Two block with masses `m_(1) = 3kg and m_(2) = 5 kg` are connected by a light string that slides over a frictionless pulley as shown in figure. Initially, m2 is held 5 m off the floor while `m_(1)` is on the floor. The system is then released. At what speed does `m_(2)` hit the floor?

Explanation

The inital and final configurations are shown in the figure. It is convenient to set `U_(g) = 0` at the floor. Initially, only `m_(2)` has potential energy. As it falls, it loses potential energy and gains kinetic nergy. At the same time, `m_(1)` gains potential energy and kinetic energy. Just before `m_(2)` lands, it has only kinetic energy. Let v the final speed of each mass. Then, using the law of conservation of mechanical energy.

Final (Just before m strikes the floor)
`K_(f)+U_(f)=K_(i)+U_(i)`
`(1)/(2)(m_(1)+m_(2))v^(2)+m_(1)gh=0+m_(2)gh`
`v^(2)=(2(m_(2)-m_(1))gh)/(m_(1)+m_(2))`
Putting `m_(1)=3kg,m_(2)=5kg,h=5mandg=10m//s^(2)`
we get `v^(2)=(2(5-3)(10)(5))/(5+3)orv=5m//s`