Answer:
The the recoil velocity of the hunter is 0.056 m/s in opposite direction of the bullet.
Explanation:
Given;
mass of bullet, m₁ = 4.2 g = 0.0042 kg
mass of hunter + gun = 72.5 kg
velocity of the bullet, u = 965 m/s
Momentum of the bullet when it was fired;
P = mv
P = 0.0042 x 965
P = 4.053 kg.m/s
Determine the recoil velocity of the hunter.
Total momentum = sum of the individual momenta
Total momentum = momentum of the bullet + momentum of the hunter
Apply the principle of conservation of momentum, sum of the momentum is equal to zero.
[tex]P_{hunter} + P_{bullet} = 0\\\\P_{hunter} = -P_{bullet}\\\\72.5v = -4.053\\\\v = \frac{-4.053}{72.5} \\\\v = - 0.056 \ m/s\\\\Thus, the \ recoil \ velocity \ of \ the \ hunter \ is \ 0.056 \ m/s, \ in \ opposite \ direction \ of \ the \ bullet.[/tex]
Therefore, the the recoil velocity of the hunter is 0.056 m/s in opposite direction of the bullet.
