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.
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