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Two block of masses $m_1$ and $m_2$ connected with the help of a spring of spring constant $k$ initially to natural length as shown. A sharp impulse is given to mass $m_2$ so that it acquires a velocity $v_0$ towards right. If the system is kept an smooth floor then find the maximum elongation that the spring will suffer
${v_0}\sqrt {\frac{{{m_1}{m_2}}}{{k({m_1} + {m_2})}}} $
${v_0}\sqrt {\frac{{({m_1} + {m_2})}}{{k{m_1}{m_2}}}} $
${v_0}\sqrt {\frac{{2{m_1}{m_2}}}{{k({m_1} + {m_2})}}} $
$2{v_0}\sqrt {\frac{{{m_1}{m_2}}}{{k({m_1} + {m_2})}}} $
Solution
$\mathrm{F}=0 ; \mathrm{p}=\mathrm{constant}$
$\mathrm{m}_{2} \mathrm{v}_{0}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{v}$
$\& \frac{1}{2} \mathrm{m}_{2} \mathrm{v}_{0}^{2}=\frac{1}{2} \mathrm{m}_{1} \mathrm{v}^{2}+\frac{1}{2} \mathrm{m}_{2} \mathrm{v}^{2}+\frac{1}{2} \mathrm{kx}_{0}^{2}$
$\mathrm{v}_{\mathrm{cm}} \Rightarrow$ remain unchanged
$\& \frac{1}{2} \mathrm{m}_{2} \mathrm{v}_{0}^{2}=\frac{1}{2}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)\left(\frac{\mathrm{m}_{2} \mathrm{v}_{0}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right)^{2}+\frac{1}{2} \mathrm{k} \mathrm{x}_{0}^{2}$
$\mathrm{m}_{2} \mathrm{v}_{0}^{2}\left[1-\frac{\mathrm{m}_{1}+\mathrm{m}_{2}}{\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)^{2}} \times \mathrm{m}_{2}\right]=\mathrm{Kx}_{0}^{2}$
$\mathrm{m}_{2} \mathrm{v}_{0}^{2}\left[\frac{\mathrm{m}_{1}+\mathrm{m}_{2}-\mathrm{m}_{2}}{\mathrm{m}_{1}+\mathrm{m}_{2}}\right]=\mathrm{kx}_{0}^{2}$
$\mathrm{v}_{0}^{2} \frac{\mathrm{m}_{1} \mathrm{m}_{2}}{\mathrm{m}_{1}+\mathrm{m}_{2}} \times \frac{1}{\mathrm{k}}=\mathrm{x}_{0}^{2}$
$\Rightarrow \mathrm{x}_{0}=\mathrm{v}_{0} \sqrt{\frac{\mathrm{m}_{1} \mathrm{m}_{2}}{\mathrm{k}\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)}}$
