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A bullet of mass $25 \mathrm{~g}$ moves horizontally at a speed of $250 \mathrm{~m} / \mathrm{s}$ is fired into a wooden block of mass $1 \mathrm{~kg}$ suspended by a long string. The bullet crosses the block and emerges on the other side. If the centre of the mass of the block rises through a height of $20 \mathrm{~cm}$. The speed of the bullet as it emerges from the block is (take, $g=10 \mathrm{~m} / \mathrm{s}^2$ )
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The correct answer is:
170 m/s
The given situation is shown in the following figure
Let $v_1$ and $v_2$ are final velocities of bullet and block, respectively, as block rises upto height $h$. Hence, according to conservation of energy,
$\frac{1}{2} M v_2^2=M g h \Rightarrow v_2=\sqrt{2 g h}$
Given, $h=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}$
$\Rightarrow \quad v_2=\sqrt{2 \times 10 \times 20 \times 10^{-2}}=2 \mathrm{~m} / \mathrm{s}$
Now, conservation of momentum gives
$\begin{aligned}
m u & =m v_1+M v_2 \\
\frac{25}{1000} \times 250 & =\frac{25}{1000} \times v_1+1 \times 2 \\
\frac{25}{4} & =\frac{25}{1000} \times v_1+2 \Rightarrow v_1=170 \mathrm{~ms}^{-1}
\end{aligned}$
Let $v_1$ and $v_2$ are final velocities of bullet and block, respectively, as block rises upto height $h$. Hence, according to conservation of energy,
$\frac{1}{2} M v_2^2=M g h \Rightarrow v_2=\sqrt{2 g h}$
Given, $h=20 \mathrm{~cm}=20 \times 10^{-2} \mathrm{~m}$
$\Rightarrow \quad v_2=\sqrt{2 \times 10 \times 20 \times 10^{-2}}=2 \mathrm{~m} / \mathrm{s}$
Now, conservation of momentum gives
$\begin{aligned}
m u & =m v_1+M v_2 \\
\frac{25}{1000} \times 250 & =\frac{25}{1000} \times v_1+1 \times 2 \\
\frac{25}{4} & =\frac{25}{1000} \times v_1+2 \Rightarrow v_1=170 \mathrm{~ms}^{-1}
\end{aligned}$
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