as per the given figure two blocks each of mass 50 gram connected to a spring of spring constant 2 ...............................

Question

Correct option is (B) v2v2 

sir/mam i am unable to get this solution please sir / mam send me the complete or fully explained mam please send it mam

Answer 1

Change in kinetic energy = potential energy stored in spring

\(\frac 12 m_2v_0^2 - \frac 12 (m_1 + m_2) \left(\frac{m_2v_0}{m_1 + m_2}\right)^2 = \frac 12 kx^2\)

\(\frac 12 m_2v_0^2 - \frac 12 (m_1 + m_2)\frac{m_2^2v_0^2}{(m_1 + m_2)^2} = \frac 12 kx^2 = \frac 12 kx^2\)

\(\frac 12 m_2v_0^2 - \frac 12 \frac{m_2^2v_0^2}{(m_1 + m_2)} = \frac 12 kx^2\)

\(x = \left[\frac{m_1m_2}{k(m_1 + m_2)}\right]^{1/2}V_0\)

\(x = \left[\frac{0.050\times0.050}{0\times(0.050 + 0.050)}\right]^{1/2}V_0\)

\(x = \left[\frac{0.0025}{2\times0.1}\right]^{1/2}V_0\)

\(x = \left[\frac{0.0025}{0.2}\right]^{1/2}V_0\)

\(x = (0.0125)^{1/2}V_0\)

\(x = 0.111 V_0\)