Answer:
If the rifle is held loosely away from the shoulder, the recoil velocity will be of -8.5 m/s, and the kinetic energy the rifle gains will be 81.28 J.
Explanation:
By momentum conservation, and given the bullit and the recoil are in a straight line, the momentum analysis will be unidimentional. As the initial momentum is equal to zero (the masses are at rest), we have that the final momentum equals zero, so
[tex]0=P_{f}=m_{b} *v_{b}+m_{r}*v_{r}[/tex]
now we clear [tex]v_{r}[/tex] and use the given data to get that
[tex]v_{r}=-8.5\frac{m}{s}[/tex]
But we have to keep in mind that the bullit accelerate from rest to a speed of 425 m/s, then if the rifle were against the shoulder, the recoil velocity would be a fraction of the result obtained, but, as the gun is a few centimeters away from the shoulder, it is assumed that the bullit get to its final velocity, so the kick of the gun, gets to its final velocity [tex]\bold{v_{r}}[/tex] too.
Finally, using [tex]v_{r}[/tex] we calculate the kinetic energy as
[tex]K=\frac{1}{2}m_{r}v_{r}^{2}=81.28J[/tex]
