Two blocks of mass m₁ and m₂ are moving along a smooth horizontal floor. A non-ideal spring is attached at the back of mass m₂. Initial velocities of the blocks are u₁ and u₂ as shown; with u₁ > u₂. After collision the two blocks were found to be moving with velocities V₁ and V₂ respectively. Find the ratio of impulse (on each block) during the deformation phase of the spring and that during its restoration phase. [By non ideal spring we mean that it does not completely regain its original shape after deformation. You can neglect the mass of the spring.]

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$\frac{u_{2} - u_{1}}{V_{2} - V_{1}}$

$\frac{u_{1} - u_{2}}{V_{1} - V_{2}}$

$\frac{u_{1} - u_{2}}{V_{2} - V_{1}}$

$\frac{2 u_{2}}{V_{2} - V_{1}}$

answer is B.

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Detailed Solution

before collision
Maximum Deformation
After collision
Impulse on m₂ during deformation phase is given by- $J_{D} = m_{2} u - m_{2} u_{2}$

Where u = Common velocity in maximum deformation state
Impulse on m2 during restoration phase is given by

$∴ \frac{J_{D}}{J_{R}} = \frac{u - u_{2}}{V_{2} - u} But u = \frac{m_{1} u_{1} + m_{2} u_{2}}{m_{1} + m_{2}} = \frac{m_{1} V_{1} + m_{2} V_{2}}{m_{1} + m_{2}} ∴ \frac{J_{D}}{J_{R}} = \frac{\frac{m_{1} u_{1} + m_{2} u_{2}}{m_{1} + m_{2}} - u_{2}}{V_{2} - \frac{m_{1} V_{1} + m_{2} V_{2}}{m_{1} + m_{2}}} = \frac{u_{1} - u_{2}}{V_{2} - V_{1}}$