In the arrangement shown in figure the mass of body 1 is `eta=4.0` times as great as that of body 2. The height `h=20cm`. The masses of the pulleys and the threads, as well as the friction, are negligible. At a certain moment body 2 is released and the arrangement set in motion. What is the maximum height that body 2 will go up to?

Explanation

Using Newton's second law in projection form along x-axis for the body 1 and along negative x-axis for the body 2 respectively, we get
`m_1g-T_1=m_1w_1` (1)
`T_2-m_2g=m_2w_2` (2)
For the pulley lowering in downward direction from Newton's law along x axis,
`T_1-2T_2=0` (as pulley is mass less)
or, `T_1=2T_2` (3)
As the length of the thread is constant so,
`w_2=2w_1` (4)
The simultaneous solution of above equations yields,
`w_2=(2(m_1-2m_2)g)/(4m_2+m_1)=(2(eta-2))/(eta+4)` (as `m_1/m_2=eta`) (5)
Obviously during the time interval in which the body 1 comes to the horizontal floor covering the distance h, the body 2 moves upward the distance `2h`. At the moment when the body 2 is at the height `2h` from the floor its velocity is given by the expression:
`v_2^2=2w_2(2h)=2[(2(eta-2)g)/(eta+4)]2h=(8h(eta-2)g)/(eta+4)`
After the body `m_1` touches the floor the thread becomes slack or the tension in the thread zero, thus as a result body 2 is only under gravity for it's subsequent motion.
Owing to the velocity `v_2` at that moment or at the height `2h` from the floor, the body 2 further goes up under gravity by the distance,
`h^'=(v_2^2)/(2g)=(4h(eta-2))/(eta+4)`
Thus the sought maximum height attained by the body 2:
`H=2h+h^'=2h+(4h(eta-2))/((eta+4))=(6etah)/(eta+4)`