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A bullet of mass $\mathrm{m}$ is fired horizontally into a large sphere of mass $\mathrm{M}$ and radius $\mathrm{R}$ resting on a smooth horizontal table.
The bullet hits the sphere at a height $\mathrm{h}$ from the table and sticks to its surface. If the sphere starts rolling without slippng immediately on impact, then
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The bullet hits the sphere at a height $\mathrm{h}$ from the table and sticks to its surface. If the sphere starts rolling without slippng immediately on impact, then
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Verified Answer
The correct answer is:
$\frac{\mathrm{h}}{\mathrm{R}}=\frac{10 \mathrm{~m}+7 \mathrm{M}}{5(\mathrm{~m}+\mathrm{M})}$
Apply conservation of linear momentum
$$
\begin{array}{l}
m v=(m+M) v_{0} \\
m v \sin \theta R=\left(\frac{2}{5} M R^{2}+m R^{2}\right) \omega_{0} \\
m v\left(\frac{h-R}{R}\right) R=\frac{(2 M+5 M)}{5} \omega_{0} R^{2} \\
(m+M)(h-R) \omega_{0} R=\frac{(2 M+5 M)}{5} \omega_{0} R^{2} \\
\frac{h}{R}=\frac{(10 m+7 M)}{5(m+M)}
\end{array}
$$
$$
\begin{array}{l}
m v=(m+M) v_{0} \\
m v \sin \theta R=\left(\frac{2}{5} M R^{2}+m R^{2}\right) \omega_{0} \\
m v\left(\frac{h-R}{R}\right) R=\frac{(2 M+5 M)}{5} \omega_{0} R^{2} \\
(m+M)(h-R) \omega_{0} R=\frac{(2 M+5 M)}{5} \omega_{0} R^{2} \\
\frac{h}{R}=\frac{(10 m+7 M)}{5(m+M)}
\end{array}
$$
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