Two blocks of masses m_1 and m_2 are connecte | vedclass

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(A) Since there is no external horizontal force,the linear momentum of the system is conserved.Let $v$ be the common velocity of both blocks at the mo

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Since there is no external horizontal force,the linear momentum of the system is conserved.Let $v$ be the common velocity of both blocks at the moment of maximum elongation.By conservation of linear momentum: $m_2 v_0 = (m_1 + m_2) v \implies v = \frac{m_2 v_0}{m_1 + m_2}$.By conservation of mechanical energy,the initial kinetic energy of the system equals the sum of the final kinetic energy and the potential energy stored in the spring at maximum elongation $x_0$:$\frac{1}{2} m_2 v_0^2 = \frac{1}{2} (m_1 + m_2) v^2 + \frac{1}{2} k x_0^2$.Substituting $v$ into the energy equation:$\frac{1}{2} m_2 v_0^2 = \frac{1}{2} (m_1 + m_2) \left( \frac{m_2 v_0}{m_1 + m_2} \right)^2 + \frac{1}{2} k x_0^2$.$m_2 v_0^2 = \frac{m_2^2 v_0^2}{m_1 + m_2} + k x_0^2$.$k x_0^2 = m_2 v_0^2 \left( 1 - \frac{m_2}{m_1 + m_2} \right) = m_2 v_0^2 \left( \frac{m_1}{m_1 + m_2} \right)$.$x_0^2 = \frac{m_1 m_2 v_0^2}{k(m_1 + m_2)}$.$x_0 = v_0 \sqrt{\frac{m_1 m_2}{k(m_1 + m_2)}}$.

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