Question

Two blocks of masses $m_1$ and $m_2$ connected by a massless spring of spring constant 'k' rest on a smooth horizontal plane as shown in figure. Block 2 is shifted a small distance 'x' to the left and the released. Find the velocity of centre of mass of the system after block 1 breaks off the wall.

• A. $\frac{\sqrt{m_{2}K}}{m_1+m_2}$x
• B. $\frac{\sqrt{m_{1}K}}{m_1+m_2}$x
• C. $\frac{\sqrt{(m_1+m_2)K}}{m_1+m_2}x$
• D. zero

Answer 1

The correct option is A $\frac{\sqrt{m_{2}K}}{m_1+m_2}$x

If $m_2$ is shifted by a distance x and released, the mass $m_1$ will break off from the wall when the spring restores its natural length and $m_2$ will start going towards right. At the time of breaking $m_1$, $m_2$ will be going towards right with a velocity v, which is given as
$\frac{1}{2}k{x}^2=\frac{1}{2}{m}_2{v}^2\Rightarrow v=\left(\sqrt{\frac{k}{m_2}}\right)x$
and the velocity of the centre of mass at this instant is
$v_{CM}=\frac{m_1\times 0+m_2\times v}{m_1+m_2}=\frac{\left(\sqrt{m_{2}k}\right)x}{m_1+m_2}$