question_answer

A bullet of mass 4 g when fired with a velocity of 50 ms -1, can enter a wall up to a depth of 10 cm. How much will be the average resistances offered by the wall?

Answer:

                               The hindrance offered by the wall to the motion of bullet is called the resistance offered by wall. Mass of the bullet, \[m=4\times {{10}^{-3}}kg\] Initial velocity, \[u=50\,m{{s}^{-1}}\] Depth,        \[s=10\,cm=\frac{1}{10}m\] Final velocity, v = 0, Force,         F= ? \[{{\upsilon }^{2}}={{u}^{2}}+2as\]                                                          \[{{\upsilon }^{2}}-{{u}^{2}}=2as\] \[0-{{(50)}^{2}}=2a\times \frac{1}{10}\] or\[-2500=\frac{a}{5}\Rightarrow a=-12500\,\text{m}{{\text{s}}^{-2}}\] Force, \[F=ma=4\times {{10}^{-3}}\times (-12500)=-50\text{N}\] Thus, the average resistance offered = 50 N. Negative sign indicates that the force is acting opposite to the motion.