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 empulse so that it acquires a velocity v₀towards right. Find the maximum elongation that the spring will suffer.

${[\frac{m_{1} m_{2}}{m_{1} + m_{2}}]}^{\frac{1}{2}}$ v₀

$(\frac{m_{1} + m_{2}}{m_{1} - m_{2}})$ v₀

${[\frac{m_{1} + m_{2}}{m_{1} - m_{2}}]}^{\frac{1}{2}}$ v₀

${[\frac{2 m_{1} + m_{2}}{m_{1} m_{2}}]}^{\frac{1}{2}}$ v₀

Solution

${[\frac{m_{1} m_{2}}{m_{1} + m_{2}}]}^{\frac{1}{2}}$ v₀

The centre of mass is the location of particles within a system where the total mass of the system can be considered concentrated. When the system of particles is moving, the center of mass moves along with it.

The centre of mass of velocity equation is the sum of each particle's momentum ( mass times velocity ) divided by the total mass of the system.

The velocity of the centre of mass of two particles

v_cm = $\frac{m_{1} v_{1} + m_{2} v_{2}}{m_{1} + m_{2}}$

When v₁ =0 and v₂ =v₀, then

$v_{cm} = \frac{m_{2} v_{0}}{m_{1} + m_{2}}$

Now, let 'x' be the elongation in the spring.

Change in potential energy = potential energy stored in spring

⇒ $\frac{1}{2} m_{2} v_{0}^{2} - \frac{1}{2} (m_{1} + m_{2})$ ${(\frac{m_{2} v_{0}}{m_{1} + m_{2}})}^{2} = \frac{1}{2} {kx}^{2}$

⇒ m₂ $v_{0}^{2}$ $\{1 - \frac{m_{2}}{m_{1} + m_{2}}\}$ = kx²

⇒ $m_{2} v_{0}^{2} [\frac{m_{1} + m_{2} - m_{2}}{m_{1} + m_{2}}]$ = kx²

This gives

x = ${(\frac{m_{1} m_{2}}{m_{1} + m_{2}})}^{\frac{1}{2}}$ v₀

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