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Question
Two blocks of masses m1 and m2 are connected by a spring of spring constant k (See figure). The block of mass m2 is given a sharp impulse so that it acquires a velocity v0 towards right. Find (a) the velocity of the centre of mass, (b) the maximum elongation that the spring will suffer.
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Solution
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)`
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