Question

A 0.0450 kg bullet is accelerated from rest to a speed of 425 m/s in a 2.25 kg rifle (which is inititally at rest). The pain of the rifle’s kick is much worse if you hold the gun loosely a few centimeters from your shoulder rather than holding it tightly against your shoulder.

(a) Take the bullet's velocity as positive. Calculate the recoil velocity (in m/s) (with a proper sign) of the rifle if it is held loosely away from the shoulder.
(b) How much kinetic energy (in Joules) does the rifle gain?

Answer 1

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.

Step-by-step 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

`0=P_(f)=m_(b) *v_(b)+m_(r)*v_(r)`

now we clear
`v_(r)` and use the given data to get that

`v_(r)=-8.5(m)/(s)`

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
`\bold{v_(r)}` too.

Finally, using
`v_(r)` we calculate the kinetic energy as

`K=(1)/(2)m_(r)v_(r)^(2)=81.28J`