Answer :
Answer:
The initial speed of the bullet is [tex]v_{o} = 889.199\,\frac{m}{s}[/tex].
Explanation:
The collision between bullet and block is inelastic and let suppose that motion occurs on a horizontal surface, so that changes in gravitational potential energy can be neglected. Initially, the intial speed of the bullet-block system can be determined with the help of the Work-Energy Theorem and the Principle of Energy Conservation:
[tex]K = U_{k} + W_{loss}[/tex]
[tex]\frac{1}{2}\cdot (5.224\,kg)\cdot v^{2} = \frac{1}{2}\cdot \left(450\,\frac{N}{m}\right)\cdot (0.22\,m)^{2} + (0.35)\cdot (5.224\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right) \cdot (0.22\,m)[/tex]
The initial speed of the bullet-block system is:
[tex]v \approx 2.383\,\frac{m}{s}[/tex]
Now, the initial speed of the bullet is determined by applying the Principle of Momentum Conservation:
[tex](0.014\,kg)\cdot v_{o} = (5.224\,kg)\cdot \left(2.383\,\frac{m}{s} \right)[/tex]
[tex]v_{o} = 889.199\,\frac{m}{s}[/tex]