Question

A block of mass $2\text{kg}$ is placed on the top of a bigger block of mass $10\text{kg}$ as shown in figure. The system is released from rest. Assume all the surfaces are frictionless. Find the distance moved by the bigger block at the instant when the smaller block reaches the ground.

• A. $8\text{m}$
• B. $4.5\text{m}$
• C. $1.33\text{m}$
• D. $2.66\text{m}$

Answer 1

The correct option is C $1.33\text{m}$

Take leftward direction as +ve x axis.
Let the distance moved by the bigger block towards right $=x\text{m}$
Then, smaller block will move the distance = $(8-x)\text{m}$ towards left side
Given, masses $m_1=2\text{kg},m_2=10\text{kg}$

There is no external force on the system (smaller block + bigger block) in the horizontal (x) direction. Since the center of mass of the system was at rest initially, it will remain at rest in the horizontal direction.
i.e change in $x$ - coordinate of $COM$ of the system, $\Delta x_{com}=0$
$\Delta x_{com}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}=0$

$\Rightarrow \frac{2\times (8-x)+10(-x)}{2+10}=0$
$\Rightarrow 16-2x-10x=0$
$\Rightarrow 16=12x$
$\Rightarrow x=1.33\text{m}$

Distance moved by the bigger block is $1.33\text{m}$ towards right.