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. $\sqrt{\frac{2Flm}{M^2}}$
• 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 $v^2=u^2-2a_{r}l$
$0=v_0^2-\frac{2F(M+m)}{Mm}l.$
Therefore, minimum value of $v_0$ is
or $v_0=\sqrt{\frac{2Fl\left(M+m\right)}{Mm}}$