AP PHYSICS MECH.
posted by vm on .
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?

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 
Thank you :)

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