Question

A 3 mg bullet is fired from a 1 kg musket with a velocity of 3,000. m/s. What is the resulting velocity of the musket?

Answer 1

by momentum conservation we can say

`P_i = P_f`

since gun and bullet was initially at rest so initial momentum of both must be ZERO

`0 = m_1v_(1f) + m_2v_(2f)`

`0 = 3* 10^(-3)* 3000 + 1 * v_(2f)`

`0 = 9 + v_(2f)`

`v_(2f) = - 9 m/s`

so it will recoil back with speed 9 m/s