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posted by  Keppel on 7/24/2008 12:31:54 PM  |  status: Live  

Please help

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Question Details:
Mitch throws a lump of clay that is 100 g at a 500 g target that is at rest.  After impact the clay and target stick together and s;ide 2.1 meters before stopping.  If the μk =0.5, what is the initial speed of the clay?
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posted by zsm28 on 7/24/2008 2:47:50 PM  |  status: Live
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Response Details:
m = 100 g = 0.100 kg, M = 500 g = 0.500 kg, μ = 0.5, d = 2.1 m
the initial speed of the clay = v
the speed of the clay and target = v'
momentum conservation:
mv = (m + M)v'
so v' = mv/(m + M)
kinetic energy (m + M)v'2/2 changes to 0 due to work done to overcome the friction μ(m + M)g*d
(m + M)v'2/2 = μ(m + M)g*d
v' = √(2μgd)
∴mv/(m + M) = √(2μgd)
v = (m + M)√(2μgd)/m = 27.2 m/s
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