question_answer

A bullet strikes a wooden plank with a velocity of 54 kmph and emerges from it with a velocity of 36 kmph. If the thickness of the plank is 20cm, find the average resistance (in Newton) offered by it to the motion of the bullet. What is the thickness of the same plank that would have just stopped the bullet? Mass of the bullet = 15 gm.

Answer:

(i) \[u=54kmph=54\times \frac{5}{18}=15\,m/s\] \[v=36\,kmph=36\times \frac{5}{18}=10\,\,m/s\] \[S=20\,cm=\frac{20}{100}=0.2\,m\] \[m=15\,g=15\times {{10}^{-3}}kg\] \[F=?\] We know, \[W=\frac{1}{2}m{{v}^{2}}-\frac{1}{2}m{{u}^{2}}\] \[\Rightarrow F\times S=\frac{1}{2}m({{v}^{2}}-{{u}^{2}})\] \[\Rightarrow F\times S=\frac{1}{2}\times 15\times {{10}^{-3}}[{{10}^{2}}-{{15}^{2}}]\] \[=\frac{1}{2}\times 15\times {{10}^{-3}}\times (-125)\] \[\therefore F=\frac{1}{2}\times \frac{15\times {{10}^{-3}}\times (-125)}{0.2}\] \[=-0.5\times 15\times 125\times {{10}^{-3}}N=-4.687\,\,N\] The negative sign indicates the force is retarding force. (ii) If       \[u=15\,m/sv=0\,m/s\]                 \[S=?F=-4.7\,N\]              \[F\times S=\frac{1}{2}m({{v}^{2}}-{{u}^{2}})\] \[\Rightarrow S=\frac{1}{2}\times \frac{15\times {{10}^{-3}}\times (-125)}{-4.687}=360\,m\]