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Sagot :
Given data:
* The mass of the bullet is 18.8 gm.
* The initial speed of the bullet is 399 m/s.
* The initial speed of the bullet is 0 m/s.
Solution:
The net momentum of the system before the collision of the bullet with the wooden block is,
[tex]p_i=m_1u_1+m_2u_2_{}[/tex]where m_1 is the mass of bullet, m_2 is the mass of wooden block, u_1 is the initial speed of the bullet, and u_2 is the initila speed of the wooden block,
[tex]\begin{gathered} p_i=18.8\times399+0 \\ p_i=7501.2gms^{-1} \\ p_i=7.5012kgms^{-1} \end{gathered}[/tex]The final momentum of the system is,
[tex]p_f=(m_1+m_2)v[/tex]As bullet and wooden board behaves as a single body,
Substituting the known values,
[tex]\begin{gathered} p_f=(18.8+277)v \\ p_f=295.8vgms^{-1} \\ p_f=0.2958v\text{ } \end{gathered}[/tex]According to the law of conservation of momentum, the momentum of the system before and after remain same.
Thus,
[tex]\begin{gathered} p_i=p_f \\ 7.5012=0.2958v \\ v=\frac{7.5012}{0.2958} \\ v=25.36ms^{-1} \end{gathered}[/tex]Thus, the final velocity of the bullet is 25.36 meter per second.
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