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Q.

A body A of mass moving with velocity v while passing through its mean position collides in perfect inelastically with a body B of same mass which is connected to a vertical wall through a spring whose spring constant is k . After collision it sticks to B and executes S.H.M. Find the amplitude of resulting motion:

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a

$\sqrt{\frac{m}{k}} v$

b

$\sqrt{\frac{m}{2 k}} v$

c

$\frac{m}{k} \sqrt{v}$

d

$\frac{m}{2 k} \sqrt{v}$

answer is B.

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Detailed Solution

Assuming the collision lasts for a small interval. We can apply conservation of momentum and get the common velocity $= \frac{v}{2} K . E . = \frac{1}{2} (2 m) {(\frac{v}{2})}^{2} = \frac{1}{4} m v^{2}$ .

This is also the total energy of system as the spring is unstretched at this moment. If the amplitude is A, total energy $I = \frac{1}{2} k A^{2} ∴ \frac{1}{2} k A^{2} = \frac{1}{4} m v^{2}$

$∴ A = \sqrt{\frac{m}{2 k}} \cdot v$

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