A bullet of mass 2.6 g strikes a ballistic pendulum of mass 3.6 kg. The center of mass of the pendulum rises a vertical distance of 13 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.

Answer:2211.4 m/sExplanation:We are given that Mass of bullet,m=2.6 g=[tex]2.6\times 10^{-3} kg[/tex]1 kg=[tex]10^3 g[/tex] Mass of pendulum,M=3.6 kgVertical distance,y=13 cm=[tex]\frac{13}{100}=0.13 m[/tex]1 m=100 cmWe have to find the bullet's initial speed.Kinetic energy of bullet=P.E[tex]\frac{1}{2}(M+m)v^2=(m+M)gh[/tex][tex]v^2=2gh[/tex][tex]v=\sqrt{2gh}[/tex]Where [tex]g=9.8 m/s^2[/tex][tex]v=\sqrt{2\times 9.8\times 0.13}=1.596m/s[/tex]According to law of conservation of momentum[tex]mu=(m+M)v[/tex][tex]u=\frac{m+M}{m}v[/tex][tex]u=\frac{2.6\times 10^{-3}+3.6}{2.6\times 10^{-3}}\times 1.596[/tex][tex]u=2211.4 m/s[/tex]Hence, the bullet's initial speed=2211.4 m/s