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April 24, 2014

April 24, 2014

Posted by **joane** on Saturday, January 14, 2012 at 7:56am.

- physics -
**drwls**, Saturday, January 14, 2012 at 8:22amWhile 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

- physics -
**DAN**, Thursday, September 27, 2012 at 1:09pmCHICKEN NUGGETS = GOLDEN TAR

- physics -
**chikn**, Wednesday, May 1, 2013 at 12:54pmi like 44

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