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A 16 g bullet strikes and becomes embedded in a 1.10 kg block of wood placed on a horizontal surface just in front of the gun. If the coefficient of kinetic friction between the block and the surface is 0.25, and the impact drives the block a distance of 9.3 m before it comes to rest, what was the muzzle speed of the bullet?
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Added Wed, 08 Jul '15
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1-Speed of the block and the bullet after the impact
(the difference of the cinetical energie)

1/2 * (M + m)* V^2 = (M + m)* g * u * d
==> 0.5(1.10+0.15)*V^2=(1.10+0.15)* 9.8*0.25*9.5
V = Sqrt(2*g*u*d)
V = Sqrt(2 * 9.8 * 0.25 * 9.5)
V = 6.83 m/s

2- Conservation of the impulse
m v = (M + m)V
==> v = (M +m) * V/m
v = 1.25 * 6.83 / 0.15
v = 57 m/s

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Added Wed, 08 Jul '15
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