question_answer

A rifle of mass 3 kg fires a bullet of mass 0.03 kg. The bullet leaves the barrel of the rifle at a velocity of 100 m/s. If the bullet takes 0.003 s to move through its barrel, calculate the force experienced by the rifle due to its recoil.

Answer:

                                                               Here, \[{{m}_{1}}=3kg,{{m}_{2}}=0.03kg,{{u}_{1}}={{u}_{2}}=0\] \[{{\upsilon }_{1}}=?\]                   \[{{\upsilon }_{2}}=100m/s\] According to the law of conservation of momentum\[{{m}_{1}}{{\upsilon }_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{\upsilon }_{1}}+{{m}_{2}}{{\upsilon }_{2}}\] On substituting given values, we get \[0+0=3\times {{\upsilon }_{1}}+0.03\times 100\] or\[{{\upsilon }_{1}}=\frac{-100\times 0.03}{3}\] \[=-1\,m/s\] Negative sign indicates that the direction in which the rifle would move is opposite to that of the bullet.                                                  The initial and final momentum of the rifle is 0 and ?3 kg m/s and this change in momentum takes place in 0.003 s. Therefore, the rate of change of momentum or the force experience by the rifle due to its recoil would be \[F=\frac{\Delta p}{\Delta t}\] \[=\frac{-3-0}{0.003}=\frac{-3}{0.003}\] \[=-100kg\,-m/{{s}^{2}}\] \[=-1000N\] The person firing the bullet, therefore, would experience a force of 1000 N in the backward direction due to the recoil of the rifle.