Answer:
[tex]v_{0}=554.19 m/s[/tex]
Explanation:
We can use the conservation of momentum and energy.
Conservation of momentum
[tex]m_{1}v_{0}=(m_{1}+m_{2})V[/tex] (1)
Where:
- m(1) is the mass of the bullet
- m(2) is the mass of the block
- v(0) is the initial speed of the bullet
- V(f) is the final speed of the system
Conservation of energy
We have kinetic energy at the beginning and we just have potential energy at the end because the block plus the bullet swing trough a maximum angle, it means they are at rest at that point.
[tex]\frac{1}{2}(m_{1}+m_{2})V^{2}=(m_{1}+m_{2})gh[/tex] (2)
The height h will be:
[tex]h=L-Lcos(60)=L(1-cos(60))[/tex]
h is the difference height between the initial and final positions.
So, using the equation (2) we can find V.
[tex]V^{2}=2gh[/tex]
[tex]V=\sqrt{2gL(1-cos(60))}[/tex]
[tex]V=\sqrt{2*9.81*2.2*(1-cos(60))}[/tex]
[tex]V=4.65 m/s[/tex]
Finally, putting this value in the equation (1), the v(1i) will be:
[tex]v_{0}=V\frac{(m_{1}+m_{2})}{m_{1}}[/tex]
[tex]v_{0}=4.65\frac{(0.022+2.6)}{0.022}[/tex]
[tex]v_{0}=554.19 m/s[/tex]
I hope it helps you!
