from a rifle of 4 kg A bullet of mass 50 gram is fired with initial velocity of 35 metre per second calculate the initial recoil velocity of the rifle
Answers
Mass of the rifle (m1) = 4 kg
Mass of the bullet (m2) = 50 g = 0.05 kg
Recoil velocity of the rifle = v1
A bullet is fired with an initial velocity (v2) = 35 m/s
Initially, the rifle is at rest.
Thus, its initial velocity (v) = 0
Total initial momentum of the rifle and bullet system
⇒ (m1+m2) × v = 0
The total momentum of the rifle and bullet system after firing
= m1v1 + m2v2
= (4 × v1 ) + (0.05 × 35)
= 4v1 + 1.75
According to the law of conservation of momentum
Total momentum after the firing = Total momentum before the firing
4v1 + 1.75 = 0
v1= -1.75 / 4
v1 = – 0.4375 m/s
The negative sign indicates that the rifle recoils backwards with a velocity of 0.4375 m/s.
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Solution :-
- Mass of the rifle = 4 kg
- Mass of the bullet = 50 g = 0.05 kg
- Initially the system is at rest so the velocity of the system
i.e. Vr = Vb = 0
- Velocity of the bullet = 35 m/s
Initial momentum of the system ( Pi)
➽ Pi = Pr + Pb
➽ Pi = Mr x Vr + Mb x Vb
➽ Pi = 4 x 0 + 0.05 x 0
➽ Pi = 0
As no external force is acting on the system the total momentum of the system is conserved
By applying law of conservation of momentum ,
➽ Pi = Pf
➽ 0 = Mr x Vr '+ Mb x Vb '
➽ Vr' = - Mb x Vb ' / Mr
➽ Vr' = - 0.05 x 35 / 4
➽ Vr' = - 1.75 /4
➽ Vr' = - 0.4375 m /s
Negative sign denotes that the direction of the recoil velocity is opposite to the motion of the bullet
The recoil velocity of the rifle is 0.4375 m/ s