1. The problem statement, all variables and given/known data A 20.0 g bullet moving horizontally at 70.0 m/s strikes a 9.00 kg block of wood lying at rest on the pavement. The bullet embeds itself in the block of wood during the collision. Find a. the speed of the block of wood immediately after the collision, b. the frictional force between the pavement and the block of wood if the block of wood moves 2.50 m before stopping. 2. Relevant equations conservation of momentum and kinetic energy 3. The attempt at a solution Using conservation of momentun: A)MbVb+MwVw=(Mb+Mw)V MbVb+0=(Mb+Mw)V (.0200kg)70m/s=(9.00kg+.0200kg)V V=[(.0200kg)(70m/s)]/(9.02kg) V=.1552106....m/s Using conservation of Kinetic energy: .5MbVb^2+0=.5(Mb+Mw)V^2 (.0100kg)(70m/s)^2=4.51kgV^2 V^2=[.0100kg(4900m/s)]/4.51kg V^2=10.86474501.... V=3.29617...m/s Should'nt these values be the same have I done something wrong? I used both to check that my answers were correct and then I got this result. the second part should be easy once we figure ths out.