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