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April 20, 2014

April 20, 2014

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) What is the resulting speed of the block?

(b) What is the speed of the bullet-block center of mass?

(b) What is the speed of the bullet-block center of mass?

- AP PHYSICS MECH. -
**Damon**, Saturday, February 12, 2011 at 3:17pmMomentum 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 2:35pmThank you :)

- AP PHYSICS MECH. -
**Anonymous**, Sunday, December 16, 2012 at 1:04pmI like birds

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