Answer:
The initial speed of the bullet will be:
[tex]v_{ib}=331.36\: m/s[/tex]
Explanation:
Using the momentum conservation:
[tex]p_{i}=p_{f}[/tex]
[tex]m_{b}v_{ib}=Mv[/tex] (1)
Where:
Using the conservation of energy:
[tex]\frac{1}{2}Mv^{2}=\frac{1}{2}Mv_{2}^{2}+Mgh [/tex] (2)
Where v(2) is the speed of the system at 0.8 m of vertical height.
Using the forces acting on the system we can find v(2).
The forces can be equal to the centripetal force:
[tex]T-Mg*sin(\alpha)=M\frac{v_{2}^{2}}{L}[/tex]
α is the angle of T with respect to the horizontal, here α = 25.4°
So, v(2) will be:
[tex]T-Mg*sin(\alpha)=M\frac{v_{2}^{2}}{L}[/tex]
[tex]\frac{L}{M}(T-Mg*sin(\alpha))=v_{2}^{2}[/tex]
[tex]\frac{1.4}{0.76}(5-0.76*9.81*sin(25.4))=v_{2}^{2}[/tex]
[tex]v_{2}=1.82 \: m/s[/tex]
Using this value on equation (2) we will find v.
[tex]v^{2}=v_{2}^{2}+2gh [/tex]
[tex]v^{2}=1.82^{2}+2(9.81)(0.8) [/tex]
[tex]v=4.36\: m/s [/tex]
And finally using equation (1) we can find the initial speed of the bullet.
[tex]m_{b}v_{ib}=Mv[/tex]
[tex]0.01*v_{ib}=0.76*4.36[/tex]
[tex]v_{ib}=331.36\: m/s[/tex]
I hope it helps you!
