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An 8.0 g bullet is fired from a gun at 120 m/s into a 2.0 kg block of wood sitting on a fence post. If the bullet hits the wood and remains in it, what is the speed of the block after the collision?

Respuesta :

Answer:

[tex]v = 0.478 m/s[/tex]

Explanation:

As we know that bullet strike the block and fix in it

so here if bullet and box is taken as a system then there is no external force on it

so here the initial momentum of bullet and block must be equal to final momentum of bullet and block

so here we have

[tex]m_1v_{1i} + m_2v_{2i} = m_1v + m_2v[/tex]

now we have

[tex]m_1 = 8 g[/tex]

[tex]v_1 = 120 m/s[/tex]

[tex]m_2 = 2kg[/tex]

[tex]v_{2i} = 0[/tex]

now from above equation we have

[tex](0.008)(120) + 2(0) = (0.008)v + 2v[/tex]

[tex]v = 0.478 m/s[/tex]

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