question_answer

A bullet of mass w moving with velocity \[{{v}_{0}}\]hits a wooden plank \[A\] of mass \[M\] placed on a smooth horizontal surface. The length of the plank is \[\ell \]. 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) \[\sqrt{\frac{F\ell /m}{{{M}^{2}}}}\]   

B) \[\begin{align}   & \sqrt{\frac{2F\ell \,(M+m)}{Mm}} \\  &  \\ \end{align}\]

C)  \[\begin{align}   & \sqrt{\frac{2F\ell \,m}{{{M}^{2}}}} \\  &  \\ \end{align}\]     

D) \[\sqrt{\frac{F\ell \,(M+m)}{M{{m}^{{}}}}}\]

Correct Answer: B

Solution :

[b] From Newton's third law a force F acts on the block in forward direction. Acceleration of black \[{{a}_{1}}=\frac{F}{M}\] Retardation of bullet \[{{a}_{2}}=\frac{F}{m}\] Relative retardation of bullet \[{{a}_{r}}={{a}_{1}}+{{a}_{2}}=\frac{F(M+m)}{Mm}\] Applying \[{{v}^{2}}={{u}^{2}}-2{{a}_{r}}\ell \] \[0={{v}_{0}}^{2}-\frac{2F(M+m)}{Mm}.\ell \] Therefore, minimum value of \[{{V}_{0}}\]is Or \[{{V}_{0}}=\sqrt{\frac{2F\ell (M+m)}{Mm}}\]