question_answer

A bullet loses 1/20 of its velocity after penetrating a plank. How many planks are required to stop the bullet?

A) 6                                             

B) 9

C) 11                                          

D) 13

Correct Answer: B

Solution :

The final velocity after it passes the plank is \[\frac{19u}{20}.\] Let \[x\] be the thickness of the plank, the deceleration due to resistance of plank, is given by \[{{v}^{2}}={{u}^{2}}+2as\]where v is final velocity, u is initial velocity, a is acceleration and s is displacement. Here           \[v=\frac{19}{20}u\] \[\therefore \]      \[{{\left( \frac{19}{20}u \right)}^{2}}={{u}^{2}}+2\,ax\] \[\Rightarrow \]     \[2\,ax=\frac{-39}{400}{{u}^{2}}\] Suppose the bullet is stopped after passing through n such planks. Then the distance covered by bullet is \[nx\]. \[\therefore \]             \[0={{\left( \frac{19}{20} \right)}^{2}}{{u}^{2}}+2\,anx\] \[\Rightarrow \]          \[-{{\left( \frac{19}{20} \right)}^{2}}{{u}^{2}}=n\times \frac{-39}{400}{{u}^{2}}\] \[\Rightarrow \]         \[n=\frac{361}{39}\approx 9\]