Two blocks of masses m₁ and m₂ are connected by a spring of spring constant k. 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,
$v_{c m} = \frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$

$\Rightarrow v_{c m} = \frac{m_{1} \times 0 + m_{2} \times v_{0}}{m_{1} + m_{2}} \Rightarrow v_{c m} = \frac{m_{2} v_{0}}{m_{1} + m_{2}}$

(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 = v_cm

On applying the law of conservation of energy, we can write:
Change in kinetic energy = Potential energy stored in spring

$\Rightarrow \frac{1}{2} m_{2} v_{0}^{2} - \frac{1}{2} (m_{1} + m_{2}) {(\frac{m_{2} v_{0}}{m_{1} + m_{2}})}^{2} = \frac{1}{2} k x^{2} \Rightarrow m_{2} v_{0}^{2} (1 - \frac{m_{2}}{m_{1} + m_{2}}) = k x^{2}$

$\Rightarrow x = v_{0} {[\frac{m_{1} m_{2}}{(m_{1} + m_{2}) k}]}^{1 / 2}$

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Two blocks of masses $m_{1}$ and $m_{2}$ are connected by a spring of constant k (given figure).The block of mass $m_{2}$ is given a sharp impulse so that it acquires a velocity $v_{0}$ towards right.Find (a) the velocity of the centre of mass,(b) the maximum elongation that the spring will suffer.