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. A bullet of mass $0.01 \mathrm{~kg}$ travelling at a speed of $500 \mathrm{~ms}^{-1}$ strikes a block of mass $2 \mathrm{~kg}$ which is suspended by a string of length $5 \mathrm{~m}$. The centre of gravity of the block is found to rise a vertical distance of $0.1 \mathrm{~m}$. What is the speed of the bullet after it emerges from the block?
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Verified Answer
The correct answer is:
$220 \mathrm{~ms}^{-1}$
According to question, the given situation is shown in the following figure.
Mass of block, $M=2 \mathrm{~kg}$
Mass of bullet, $m_1=0.01 \mathrm{~kg}$
Speed of buIlet, $v=500 \mathrm{~ms}^{-1}$
$$
h=0.1 \mathrm{~m}
$$
If $v_1$ and $v_2$ be the velocities of the bullet and blocks after collision, then by conservation of energy,
$$
\begin{aligned}
\frac{1}{2} M v_2^2 & =M g h \\
v_2 & =\sqrt{2 g h}=\sqrt{2 \times 9.8 \times 0.1}=1.4 \mathrm{~ms}^{-1}
\end{aligned}
$$
According to the law of conservation of momentum,
$$
\begin{aligned}
m_1 v & =m_1 v_1+M v_2 \\
\Rightarrow \quad v_1 & =\frac{m_1 v-M v_2}{m_1}=\frac{0.01 \times 500-2 \times 1.4}{0.01} \\
& =\frac{5-28}{0.01}=\frac{2.2}{0.01}=220 \mathrm{~ms}^{-1}
\end{aligned}
$$
Mass of block, $M=2 \mathrm{~kg}$
Mass of bullet, $m_1=0.01 \mathrm{~kg}$
Speed of buIlet, $v=500 \mathrm{~ms}^{-1}$
$$
h=0.1 \mathrm{~m}
$$
If $v_1$ and $v_2$ be the velocities of the bullet and blocks after collision, then by conservation of energy,
$$
\begin{aligned}
\frac{1}{2} M v_2^2 & =M g h \\
v_2 & =\sqrt{2 g h}=\sqrt{2 \times 9.8 \times 0.1}=1.4 \mathrm{~ms}^{-1}
\end{aligned}
$$
According to the law of conservation of momentum,
$$
\begin{aligned}
m_1 v & =m_1 v_1+M v_2 \\
\Rightarrow \quad v_1 & =\frac{m_1 v-M v_2}{m_1}=\frac{0.01 \times 500-2 \times 1.4}{0.01} \\
& =\frac{5-28}{0.01}=\frac{2.2}{0.01}=220 \mathrm{~ms}^{-1}
\end{aligned}
$$
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