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PhysicsSHMMiscellaneousMedium2 minQB

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The left block collides inelastically with the right block and sticks to it. Find the amplitude of the resulting simple harmonic motion.

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Answer:

Solution:

Ans. $v \frac{m}{2 K}$ Velocity of system after collision $v_{1} = \frac{mv}{2 m} = \frac{v}{2}$ let amplitude of SHM be a then $\frac{1}{2} Ka^{2} = \frac{1}{2} (2 m) (\frac{v}{2})^{2} \Rightarrow a = v \frac{m}{2 K}$

Stream:JEESubject:PhysicsTopic:SHMSubtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

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