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]
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]
Solution: 2 Marks
Given, mass of the rifle, m1=5 kg,
mass of the bullet, m2=0.04 kg,
time interval, t=0.004 s
initial velocity of rifle, u1=0
initial velocity of bullet,u2=0
After firing, velocity of the bullet, v2=200 ms−1
Let velocity of the rifle be v1.
According to the law of conservation of momentum,
m1u1+m2u2=m1v1+m2v2
⟹0+0=(5×v1)+(0.04×200)
⟹v1=−1.6 ms−1
Initial momentum of the rifle =m1u1=0
Final momentum of the rifle =m1v1=5 kg×−1.6 ms−1=−8 kgms−1
Change in momentum= Final momentum-initial momentum=−8−0=−8 kgms−1
According to Newton's second law of motion,
Force, F = Rate of change of momentum=Change in momentumTime=−80.004=−2000 N.
Solution: 2 Marks
Given, mass of the rifle, m1=5 kg,
mass of the bullet, m2=0.04 kg,
time interval, t=0.004 s
initial velocity of rifle, u1=0
initial velocity of bullet,u2=0
After firing, velocity of the bullet, v2=200 ms−1
Let velocity of the rifle be v1.
According to the law of conservation of momentum,
m1u1+m2u2=m1v1+m2v2
⟹0+0=(5×v1)+(0.04×200)
⟹v1=−1.6 ms−1
Initial momentum of the rifle =m1u1=0
Final momentum of the rifle =m1v1=5 kg×−1.6 ms−1=−8 kgms−1
Change in momentum= Final momentum-initial momentum=−8−0=−8 kgms−1
According to Newton's second law of motion,
Force, F = Rate of change of momentum=Change in momentumTime=−80.004=−2000 N.
