The principle of conservation of momentum can be applied here.
when two objects interact, the total momentum remains the same provided no external forces are acting.
Consider the whole system , gun and bullet. as an isolated system, so the net momentum is constant. In particular before firing the gun, the net momentum is zero. The conservation of momentum,
assume the bullet goes to right side and the gravitational acceleration =10
so now the weight of the rifle=
this is a negative velocity to the right side. that means the rifle recoils to the left side
promise you wont victory royale somewhere? ok here we go.
it should be like 16.98 pounds of recoil
The recoil speed is 2.14 m/s
We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the bullet-rifle must be conserved before and after the explosion.
Before the explotions, both are at rest, so the total momentum is zero:
After the shot, the total momentum is:
m = 12.70 g = 0.0127 kg is the mass of the bullet
v = 430 m/s is the velocity of the bullet
M is the mass of the rifle
V is the recoil velocity of the rifle
We know the weight of the rifle, W = 25.0 N, so we can find its mass:
And momentum is conserved, we can equate the initial and final momentum and solve for V:
Where the negative sign indicates that the rifle's recoil is in the opposite direction: therefore, the speed of recoil is 2.14 m/s.
Learn more about momentum:
V = 2.14 m/s
The mass of the bullet, m = 12.7 g
= 0.0127 Kg
The velocity of the bullet, v = 430 m/s
The weight of the rifle, w = 25 N
∴ The mass of the rifle, M = 25/9.8
= 2.55 Kg
According to the law of conservation of momentum,
MV + mv = 0
Therefore, the recoil velocity of the gun
V = -mv/M
Substituting the given values in the above equation
V = -0.0127 x 430 /2.55
= -2.14 m/s
The negative sign indicates the direction of V with respect to v.
Hence, the recoil velocity of the gun is, V = 2.14 m/s