A1.72-kg block of soft wood is suspended by two strings from the ceiling. the wood is free to rotate in pendulum-like fashion when a force is exerted upon it. a 8.50-g bullet is fired into the wood. the bullet enters the wood at 431 m/s and exits the opposite side shortly thereafter. if the wood rises to a height of 13.8 cm, then what is the exit speed of the bullet?

= 98.21 m/s                    

Explanation:

The block has a final potential energy

U = mgh

U = (1.72)(9.8)(0.138)

    = 2.3261 J

How after of the collision the block's mechanical energy is conserved, then we can calculate the block's initial velocity

     2.3261 =

     2.3261 =

            v = 1.6446 m/s

In the collision is conserved the lineal momentum of the system then:

(1.72)(0) + (0.0085)(431) = (1.72)(1.6446) + (0.085)

3.6635 = 2.8287 + 0.0085

= 98.21 m/s                    

f is prop to m1xm2 double m1 double force