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Question

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As show above, a 125. gram bullet traveling at 300. m/s strike and sticks into a stationary 5.00 kg block of wood. The combined bullet + wood initially slides on a horizontal frictionless surface until they encounter a second horizontal surface with a coefficient of kinetic friction between the plane and the bottom of the block of 0.250. How far along the second surface will the block and bullet slide before coming to rest?

Explanation / Answer

The mass of the bulet m = 125g = 0.125kg

 

the mass of the block M  = 5kg

 

the speed of the bullet vi = 300m/s

 

from law of conservation of momentum

 

    mvi = (m + M)vf

 

then vf = mvi / (m + M)

 

           = (0.125)(300)/(0.125 + 5)

 

           = 7.32 m/s

 

Now from work energy theorem

 

    W = K

 

    (m + M)g *x = 1/2(m + M) v^2

 

therefore the distance traveled

 

          x  = v^2 / g = (7.32)^2 / (0.25)(9.8) = 21.87 m

 

 

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