Two block of masses m_1 and m_2 connected wit | vedclass

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Question

$\mathrm{F}=0 ; \mathrm{p}=\mathrm{constant}$

$\mathrm{m}_{2} \mathrm{v}_{0}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{v}$

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Vedclass · Accepted Answer

${v_0}\sqrt {\frac{{{m_1}{m_2}}}{{k({m_1} + {m_2})}}} $

Vedclass · Answer

$\mathrm{F}=0 ; \mathrm{p}=\mathrm{constant}$

$\mathrm{m}_{2} \mathrm{v}_{0}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}ight) \mathrm{v}$

$\& \frac{1}{2} \mathrm{m}_{2} \mathrm{v}_{0}^{2}=\frac{1}{2} \mathrm{m}_{1} \mathrm{v}^{2}+\frac{1}{2} \mathrm{m}_{2} \mathrm{v}^{2}+\frac{1}{2} \mathrm{kx}_{0}^{2}$

$\mathrm{v}_{\mathrm{cm}} \Rightarrow$ remain unchanged

$\& \frac{1}{2} \mathrm{m}_{2} \mathrm{v}_{0}^{2}=\frac{1}{2}\left(\mathrm{m}_{1}+\mathrm{m}_{2}ight)\left(\frac{\mathrm{m}_{2} \mathrm{v}_{0}}{\mathrm{m}_{1}+\mathrm{m}_{2}}ight)^{2}+\frac{1}{2} \mathrm{k} \mathrm{x}_{0}^{2}$

$\mathrm{m}_{2} \mathrm{v}_{0}^{2}\left[1-\frac{\mathrm{m}_{1}+\mathrm{m}_{2}}{\left(\mathrm{m}_{1}+\mathrm{m}_{2}ight)^{2}} 	imes \mathrm{m}_{2}ight]=\mathrm{Kx}_{0}^{2}$

$\mathrm{m}_{2} \mathrm{v}_{0}^{2}\left[\frac{\mathrm{m}_{1}+\mathrm{m}_{2}-\mathrm{m}_{2}}{\mathrm{m}_{1}+\mathrm{m}_{2}}ight]=\mathrm{kx}_{0}^{2}$

$\mathrm{v}_{0}^{2} \frac{\mathrm{m}_{1} \mathrm{m}_{2}}{\mathrm{m}_{1}+\mathrm{m}_{2}} 	imes \frac{1}{\mathrm{k}}=\mathrm{x}_{0}^{2}$

$\Rightarrow \mathrm{x}_{0}=\mathrm{v}_{0} \sqrt{\frac{\mathrm{m}_{1} \mathrm{m}_{2}}{\mathrm{k}\left(\mathrm{m}_{1}+\mathrm{m}_{2}ight)}}$

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