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synergix
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Homework Statement
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.
Homework Equations
conservation of momentum and kinetic energy
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.