Explanation:
Given,
Mass of the gun, \(M = 3\;kg\)
Mass of the bullet, \(m=10 \mathrm{~g}\)
Applying conservation of linear momentum we get, \((M-m) v_{g}=m v_{b}\)
Since, \(M>>m\)
\(M-m \approx M\)
\(\therefore M v_{g}=m v_{b}\)
\(\therefore\) Impulse supplied to the gun
\(=\) Impulse supplied to the bullet
\( = m{v_b} = \frac{{10}}{{1000}} \times 600\;kg{\rm{ - }}m{\rm{/}}s\)
\( = 6\;kg - m/s\,\,\,\,i.\,e.\,\,\,\,6\,N\,s.\)