question_answer

A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion?                      [AIEEE 2005]            

A)                     1.5 cm                

B)                    1.0 cm            

C)                     3.0 cm                

D)                   2.0 cm

Correct Answer: B

Solution :

                       Let initial velocity of the bullet = u             After penetrating 3 cm its velocity becomes \[\frac{u}{2}\]                                                                                            From \[{{v}^{2}}={{u}^{2}}-2as\]             \[{{\left( \frac{u}{2} \right)}^{2}}={{u}^{2}}-2a\,(3)\]              Þ \[6a=\frac{3{{u}^{2}}}{4}\] Þ \[a=\frac{{{u}^{2}}}{8}\]             Let further it will penetrate through distance x and stops at point C.             For distance BC, \[v=0,\,u=u/2,\,s=x,\,a={{u}^{2}}/8\]                         From \[{{v}^{2}}={{u}^{2}}-2as\]Þ\[0={{\left( \frac{u}{2} \right)}^{2}}-2\left( \frac{{{u}^{2}}}{8} \right)\,.\,x\]Þ\[x=1\] cm.