Question

For the system shown in figure, find the speed with which the $12\text{kg}$ block will hit the ground. Take $g=10{\text{m/s}}^2$.

• A. $2\sqrt{10}{\text{ms}}^{-1}$
• B. $4\sqrt{2}{\text{ms}}^{-1}$
• C. $2\sqrt{5}{\text{ms}}^{-1}$
• D. $3{\text{ms}}^{-1}$

Answer 1

The correct option is C $2\sqrt{5}{\text{ms}}^{-1}$

When the block hits ground, let its speed be $v$.
The velocity of both blocks are equal in magnitude $\left(v\right)$ and opposite in direction, so as to maintain string's length constant.

Applying mechanical energy conservation for system:
$\text{Loss in PE of}{m}_2=\text{Gain in KE of}{m}_1+\text{Gain in KE of}{m}_2+\text{Gain in PE of}{m}_1...\left(i\right)$

$\because h=2\text{m}$ is the displacement of both blocks in opposite directions, when $m_2$ hits the ground. i.e $m_2$ moves downward $\&$ $m_1$ moves upward

From Eq. $\left(i\right)$,
$m_{2}gh=\frac{1}{2}{m}_1{v}^2+\frac{1}{2}{m}_2{v}^2+m_{1}gh$
$(12\times 10\times 2)-(4\times 10\times 2)=\frac{1}{2}\times (m_1+m_2)v^2$
$\Rightarrow 160=8v^2$
$\therefore v=\sqrt{20}=2\sqrt{5}{\text{ms}}^{-1}$