Question

Two blocks of mass $m_1=2\text{kg},m_2=4\text{kg}$ connected by weightless spring of stiffness $k$ rest on a horizontal plane as shown. $m_2$ is shifted a small distance $x=1\text{cm}$ to the left and then released. Find the velocity of COM of the system just after $m_1$ break off the wall. Assume $k=4\text{N/m}$.

• A. $1\text{m/s}$
• B. $\frac{2}{3}\text{m/s}$
• C. $3\text{m/s}$
• D. $2\text{m/s}$

Answer 1

The correct option is B $\frac{2}{3}\text{m/s}$

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 $v_2$, which is given as
$\frac{1}{2}k{x}^2=\frac{1}{2}{m}_2{v}_2^2$
$v_2=\sqrt{\frac{k}{m_2}}.x=\sqrt{\frac{4}{4}}.1=1\text{m/s}$
And the velocity of the COM at that instant is,
$v_{com}=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}$
Hence, $v_{com}=\frac{2\times 0+4\times \sqrt{\frac{4}{4}}.1}{2+4}=\frac{2}{3}\text{m/s}$