Answer:
217.43298 m/s
Explanation:
[tex]m_1[/tex] = Mass of bullet = 19 g
[tex]m_2[/tex] = Mass of bob = 1.3 kg
L = Length of pendulum = 2.3 m
[tex]\theta[/tex] = Angle of deflection = 60°
u = Velocity of bullet
Combined velocity of bullet and bob is given by
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2aL(1-cos\theta)+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times (1-cos60)+0^2}\\\Rightarrow v=3.13209\ m/s[/tex]
As the momentum is conserved
[tex]m_1u=(m_1+m_2)v\\\Rightarrow u=\frac{(m_1+m_2)v}{m_1}\\\Rightarrow v=\frac{(0.019+1.3)\times 3.13209}{0.019}\\\Rightarrow v=217.43298\ m/s[/tex]
The speed of the bullet is 217.43298 m/s
