened to the http://www.rdoh.com/exams/
anyone knows? just wanted to get answers for M2 specimen paper but the site is not working.
dont know about that site but you can always try
http://www.mrhughes.net/Maths/Hidden%20Pages/login.htm
username:alevel
password:mrhughes
they should have it
http://www.mrhughes.net/Maths/Hidden%20Pages/login.htm
username:alevel
password:mrhughes
they should have it
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
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.
Some help please.
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.
Some help please.
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.
(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.
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.
(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.
Jonny W
Newton's experimental law
= Newton’s law of restitution
= What the coefficient of restitution e means
= 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?
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