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A bullet of mass $m$ enters a wooden block of length $L$ at a speed $v_1$ and emerges out of block with a speed $v_2$. If $F$ is the average force which impeded its motion through the wooden block, then (Assume uniform deceleration inside the block)
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Verified Answer
The correct answer is:
$F=\frac{m}{2 L}\left(v_2^2-v_1^2\right)$
Given that, mass of bullet $=m$
Average resistance force $=F$
i.e. retardation, $a=F / m$
Using equation of kinematics,
$$
\begin{array}{rlrl}
v^2-u^2 & =2 a s \\
\therefore \quad v_2^2-v_1^2 & =\frac{2 F}{m} L \\
\Rightarrow \quad F =\frac{\left(v_2^2-v_1^2\right)}{2 L} m
\end{array}
$$
Average resistance force $=F$
i.e. retardation, $a=F / m$
Using equation of kinematics,
$$
\begin{array}{rlrl}
v^2-u^2 & =2 a s \\
\therefore \quad v_2^2-v_1^2 & =\frac{2 F}{m} L \\
\Rightarrow \quad F =\frac{\left(v_2^2-v_1^2\right)}{2 L} m
\end{array}
$$
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