Posted by vm on Saturday, February 12, 2011 at 2:45pm.
A 5.00 g bullet moving at 648 m/s strikes a 670 g wooden block at rest on a frictionless surface. The bullet emerges, traveling in the same direction with its speed reduced to 440 m/s.
A 5.00 g bullet moving at 648 m/s strikes a 670 g wooden block at rest on a frictionless surface. The bullet emerges, traveling in the same direction with its speed reduced to 440 m/s.
(a) What is the resulting speed of the block?
(b) What is the speed of the bulletblock center of mass?
(b) What is the speed of the bulletblock center of mass?

AP PHYSICS MECH.  Damon, Saturday, February 12, 2011 at 3:17pm
Momentum before = 648 * .005
equals
Momentum after = 440*.005 + v * .670
solve for v
v = 1.55 m/s
Speed of center of mass does not change because total mass does not change
b)before : average speed = (648*.005 + 0*.670)/.675 = 4.8 m/s
c) after: average speed = (440*.005 + 1.55*.670)/.675 = 4.8 m/s sure enough 
AP PHYSICS MECH.  Anonymous, Sunday, February 13, 2011 at 2:35pm
Thank you :)

AP PHYSICS MECH.  Anonymous, Sunday, December 16, 2012 at 1:04pm
I like birds