A bullet of mass $m$ hits a target of mass $m$ hanging by a string 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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$υ = \sqrt{2 g h}$

$υ = \sqrt{2 g h} [1 + \frac{m}{M}]$

$υ = \sqrt{2 g h} [1 + \frac{M}{m}]$

$υ = \sqrt{2 g h} [1 - \frac{m}{M}]$

answer is C.

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Detailed Solution

Let $v'$ is the velocity of $(M + m)$ after collision.

Then from conservation of linear momentum we have,

$m v = (M + m) v' ∴ v' = (\frac{m}{M + m}) v$

Now, $v' = \sqrt{2 g h}$

$∴ (\frac{m}{M + m}) v = \sqrt{2 g h} o r v = \sqrt{2 g h} (\frac{M + m}{m})$

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