**1. The problem statement, all variables and given/known data**

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)

**2. Relevant 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}

**3. 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…?

http://ift.tt/1jUH8w1