Question

In the arrangement shown in the figure, mass of the block $A$ is $4$ times that of block $B$. The height $h=20\text{cm}$. At a certain instant, the block $A$ is released and the system is set in motion. What is the maximum height (in $\text{cm})$ that the body $B$ will go up? Assume enough space allowed for $B$ and $A$ (i.e $A$ and $B$ are of small size) and take $g=10{\text{m/s}}^2$

• A. $40\text{cm}$
• B. $60\text{cm}$
• C. $20\text{cm}$
• D. $0\text{cm}$

Answer 1

The correct option is B $60\text{cm}$

Given, initial speed of both masses i.e $u=0$


Assume $a$ and $b$ to be the accelerations of the masses $A$ and $B$ respectively.
Assume tension in the string connected with block $B$ is $T$.
So, the tension in the string connected with mass $A$ will be $2T$.
(because pulley is massless, $\sum F=0$)

Given, $m_A=4m_B$
F.B.D. of mass $B$:


When block B just starts the motion, it loses contact with the surface, so $N=0$,
$\therefore T-m_{B}g=m_{B}b...\left(1\right)$
F.B.D of mass $A$:


$m_{A}g-2T=m_{A}a$
As given $m_A=4m_B$
$\Rightarrow 4m_{B}g-2T=4m_{B}a...\left(2\right)$

Applying string constraint on the string:
$l_1+l_2+l_3=\text{constant}$
Differentiating twice with respect to time,
$+a+a-b=0$
($\because$Length of string $l_1\&l_2$ is increasing, and for $l_3$ length is decreasing.)
$\Rightarrow b=2a...\left(3\right)$

Solving Eq. $\left(1\right),\left(2\right)\&\left(3\right)$ we get,
$a=\frac{g}{4}$, $b=\frac{g}{2}$

$\because b=2a$, $h_B=2\times h_A$ i.e when block $A$ reaches the horizontal floor covering a distance of $h=0.20\text{m}$, the block$B$ moves upward by a distance $2h=0.40\text{m}$.
When block $B$ is at the height $2h=0.4\text{m}$ from the floor, its velocity will be,
$v^2=u^2+2as$
$v^2=0+2\times \frac{g}{2}\times 0.4$
$v^2=4$
or $v=2\text{m/s}$

After the block $A$ touches the floor, the tension in the string becomes zero, as a result block $B$ is only under gravity for its subsequent motion.
$\Rightarrow$Maximum height covered by the block $B$,
$v_B^2-u_B^2=2a{s}_B$
{$v_B=0$ at maximum height, $u_B=v=2\text{m/s},a_B=-g{\text{m/s}}^2$}
$\Rightarrow 0-2^2=2(-10)\times s_B$
$\Rightarrow s_B=0.2\text{m}$

Therefore, total distance covered by the block $B$ is given by,
$H=2h+s_B$
$\therefore H=0.4+0.2=0.6\text{m}=60\text{cm}$