Question

In the system shown, find the speed with which the $12\text{kg}$ block hits the ground. Consider the pulley and string to be massless and friction is absent everywhere. [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 $2\sqrt{5}{\text{ms}}^{-1}$

From conservation of mechanical energy,
$\Delta KE+\Delta PE=0$

Since initial $KE=0$ (both the blocks are at rest), we can write,
$K{E}_f+P{E}_f=P{E}_i$
$\Rightarrow \frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2+m_{1}g{h}_1+m_{2}g\left(0\right)=m_{1}g\left(0\right)+m_{2}g{h}_2$
(Taking datum at ground level)

Distance moved up by $m_1=$ distance moved down by $m_2$ (due to string constraint)
i.e $h_1=h_2$
Also, $v_1=v_2$ by the same reasoning.

$\Rightarrow \frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2=m_{2}gh-m_{1}gh$
$\Rightarrow 8\times 10\times 2=\frac{1}{2}\times 16v^2$
$\therefore v=2\sqrt{5}{\text{ms}}^{-1}$ is the speed of block $m_2$ when it hits the ground.