Q119 / 180 · JIPMER 2015 · Physics

Two blocks of masses $m_1$ and $m_2$ are connected by a spring of spring constant $k$. The block of mass $m_2$ is given a sharp empulse so that it acquires a velocity $v_0$ towards right. Find the maximum elongation that the spring will suffer. Figure

  • Correct answer
    A
    $\left[\frac{m_1 m_2}{m_1+m_2}\right]^{\frac{1}{2}} v_0$
  • B
    $\left(\frac{m_1+m_2}{m_1-m_2}\right) v_0$
  • C
    $\left[\frac{m_1+m_2}{m_1-m_2}\right]^{\frac{1}{2}} v_0$
  • D
    $\left[\frac{2 m_1+m_2}{m_1 m_2}\right]^{\frac{1}{2}} v_0$

Solution

The velocity of centre of mass $v_{\mathrm{cm}}=\frac{m_1 v_1+m_2 v_2}{m_1+m_2}$ When $v_1=0$ and $v_2=v_0$, then $v_{c m}=\frac{m_2 v_0}{m_1+m_2}$ Now, let $x$ be the elongation in the spring. Change in potential energy $=$ potential energy stored in spring $\begin{gathered} \Rightarrow \quad \frac{1}{2} m_2 v_0^2-\frac{1}{2}\left(m_1+m_2\right) \\ \left(\frac{m_2 V_0}{m_1+m_2}\right)^2=\frac{1}{2} k x^2 \end{gathered}$ This gives, $x=\left(\frac{m_1 m_2}{m_1+m_2}\right)^{\frac{1}{2}} v_0$

Hence, the correct answer is option: 'A'.

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