Military rifles have a mechanism for reducing the recoil forces of the gun on the person firing it. An internal part recoils over a relatively large distance and is stopped by damping mechanisms in the gun. The larger distance reduces the average force needed to stop the internal part. (a) Calculate the recoil velocity of a 1.00-kg plunger that directly interacts with a 0.0200-kg bullet fired at 600 m/s from the gun. (b) If this part is stopped over a distance of 20.0 cm, what average force is exerted upon it by the gun

The recoil velocity of the plunger is 12 m/s.

The average force exerted upon the bullet by the gun is 18,000 N.

The given parameters;

The recoil velocity of the plunger is calculated by applying the principle of conservation of linear momentum;

m₁u₁ = m₂u₂

The acceleration of the bullet  before stopping in 20 cm is calculated as follows;

The average force exerted by the bullet on the gun is calculated as;

F = ma

F = (0.02)(-900,000)

F = - 18,000 N

Thus, the average force exerted upon the bullet by the gun is 18,000 N.

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(a) The velocity is 12 m/s.

(b) The average force is 18 000N.

Explanation:

(a) The principle of conservation of momentum,

total momentum before = total momentum after

 -   = +

But, the initial velocities equal zero,

               =    =  0

So that,

          0 = +

Substituting the values of variables given in the question,

        0 = 1 ×      +     0.02 × 600

        0 =  + 12

    = - 12 m/s

Thus, the velocity is 12 m/s.

(b)    F = ma

From the third equation of motion,

     =    + 2as

        a = -

        =  

        =  -900, 000 m/s^2

   a   =  900, 000 m/s^2

The acceleration is 900 Km/s^2.

 F = 0.02 × 900000

    = 18 000 N

Thus, the average force is 18 000N.

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

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