Answer:
2211.4 m/s
Explanation:
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 kg
Vertical distance,y=13 cm=[tex]\frac{13}{100}=0.13 m[/tex]
1 m=100 cm
We 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
