A rifle of mass 5 kg fires a bullet of mass 40 g. The bullet leaves the barrel of the rifle with a velocity 200 $m s^{- 1}$ . If the bullet takes 0.004 s to leave the barrel, calculate the following:
(i) recoil velocity of the rifle and
(ii) the force experienced by the rifle due to its recoil. [2 MARKS]

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Solution

Solution: 2 Marks
Given, mass of the rifle, $m_{1} = 5 k g$ ,
mass of the bullet, $m_{2} = 0.04 k g$ ,
time interval, $t = 0.004 s$
initial velocity of rifle, $u_{1} = 0$
initial velocity of bullet, $u_{2} = 0$
After firing, velocity of the bullet, $v_{2} = 200 m s^{- 1}$
Let velocity of the rifle be $v_{1}$ .
According to the law of conservation of momentum,
$m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
$⟹ 0 + 0 = (5 \times v_{1}) + (0.04 \times 200)$
$⟹ v_{1} = - 1.6 m s^{- 1}$
Initial momentum of the rifle $= m_{1} u_{1} = 0$
Final momentum of the rifle $= m_{1} v_{1} = 5 k g \times - 1.6 m s^{- 1} = - 8 k g m s^{- 1}$
Change in momentum= Final momentum-initial momentum $= - 8 - 0 = - 8 k g m s^{- 1}$
According to Newton's second law of motion,
Force, F = Rate of change of momentum= $\frac{C h a n g e i n m o m e n t u m}{T i m e} = \frac{- 8}{0.004} = - 2000 N$ .

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A rifle of mass 5 kg fires a bullet of mass 40 g. The bullet leaves the barrel of the rifle with a velocity 200 ms−1. If the bullet takes 0.004 s to leave the barrel, calculate the following: (i) recoil velocity of the rifle and (ii) the force experienced by the rifle due to its recoil. [2 MARKS]