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A uniform disc of mass m and radius R rotates about a fixed vertical axis passing through its centre with angular velocity $ω .$ A particle of same mass m and moving horizontally in the plane of disc with velocity $2 ω R$ towards centre of the disc collides with the disc and sticks to its rim. If the impulse on the particle due to disc is $\frac{\sqrt{37}}{z} m ω R$ , find the value of z.

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answer is 3.

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Detailed Solution

Applying law of conservation of angular momentum about center of disc,

$\frac{{mR}^{2}}{2} ω = (\frac{{mR}^{2}}{2} + {mR}^{2}) ω^{1} \Rightarrow ω^{1} = \frac{ω}{3}$

For the particle, intital momentum $P_{i} = 2 m R ω (\leftarrow)$

Final momentum $P_{f} = {mv}^{1} = \frac{mRω}{3} (↑)$

Impulse on the particle = change in momentum of the particle

$= \sqrt{(2 mRω)^{2} + {(\frac{mRω}{3})}^{2}} = \frac{\sqrt{37}}{3} mRω$

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