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)

• Correct answer
  A

  $F=\frac{m}{2 L}\left(v_2^2-v_1^2\right)$
• B

  $F=\frac{m}{4 L}\left(v_2^2+v_1^2\right)$
• C

  $F=2 \frac{m}{L}\left(v_2-v_1\right)^2$
• D

  $F=\frac{m}{2 L}\left(v_2+v_1\right)^2$

Solution

Given that, mass of bullet $=m$
Average resistance force $=F$ i.e. retardation, $a=F / m$ Using equation of kinematics, $$ \begin{array}{rlrl} v^2-u^2 & =2 a s \\ \therefore \quad v_2^2-v_1^2 & =\frac{2 F}{m} L \\ \Rightarrow \quad F =\frac{\left(v_2^2-v_1^2\right)}{2 L} m \end{array} $$

Hence, the correct answer is option: 'A'.