M1 Impulse/Momentum

Particle A of mass 1kg is at rest 0.2m from the edge of a smooth horizontal 0.8m high table. It is connected by a light inextensible string of length 0.7m to a particle B of mass 0.5kg. Particle B is initially at rest at the edge of the table closest to A but then falls off. Assuming B's initial horizontal velocity to be zero find the speed with which A begins to move.

to make it much easier to solve this kind of problems , you should analyse what you have.
1.they both move in the same direction so both velocitys are +
2.the speed of A when the string becomes taut , is the initial speed.
now standard suvat protocool:
u = 0
s = 0.7-0.2 = 0.5
a = 9.8
v = ?
v² = u² + 2(as) (sub values)
v² = 2(0.5 * 9.8) = 9.8
v = sqr(9.8)
now standard conversion of momentum protocool
m₁ = 1kg
m₂ = 0.5kg
u₁ = 0
u₂ = sqr(9.8)
v₁ = v₂
all speeds are in the same direction and both final speeds are equal.
m₁u₁ + m₂u₂ = m₁v + m₂v (sub values)
1*0 + 0.5*sqr(9.8) = 1*v + 0.5*v (simplify)
1.5v = 0.5*sqr(9.8)
v = (0.5*sqr(9.8)) / 1.5 = 1.04 (3 s.f)
hope this helps.