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PhysicsW.E.P.Conservative force & potential energyMedium2 minPYQ_2020

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The figure shows two blocks of mass $m$ each, connected by an ideal unstretched spring and then placed on a frictionless floor. If the blocks are given velocities $v_{0}$ and $2 v_{0}$ as shown, then the maximum extension in the spring is

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Answer:

Solution:

$2 m v = m 2 v_{0} - m v_{0}$

$v = \frac{v_{0}}{2}$

$w_{s} = k_{f} - k_{i}$

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

$= m {(\frac{v_{0}}{2})}^{2} - \frac{5 m v_{0}^{2}}{2}$

$\Rightarrow - \frac{k x^{2}}{2} = - \frac{9 m v_{0}^{2}}{4}$

$x = \sqrt{\frac{9}{2} \frac{m v_{0}^{2}}{k}}$

Stream:JEESubject:PhysicsTopic:W.E.P.Subtopic:Conservative force & potential energy

⏱ 2mℹ️ Source: PYQ_2020

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