velocity and average force of a bullet fired?

Question

A gun fires a bullet of mass 26 grams out of a barrel 32 cm long. The gun is attached to a spring. From the recoil of the spring and the masses of the gun and the spring we determine that the gun recoiled with a total momentum of 6.2 kg m/s.

* How do I find the velocity of the bullet as it exits the barrel?
* What was the avg force exerted on the bullet as it accelerated along the length of the barrel?
* What average force would be felt by the individual holding the gun for the time the bullet accelerates along the length of the barrel?

Answer 1

The explosion of the gun powder creates the equivalent of a non-elastic collision so energy is not conserved. However, the momentum of the bullet and the rifle are conserved, so the bullet exiting the muzzle has the same momentum as the recoil of the gun. Since you are given 6.2 kg m/s, and using
0=mb*vb+mg*vg and that the velocities are vectors, then
.026 * vb-6.2=0
vb=6.2/.026 m/s
238 m/s

The second two parts are about impulse. The average force exerted on the bullet will deliver a speed of 238 m/s after traveling 0.32 m

the average force is related to velocity as
f*d=.5*m*v^2
f=.5*.026*238^2/.32
f=2301 N

The reverse force is tricky since the barrel is allowed to recoil against the spring in the opposite direction of the bullet. That means that the barrel is allowed to accelerate as the spring compresses, and f=m*a, where f is the net force, so the force on the individual is proportional to the compression of distance of the spring while the spring compresses. If we assume that the barrel accelerates into the spring while the bullet is in the barrel, and the spring uncompresses after wards, the impulse is lengthened in time, and therefore the average force on the individual is reduced. I would assume it is roughly half.

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Answer 2

First, you're able to desire to decide the acceleration of the bullet. v_f^2 = v_0^2 + 2a(delta x) 250^2 = a million.5a sixty two,500 = a million.5a a = 40-one,667 m/s^2 Now, you will discover the tension utilized to the bullet. F = ma = 0.0.5 * 40-one,667 = 1042 N