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A bullet of mass $m$ hits a mass $M$ and gets embedded in it. If the block rises to a height $h$ as a result of this collision, the velocity of the bullet before collision is
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Verified Answer
The correct answer is:
$v=\sqrt{2 g h}\left(1+\sqrt{\frac{M}{m}}\right)$
Let the velocity of the bullet before collision be $v$, then according to law of conservation of linear momentum,
$$
m v=(m+M) v^{\prime}
$$
As the mass $m$ of the bullet gets embedded in the wall, hence velocity of bullet + target just after collision will be same,
$$
\begin{aligned}
\quad \frac{1}{2}(M+m) v^{r^2} & =(m+M) g h \\
\Rightarrow \quad v^{\prime} & =\sqrt{2 g h}
\end{aligned}
$$
Putting the value of $v^{\prime}$ in Eq. (i), we get
$$
\Rightarrow \quad v=\sqrt{2 g h} \frac{(m+M)}{m}=\sqrt{2 g h}\left(1+\frac{M}{m}\right)
$$
$$
m v=(m+M) v^{\prime}
$$
As the mass $m$ of the bullet gets embedded in the wall, hence velocity of bullet + target just after collision will be same,
$$
\begin{aligned}
\quad \frac{1}{2}(M+m) v^{r^2} & =(m+M) g h \\
\Rightarrow \quad v^{\prime} & =\sqrt{2 g h}
\end{aligned}
$$
Putting the value of $v^{\prime}$ in Eq. (i), we get
$$
\Rightarrow \quad v=\sqrt{2 g h} \frac{(m+M)}{m}=\sqrt{2 g h}\left(1+\frac{M}{m}\right)
$$
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