Energy

[kingyof2thejring]

Iam having problems with this question i think my technique is right but i,ve made a mistake somewhere

Particles of mass m and 2m respectively are connected by a light inextensible string. Initially, the particles lie at opposite edges of a smooth horizontal table with the string just taut. One of the particles is then nudged over the edge of the table. Find the ratio between the two possible speeds of the system when the other particle reaches the edge of the table.

i've let the level of potential energy be the table
m rep A
2m rep B
i've stated of nudging B and keeping A on the table
energy at start = 0
energy at finish = KE of A and B and PE of B

1/2*m*v^2 + 1/2*2m*v^2 + 2m*g*-h
Energy at start = Energy at finish
1/2mv^2(1/2 + 1) = 2mgh
v^2(1/2 + 1) =2gh
v^2 = 2gh/1.5
v = (4/3gh)^-1

now if we nudge A and keep B on the table

Energy at start = 0
Energy at finish = KE of A and B + PE of A

1/2*m*v^2 + 1/2*2m*v^2 + m*g*-h
Energy at start = Energy at finish
1/2mv^2 +mv^2 = mgh
mv^2(1/2 +1) = mgh
v^2 = gh/1.5
v = (2/3gh)^-1

(4/3gh)^-1 : (2/3gh)^-1

sub in random value for h= 5

i get a ratio of

1.4 : 1

which is an answer i don't have much faith in.
i could have made mistake in my technique or silly error
Could anyone help me out please!
Thanks in advance.

[gnome]

Check your math

1/2*m*v^2 + 1/2*2m*v^2 + 2m*g*-h
Energy at start = Energy at finish
1/2mv^2(1/2 + 1) = 2mgh ...........................here
v^2(1/2 + 1) =2gh......................................here
v^2 = 2gh/1.5
v = (4/3gh)^-1.....................................and here

and similarly in the second part.

[kingyof2thejring]

cheers
v = (4/3gh)^-1
should be v = (4/3gh)^1/2, i've got that, but this does not solve the problem

1/2mv^2(1/2 + 1) = 2mgh
should be mv^2(1/2 + 1) = 2mgh

v^2(1/2 + 1) =2gh this is correct

these are typing errors on my behalf
the ratio i get is 1.4 : 1

[gnome]

v = (4/3gh)^(1/2)
and
v = (2/3gh)^(1/2)
so the ratio is sqrt(2):1

[kingyof2thejring]

cheers ta thx