Answer:
[tex]v_1 = 91.3 m/s[/tex]
Explanation:
By energy conservation and work energy theorem we can say that after bullet hits the block, it will move on the rough floor and comes to rest
so here work done by frictional force = change in kinetic energy
so we know that
[tex]W_f = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2[/tex]
[tex]-\mu mg d = \frac{1}{2}m(v_f^2 - v_i^2)[/tex]
[tex]-0.650(9.81)(7.5) = \frac{1}{2}(0 - v_i^2)[/tex]
[tex]47.8 = \frac{1}{2}v^2[/tex]
[tex]v = 9.78 m/s[/tex]
now by momentum conservation we have
[tex]m_1 v_1 = (m_1 + m_2) v[/tex]
[tex]12 v_1 = (100 + 12) 9.78[/tex]
[tex]v_1 = 91.3 m/s[/tex]
