Question

In the arrangement shown in figure above 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. Find the maximum height that body $2$ will go up to in cm.

Answer 1

Let $w_1$ be the acceleration of mass $m_1$ and $w_2$ is the acceleration of mass $m_2$.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
$m_{1}g-T_1=m_1{w}_1$ (1)
$T_2-m_{2}g=m_2{w}_2$ (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 equation yields,
$w_2=\frac{2(m_1-2m_2)g}{4m_2+m_1}=\frac{2(\eta -2)}{\eta +4}$ (as $\frac{m_1}{m_2}=\eta$)
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:
$v_2^2=2w_2\left(2h\right)=2\left[\frac{2(\eta -2)}{\eta +4}\right]2h=\frac{8h(\eta -2)g}{\eta +4}$
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,
$h^{\prime }=\frac{v_2^2}{2g}=\frac{4h(\eta -2)}{\eta +4}$
Thus the sought maximum height attained by the body 2:
$H=2h+h^{\prime }=2h+\frac{4h(\eta -2)}{\eta +4}=\frac{6\eta h}{\eta +4}$