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    Impulse-Momentum Theorem of a Bullet

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    A gun fires a bullet of mass 17 grams out of a barrel 24 cm long. The bullet is accelerated down the length of the barrel with an average force of 16 Newtons.

    Using the Impulse-Momentum Theorem, determine the velocity with which the bullet exits the barrel.

    What average force is exerted by the individual holding the gun?

    The individual firing the gun has mass 63 kg and is standing on a nearly frictionless skateboard, initially at rest. With what velocity will the individual be moving immediately after firing the gun?

    © BrainMass Inc. brainmass.com December 24, 2021, 9:36 pm ad1c9bdddf
    https://brainmass.com/physics/velocity/impulse-momentum-theorem-bullet-396775

    SOLUTION This solution is FREE courtesy of BrainMass!

    A gun fires a bullet of mass 17 grams out of a barrel 24 cm long.   The bullet is accelerated down the length of the barrel with an average force of 16 N.  

    a) Using the Impulse-Momentum Theorem, determine the velocity with which the bullet exits the barrel.

    b) What average force is exerted by the individual holding the gun?

    c) The individual firing the gun has mass 63 kg and is standing on a nearly frictionless skateboard, initially at rest.  With what velocity will the individual be moving immediately after firing the gun?

    Solution: a) As per impulse-momentum theorem, impulse applied on a body is equal to change in its momentum.

    FΔt = Δ(mv) = mΔv

    Force acting on the bullet = F = 16 N

    Acceleration of the bullet = a = Force/Mass = 16/17x10^-3 = 941.18 m/s^2

    As the bullet starts from rest, its initial velocity u = 0

    Distance travelled s = 0.24 m

    Using the kinematic equation: s = ut + ½ at^2 and substituting values we get:

    0.24 = 0 + ½ x 941.18 x t^2
    ______________
    t = √(0.24 x 2/941.18) = 0.0226 sec

    Time for which the force acts on the bullet Δt = 0.0226 sec

    Impulse applied on the bullet = FΔt = 16 x 0.0226 = 0.3616 N.s

    By impulse-momentum theorem: Change in momentum of the bullet mΔv = 0.3616 kg.m/s

    Change in velocity Δv = 0.3616/m = 0.3616/(17x10^-3) = 21.27 m/s

    As initial velocity is zero, the bullet exits the barrel with a velocity of 21.27 m/s

    b) The force exerted by the gun on the bullet is 16 N. By Newton's third law of motion, the bullet exerts an equal and opposite force on the gun i.e.16 N (recoil force). To hold the gun in place, the individual holding the gun must apply a forward force of 16 N on the gun.

    c) By the law of conservation of linear momentum:

    Sum of momentum of the individual with gun and the bullet immediately after firing =
    Sum of momentum of the individual with gun and the bullet immediately before firing ..........(1)

    Assuming the individual holding the gun is at rest before its firing, the right hand side of the above equation is zero.

    Momentum of the bullet after firing = 0.3616 kg.m/s

    Momentum of the individual with gun = MV

    Substituting in (1): MV + 0.3616 = 0 where M = Mass of the individual with gun, V = Velocity of the individual immediately after firing

    V = - 0.3616/M = - 0.3616/63 = - 5.74 x 10^-3 m/s or - 0.574 cm/sec

    The individual will move backwards with a speed of 0.574 cm/sec.

    Note: The problem specifies mass of the individual firing the gun as 63 kg. Mass of the gun and mass of the skate board have not been specified. Hence, in the above solution 63 kg has been assumed to include mass of the gun and the skate board also.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 9:36 pm ad1c9bdddf>
    https://brainmass.com/physics/velocity/impulse-momentum-theorem-bullet-396775

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