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.