Posted by joane on Saturday, January 14, 2012 at 7:56am.
While swinging up H = 0.45 m elevation, the block and bullet convert kinetic energy to potential energy, without heat loss. This tells you that
V = sqrt(2gH) = 2.97 m/s
is the velocity of the block and bullet just after impact.
Next, apply conservation of momentum to the inelastic bullet-impact process, assuming that the pendulum does not swing during that brief interval. Let v be the bullet speed, m the bullet mass and M be the block's mass. Momentum conservation tells you that
m v = (M + m) V
Solve for v
CHICKEN NUGGETS = GOLDEN TAR
i like 44
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