Respuesta :
Answer:
v₂ = 1.8 [m/s]
Explanation:
To be able to solve this problem we must raise the principle of conservation and quantity of movement. It is going to study the momentum before and after the firing of the rifle.
P = m*v
where:
P = momentum [kg*m/s]
m = mass [kg]
v = velocity [m/s]
Now we need to analyze the moment before and after the shot.
[tex](m_{1}+m_{2})*v_{1}=-(m_{1}*v_{2})+(m_{2}*v_{3})[/tex]
where:
m₁ = mass of the rifle = 5 [kg]
v₁ = velocity of the rifle and the bullet before firing = 0 (no movement)
m₂ = mass of the bullet = 15 [g] = 0.015 [kg]
v₂ = velocity of recoil [m/s]
v₃ = velocity of the bullet after firing = 600 [m/s]
Now replacing:
[tex](5+0.015)*0 = -(5*v_{2})+(0.015*600)\\9 = (5*v_{2})\\v_{2}=1.8 [m/s][/tex]
Momentum is the product of mass and velocity. The recoil velocity of the rifle weighing 5 kg is 1.8 m/s.
What is conservation of momentum?
According to the law of conservation of momentum if two objects are in contact with each other then for that system of the two objects the momentum of the object before will be equal to the momentum of the object after the collision.
GIven to us
Mass of the rifle, M = 5 kg
Mass of the bullet, m = 15 g = 0.015 kg
Velocity of the bullet afterwards, v₂ = 600 m/s
We know that according to the law of conservation of momentum, the momentum of the bullet and riffle before shooting will be equal to the momentum of the bullet and riffle after shooting.
[tex]Mu_1 + mu_1 = Mv_1 + mv_2[/tex]
Since, before shooting the bullet was in the riffle and the velocity of both was 0,
[tex]M(0) + m(0) = Mv_1 + mv_2\\\\-Mv_1 = mv_2\\\\v_1 = -\dfrac{m \times v_2}{M}[/tex]
Substitute the values,
[tex]v_1 = -\dfrac{0.015 \times 600}{5}\\\\v_1 = -1.8\rm\ m/s[/tex]
Thus, the recoil of the rifle is -1.8 m/s, the negative sign denotes that the velocity of recoil was in opposite direction to the bullet.
Hence, the recoil velocity of the rifle weighing 5 kg is 1.8 m/s.
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