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A 28.5 g bullet is fired horizontally into a 1.13 kg wooden block resting on a horizontal surface with coefficient of friction 0.227. The bullet goes through the block and comes out with a speed of 394 m/s. If the block travels 3.42 m before coming to rest, what was the initial speed of the bullet? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s

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Answer:

u = 548.667 m/s

Explanation:

First of all we correlate the data we have, as well

[tex]m_b=28.5g \\m_w=1.13Kg\\\mu_f=0.227\\V_f=394m/s\\d=3.42m\\g=9.8m/s^2\\[/tex]

Thus we proceed to the following,

[tex]V = \sqrt{2 a d}[/tex]

[tex]V= \sqrt{2*\mu g*d}[/tex]

[tex]V= \sqrt{2*0.227*9.8*3.42} = 3.9009 m/s[/tex]

Momentum conservation,

[tex]m_b u = m_w V + m_b V_f[/tex]

[tex]0.0285*u = 1.13*3.9009+0.0285*394[/tex]

[tex]u = 548.667 m/s[/tex]

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