Collision of boulder and ice-fishing shack

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The discussion centers on a physics problem involving a 55.6 kg boulder rolling down a mountain and colliding with a 204 kg ice-fishing shack. The boulder, located 14.6 meters above the lake, converts potential energy into kinetic energy upon impact. The equation m1v1 + m2v2 = v' (m1 + m2) is used to find the velocity after the collision, while the coefficient of friction (0.392) is essential for calculating the distance the shack and boulder slide across the ice. The correct approach involves using energy conservation and friction to determine the sliding distance.

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A 55.6 kg boulder sat on the side of mountain beside a lake. The boulder was 14.6m above the surface of lake. One winter night, the boulder rolled down the mountain, directly into a 204 kg ice-fishing shack that was on a frozen lake. What was the velovity of the boulder and the shack at the instant that they began to slide across the ice? If the coefficient of friction between shack and rough ice is 0.392, how far did the shack and boulder slide?
:rolleyes:

Ok I figured that
m1v1 + m2v2= v' (m1 + m2)

then mgh = 0.5mv^2
(55.6*9.8*14.6) = 0.5*(204+55.6)v^2

I don't know why they are giving us friction for.

Someone please help soon. I have test tommorow!
 
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You need the friction to calc out the distance traveled.
 
civil_dude said:
You need the friction to calc out the distance traveled.

I don't know how? Do we use d=E/f
 

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