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A bullet of mass $m$ moving with velocity just grazes the top of a solid cylinder of mass $M$ and radius $R$ resting on a rough horizontal surface with its curved surface in con tact with the horizontal surface . Assuming that the cylinder rolls without slipping, find the angular velocity of the cylinder and the final velocity of the bullet.

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$ω = \frac{4 m u}{(8 m + 3 M) R}, v = \frac{(m u)}{(8 m + 3 M)}$

$ω = \frac{m u}{(8 m + 3 M) R}, v = \frac{(8 m u)}{(8 m + 3 M)}$

$ω = \frac{4 m u}{(8 m + 3 M) R}, v = \frac{(8 m u)}{(8 m + 3 M)}$ ,

$ω = \frac{4 m u}{(m + 3 M) R}, v = \frac{(8 m u)}{(8 m + 3 M)}$

answer is C.

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Detailed Solution

We can assume the rolling of the cylinder as pure rotation about an axis passing through $O$ .

Using conservation of angular momentum about $O$

$(m u) \times 2 R = (m v) \times 2 R + I_{0} ω$

and $v = ω \times 2 R$ and $I_{O} = \frac{M R^{2}}{2} + M R^{2}$

After substituting these values in above equations, we get

$ω = \frac{4 m u}{(8 m + 3 M) R}$ and $v = \frac{(8 m u)}{(8 m + 3 M)}$

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