Question

A hunter stands on a frozen pond (frictionless) and fires a 4.20g bullet at 965m/s horizontally.The mass of hunter + gun is 72.5kg. Find the recoil velocity of the hunter.

Answer 1

The the recoil velocity of the hunter is 0.056 m/s in opposite direction of the bullet.

Step-by-step 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.

`P_(hunter) + P_(bullet) = 0\\\\P_(hunter) = -P_(bullet)\\\\72.5v = -4.053\\\\v = (-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.`

Therefore, the the recoil velocity of the hunter is 0.056 m/s in opposite direction of the bullet.