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    Analysis of a bullet undergoing constant acceleration

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    A gun fires a bullet of mass 22 grams out of a barrel 36 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 4.7 kg m/s.

    - With what velocity did the bullet exit the barrel?
    - Assuming that the bullet accelerated uniformly from rest along the length of the barrel, how long did it take the bullet to accelerate from rest down the length of the barrel?
    - What was the average 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?

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    Solution Preview

    - Conservation of linear momentum applies

    Let momentum of Gun/Spring assembly p(g) = 4.7 kg.m/s

    Momentum of bullet p(b) is the product of its exit velocity v(b) and mass m(b) = 22g = 0.022 kg

    p(b) = m(b)*v(b) = p(g)

    v(b) = p(g)/m(b) = 4.7/0.022 = 213.64 m/s

    - For linear ...

    Solution Summary

    Linear kinematic equations (motion under constant acceleration) are used to determine some dynamics about the firing of a bullet including its exit velocity from the barrel of a gun, how long it takes to accelerate down the barrel of the gun, the average force exerted on the bullet during its firing and the recoil force