A70 kg hunter, standing on frictionless ice, shoots a 42 g bullet horizontally at a speed of 590 m/s . part a part complete what is the recoil speed of the hunter?

The recoil velocity is 0.354 m/s.

Explanation:

Given that,

Mass of hunter = 70 kg

Mass of bullet = 42 g = 0.042 kg

Speed of bullet = 590 m/s

We need to calculate the recoil speed of hunter

Using conservation of momentum

Where, = mass of hunter

= mass of bullet

u = initial velocity

v = recoil velocity

Put the value in the equation

Hence, The recoil velocity is 0.354 m/s.

the answer would be 32*f