# Tow blocks $m_1$ and $m_2$ are placed on a smooth horizontal surface and are joined together with a spring of stiffness k. Suddenly, block $m_2$ receives a horizontal velocity $v_0$, then the maximum extension $x$ in the spring is

$\begin {array} {1 1} (a)\;v_o \sqrt {\frac{m_1m_2}{v(m_1+m_2)}} & \quad (b)\;v_o \sqrt {\frac{2m_1m_2}{(m_1+m_2)R}} \\ (c)\;v_o \sqrt {\frac{m_1m_2}{2(m_1+m_2)k}} & \quad (d)\;v_o \sqrt {\frac{m_1m_2}{(m_1+m_2)k}} \end {array}$

## 1 Answer

$V_{c}=\large\frac{m_2V_0}{(m_1+m_2)}$
At maximum extension, velocity of both $m_1$ and $m_2$ are equal and equal to velocity of Centre of mass .
By energy conservation
$\large\frac{1}{2}$$m_2 V_0^2=\large\frac{1}{2}$$(m_1+m_2)V_{c}^2+\large\frac{1}{2}$$kx^2 \large\frac{1}{2}$$m_2 V_0^2=\large\frac{1}{2} \frac{(m_1+m_2)m_2^2V_0^2}{(m_1+m_2)^2}$
$\qquad+ \large\frac{1}{2}$$kx^2 \large\frac{m_1m_2V_0^2}{(m_1+m_2)}$$=kx^2$
$x =v_{0}\sqrt {\large\frac{m_1m_2}{k(m_1+m_2)}}$
answered Nov 22, 2013 by
edited Jun 17, 2014

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