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Homework Statement
A steel ball of mass m is fastened to a cord of length L and released when the cord is horizontal. At the bottom of the path, the ball strikes a hard plastic block of mass M = 4m at rest on a frictionless surface. The collision is elastic.
Find the speed of the block immediately after the collision
Answer: (2/5)*[itex]\sqrt{}[/itex](2gL)
Homework Equations
KE + PE = KE + PE; (1/2)mv^{2} + mgh = (1/2)mv'^{2} + mgh'
m_{1}v_{1} + m_{2}v_{2} = m_{1}v'_{1} + m_{2}v'_{2}
The Attempt at a Solution
COE
mgL = (1/2)mv^{2}
v = √(2gL)
COM
mv_{1} = mv'_{1} + 4mv'_{2}
m√(2gL) = m(v'_{1} + 4v'_{2})
√(2gL) = v'_{1} + 4v'_{2}
??
The textbook cites an example that wasn't even relevant. It said that the example was elastic and one object was at rest, but the example was actually the opposite.
It says v'_{1} = v_{1}(m_{1}  m_{2})/(m_{1} + m_{2}) and v'_{2} = 2m_{1}/(m_{1} + m_{2}). How can I prove this?
Also, in the diagram, it says m becomes 2m...?
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