Question

Two smooth blocks of masses $m_1$ and $m_2$ attached with an ideal spring of stiffness $k$ and kept on a horizontal surface. If $m_1$ is projected with a horizontal velocity $v_0$. Find the maximum compression of the spring.

• A. $\left[\sqrt{\frac{m_1{m}_2}{(m_1+m_2)k}}\right]v_0$
• B. $\left[\sqrt{\frac{m_1{m}_2}{(m_1-m_2)k}}\right]v_0$
• C. $\left[\sqrt{\frac{m_1{m}_2}{(m_1+m_2)k}}\right]$
• D. $\left[\sqrt{\frac{m_1{m}_2}{(m_1+m_2)}}\right]v_0$

Answer 1

The correct option is A $\left[\sqrt{\frac{m_1{m}_2}{(m_1+m_2)k}}\right]v_0$

As there is no external force acting on the system.
We can apply work-energy theorem from $CM$ frame.
$W_{\text{ext}}+W_{\text{Int}}=\Delta K_{\text{cm}}=(K_f-K_1{)}_{\text{cm}}$ ...(i)
Initial kinetic energy of system w.r.t. $CM$
$(K_i{)}_{\text{cm}}=\frac{1}{2}\left(\frac{m_1{m}_2}{m_1+m_2}\right)(v_0-0{)}^2$
$\Rightarrow (K_i{)}_{\text{cm}}=\frac{1}{2}\left(\frac{m_1{m}_2}{m_1+m_2}\right)v_0^2$
At maximum compression, both the blocks move with common velocity hence, $(v_{\text{rel}}{)}_{\text{final}}=0$ which gives, final kinetic energy of system w.r.t $CM$
$(K_f{)}_{\text{cm}}=\frac{1}{2}\mu (v_{\text{rel}}{)}_{\text{final}}^2=0$

Energy stored in the spring at maximum compression is
$W_{int}=-\frac{1}{2}k{x}^2$
Hence from (i)
$0+(-\frac{1}{2}k{x}^2)=0-\frac{1}{2}\left(\frac{m_1{m}_2}{m_1+m_2}\right)v_0^2$
$\Rightarrow x=\left[\sqrt{\frac{m_1{m}_2}{(m_1+m_2)k}}\right]v_0$