Laws Of Motion Question 130
Question: 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
Options:
A) $ \sqrt{\frac{F\ell /m}{M^{2}}} $
B) $ \begin{aligned} & \sqrt{\frac{2F\ell (M+m)}{Mm}} \&\ \end{aligned} $
C)$ \begin{aligned} & \sqrt{\frac{2F\ell m}{M^{2}}} \&\ \end{aligned} $
D) $ \sqrt{\frac{F\ell (M+m)}{M{{m}^{{}}}}} $
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Answer:
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}-2a _{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}} $
BETA
