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A bullet of mass 20 g and moving with $600 \mathrm{~m} / \mathrm{s}$ collides with a block of mass 4 kg hanging with the string. What is velocity of bullet when it comes out of block, if block rises to height 0.2 m after collision?
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The correct answer is:
$200 \mathrm{~m} / \mathrm{s}$
According to conservation of linear momentum,
$m_1 v_1=m_1 v+m_2 v_2$
where $v_1$ is velocity of bullet before collision, $v$ is velocity of bullet after the collision and $v_2$ is the velocity of block.
$\therefore \quad 0.02 \times 600=0.02 v+4 v_2$
Here, $\quad v_2=\sqrt{2 \mathrm{gh}}=\sqrt{2 \times 10 \times 0.2}=2 \mathrm{~m} / \mathrm{s}$
$\therefore \quad 0.02 \times 600=0.02 v+4 \times 2$
$\Rightarrow \quad 0.02 v=12-8$
$\Rightarrow \quad v=\frac{4}{0.02}=200 \mathrm{~m} / \mathrm{s}$
$m_1 v_1=m_1 v+m_2 v_2$
where $v_1$ is velocity of bullet before collision, $v$ is velocity of bullet after the collision and $v_2$ is the velocity of block.
$\therefore \quad 0.02 \times 600=0.02 v+4 v_2$
Here, $\quad v_2=\sqrt{2 \mathrm{gh}}=\sqrt{2 \times 10 \times 0.2}=2 \mathrm{~m} / \mathrm{s}$
$\therefore \quad 0.02 \times 600=0.02 v+4 \times 2$
$\Rightarrow \quad 0.02 v=12-8$
$\Rightarrow \quad v=\frac{4}{0.02}=200 \mathrm{~m} / \mathrm{s}$
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