Two blocks of masses m1 and m2 are connected with an ideal spring

Given,
Velocity of mass, m2 = v0
Velocity of mass, m1 = 0

(a) Velocity of centre of mass is given by,

\[v_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}\]

\[\Rightarrow v_{cm} = \frac{m_1 \times 0 + m_2 \times v_0}{m_1 + m_2}\]

\[ \Rightarrow v_{cm} = \frac{m_2 v_0}{m_1 + m_2}\]

(b) Let the maximum elongation in spring be x.The spring attains maximum elongation when velocities of both the blocks become equal to the velocity of centre of mass.

i.e. v1 = v2 = vcm

On applying the law of conservation of energy, we can write:

Change in kinetic energy = Potential energy stored in spring

\[\Rightarrow \frac{1}{2} m_2 v_0^2 - \frac{1}{2}( m_1 + m_2 ) \left( \frac{m_2 v_0}{m_1 + m_2} \right)^2 = \frac{1}{2}k x^2 \]

\[ \Rightarrow m_2 v_0^2 \left( 1 - \frac{m_2}{m_1 + m_2} \right) = k x^2\]

`Rightarrow = V_o[(m_1m_2)/((m_1+m_2)K)]^(1/2)`

Two blocks of masses m1 and m2 connected by a massless spring of spring constant 'k' rest on a smooth horizontal plane as shown in figure. Block 2 is shifted a small distance 'x' to the left and the released. Find the velocity of centre of mass of the system after block 1 breaks off the wall.

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Two smooth blocks of masses m1 and m2 attached with an ideal spring of stiffness k and kept on a horizontal surface. If m1 is projected with a horizontal velocity v 0. Find the maximum compression of the spring.

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