Question

A 1.72-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?

Answer 1

Answer:

$Vf_{2}$ = 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 = $\frac{mv^{2} }{2}$

     2.3261 = $\frac{1.72v^{2} }{2}$

            v = 1.6446 m/s

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

$m_{1}Vo_{1} + m_{2}Vo_{2} = m_{1}Vf_{1} + m_{2}Vf_{2}$

(1.72)(0) + (0.0085)(431) = (1.72)(1.6446) + (0.085)$Vf_{2}$

3.6635 = 2.8287 + 0.0085$Vf_{2}$

$Vf_{2}$ = 98.21 m/s