Question

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

Answer 1

Answer:

(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

$m_{1}u_{1}$  -  $m_{2}u_{2}$ = $m_{1}v_{1}$ + $m_{2}v_{2}$

But, the initial velocities equal zero,

           $u_{1}$    =  $u_{2}$  =  0

So that,

          0 = $m_{1}v_{1}$ + $m_{2}v_{2}$

Substituting the values of variables given in the question,

        0 = 1 × $v_{1}$      +     0.02 × 600

        0 = $v_{1}$  + 12

    $v_{1}$ = - 12 m/s

Thus, the velocity is 12 m/s.

(b)    F = ma

From the third equation of motion,

    $v^{2}$  =  $u^{2}$  + 2as

        a = - $\frac{u^{2}}{2s}$

        =  $-\frac{600^{2} }{2(0.2)}$

        =  -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.