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

`Vf_(2)` = 98.21 m/s

Step-by-step 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 =
`(mv^(2) )/(2)`

2.3261 =
`(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