Question

Two blocks of masses $m_1$ and $m_2$ are connected by a weightless spring of stiffness $k$, rests on a smooth horizontal plane. Mass $m_2$ is shifted by a small distance $x_0$ to the left and released. The velocity of the center of mass of the system when $m_1$ breaks off the wall is,

• A. $\left(\frac{\sqrt{k{m}_2}}{m_1+m_2}\right)x_0$
• B. $\left(\frac{\sqrt{k{x}_0{m}_2}}{m_1+m_2}\right)$
• C. $\left(\frac{m_1+m_2}{m_2}\right)x_{0}k$
• D. $\left(\frac{\sqrt{k{m}_1}}{m_1+m_2}\right)x_0$

Answer 1

The correct option is A $\left(\frac{\sqrt{k{m}_2}}{m_1+m_2}\right)x_0$

Since, the system is free from external forces, hence momentum is conserved.

Using the relation for velocity of COM we get,

$v_{com}=\frac{m_1{v}_1+m_2{v}_2}{m_1+m_2}$

mass $m_1$ is at rest. $\therefore v_1=0$

$v_{com}=\frac{m_2{v}_2}{m_1+m_2}......\left(1\right)$

Using conservation of energy,

$E_i=E_f$

$\frac{1}{2}k{x}_0^2=\frac{1}{2}{m}_2{v}_2^2$

$\therefore v_2=x_0\sqrt{\frac{k}{m_2}}$

Putting $v_2$ in eq. $\left(1\right)$ we get,

$v_{com}=\left(\frac{\sqrt{k{m}_2}}{m_1+m_2}\right)x_0$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, option $\left(A\right)$ is the correct answer.