Question

A 70 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?

Answer 1

The recoil velocity is 0.354 m/s.

Step-by-step 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

`m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)+m_(2)v_(2)`

Where,
`m_(1)` = mass of hunter

`m_(2)` = mass of bullet

u = initial velocity

v = recoil velocity

Put the value in the equation

`0+0=70* v_(1)+0.042*590`

`v_(1)=-(0.042*590)/(70)`

`v=-0.354\ m/s`

Hence, The recoil velocity is 0.354 m/s.