Question

A bullet of mass m moving with velocity $v_{0}$ hits a wooden plank $A$ of mass A/placed on a smooth horizontal surface. The length of the. plank is $L$ The bullet experiences a constant resistive force $F$ inside the block. The minimum value of $v_{0}$ such that it is able to come out of the plank is

• A. $\sqrt{\frac{Fl /m}{M^{2}}}$
• B. $\sqrt{\frac{2Fl \left(M+m\right)}{Mm}}$
• C. $\sqrt{\frac{2Fl m}{M^{2}}}$
• D. $\sqrt{\frac{Fl\left(M+m\right)}{Mm}}$

Answer 1

Answer: (B)

From Newton’s third law $a$ force $F$ acts on the block in forward direction.
Acceleration of block $a_{1} =\frac{F}{M}$
Retardation of bullet $a_{2} =\frac{F}{m}$
Relative retardation of bullet
$a_{r} =a_{1}+a_{2}=\frac{F(M+m)}{Mm}$
Applying $v^{2} =u^{2} -2a_{r}l$
$0=v^{2}_{0} -\frac{2F(M+m)}{Mm}l.$
Therefore, minimum value of $v_{0}$ is
or $v_{0}=\sqrt{\frac{2Fl\left(M+m\right)}{Mm}}$