After falling from rest through a height h a body of mass m begins to raise a body of mass M(Mgtm) connected to it through a pulley. (a) Determine the time it will take for body of mass M to return to its original position. (b) Find the fraction of kinetic energy lost when the body of mass Mis jerked into motion.

After falling from rest through a height h a body of mass m begins to

1.	After falling from rest through a height h a body of mass m begins to raise a body of mass M(Mgtm) connected to it through a pulley. (a) Determine the time it will take for body of mass M to return to its original position. (b) Find the fraction of kinetic energy lost when the body of mass Mis jerked into motion.
Answer» Solution :(a) The SPEED of the body `B` just before the string becomes taut is `v=sqrt(2gh)`. When the string is jerked, large impulsive reactions are generated in the string. At this moment effect of gravity is negligible. So momentum of the system is conserved at this INSTANT. Let `v'` be the common speed of the two bodies after they are jerked into motion. From conservation of momentum, we have `MV=(M+m)v'` or `v'=(m)/(M+m)v` The acceleration of the system is `SigmaF=Mg-mg=(M+m)a` or `a=-(M-m)/(M+m)g` The acceleration is negative, (opposite to `v'`) Let the system return to origianl POSITION at time `t`. `0=v't+(1)/(2)at^(2)` or `t=-(2V')/(a)=(2m)/(M-m)sqrt((2h)/(g))` (b) The fractional loss of kinetic energy is `((1)/(2)mv^(2)-(1)/(2)(M+m)v'^(2))/((1)/(2)mv^(2))=(M)/(M+m)`