Question

In the arrangement shown in Fig. the mass of body 1 is $\eta=4.0$ times as great as that of body 2. The height h = 20 cm. 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?

Answer 1

60 cm

Answer 2

Let $m_1$ and $m_2$ be the masses of body 1 and body 2, respectively. We are given $m_1 = \eta m_2$, with $\eta = 4.0$.
Let $a_1$ be the downward acceleration of body 1 and $a_2$ be the upward acceleration of body 2.
From the arrangement of the pulleys, if body 1 moves down by a distance $x$, the movable pulley moves up by $x$. The string passing over the movable pulley has one end fixed and the other end attached to body 2. If the movable pulley moves up by $x$ and body 2 moves up by $y$, the length of the string remains constant. The change in the length of the string is the sum of the changes in the lengths of the two vertical segments connected to the movable pulley. The length of the segment between the fixed support and the pulley decreases by $x$. The length of the segment between the pulley and body 2 changes by $(y-x)$. The total change in length is $-x + (y-x) = y - 2x$. Since the length is constant, $y - 2x = 0$, so $y = 2x$. Thus, the upward displacement of body 2 is twice the downward displacement of body 1. Differentiating twice with respect to time, we get $a_2 = 2a_1$.

Let $T_1$ be the tension in the string connected to body 1, and $T_2$ be the tension in the string passing over the movable pulley.
For body 1 (moving downwards): $m_1 g - T_1 = m_1 a_1$.
For body 2 (moving upwards): $T_2 - m_2 g = m_2 a_2$.
For the movable pulley (mass is negligible): The net force on the pulley is zero. The upward force is $T_1$ and the downward forces are $2T_2$. So, $T_1 - 2T_2 = 0$, which means $T_1 = 2T_2$.

Substitute $T_1 = 2T_2$ into the equation for body 1: $m_1 g - 2T_2 = m_1 a_1$.
From the equation for body 2, $T_2 = m_2 g + m_2 a_2$.
Substitute $T_2$ into the modified equation for body 1: $m_1 g - 2(m_2 g + m_2 a_2) = m_1 a_1$.
$m_1 g - 2m_2 g - 2m_2 a_2 = m_1 a_1$.
Substitute $a_2 = 2a_1$: $m_1 g - 2m_2 g - 2m_2 (2a_1) = m_1 a_1$.
$m_1 g - 2m_2 g - 4m_2 a_1 = m_1 a_1$.
$(m_1 - 2m_2) g = (m_1 + 4m_2) a_1$.
$a_1 = \frac{(m_1 - 2m_2) g}{m_1 + 4m_2}$.
Given $m_1 = \eta m_2 = 4m_2$,
$a_1 = \frac{(4m_2 - 2m_2) g}{4m_2 + 4m_2} = \frac{2m_2 g}{8m_2} = \frac{g}{4}$.
The downward acceleration of body 1 is $a_1 = g/4$.
The upward acceleration of body 2 is $a_2 = 2a_1 = 2(g/4) = g/2$.

Initially, body 1 is at height $h = 20$ cm $= 0.20$ m above the floor, and body 2 is on the floor. The system starts from rest.
Body 1 moves down with acceleration $a_1 = g/4$. It hits the floor when it has moved a distance $h$.
Using the kinematic equation $s = ut + \frac{1}{2} at^2$, with initial velocity $u=0$, displacement $s=h$, and acceleration $a=a_1$:
$h = 0 \cdot t + \frac{1}{2} a_1 t^2 = \frac{1}{2} \frac{g}{4} t^2 = \frac{g}{8} t^2$.
The time taken for body 1 to hit the floor is $t = \sqrt{\frac{8h}{g}}$.

During this time $t$, body 2 moves upwards with acceleration $a_2 = g/2$. The distance moved by body 2 is $y_2 = ut + \frac{1}{2} a_2 t^2$, with initial velocity $u=0$:
$y_2 = 0 \cdot t + \frac{1}{2} a_2 t^2 = \frac{g}{4} t^2$.
Substitute $t^2 = \frac{8h}{g}$:
$y_2 = \frac{g}{4} \left(\frac{8h}{g}\right) = 2h$.
So, when body 1 hits the floor, body 2 has moved up by a distance $2h$. The height of body 2 at this moment is $2h$.

At the moment body 1 hits the floor, the velocity of body 2 is $v_2 = u + a_2 t = 0 + a_2 t = \frac{g}{2} \sqrt{\frac{8h}{g}} = \frac{g}{2} \frac{2\sqrt{2h}}{\sqrt{g}} = \sqrt{2gh}$.

After body 1 hits the floor, the string connected to it becomes slack, and body 2 is no longer connected to body 1. Body 2 is now moving upwards with initial velocity $v_2 = \sqrt{2gh}$ and is under the influence of gravity only. Its acceleration is now $-g$ (downwards).
Body 2 will continue to move upwards until its velocity becomes zero. Let the additional height it goes up be $\Delta h_2$.
Using the kinematic equation $v^2 = u^2 + 2as$, with final velocity $v=0$, initial velocity $u=v_2$, acceleration $a=-g$, and displacement $s=\Delta h_2$:
$0^2 = v_2^2 + 2(-g) \Delta h_2$.
$0 = (2gh) - 2g \Delta h_2$.
$2g \Delta h_2 = 2gh$.
$\Delta h_2 = h$.

The total maximum height that body 2 will go up to is the height it reached when body 1 hit the floor plus the additional height it goes up afterwards.
Maximum height $= y_2 + \Delta h_2 = 2h + h = 3h$.

Given $h = 20$ cm, the maximum height is $3 \times 20$ cm $= 60$ cm.