Answer:
[tex]Vf_{2}[/tex] = 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 = [tex]\frac{mv^{2} }{2}[/tex]
2.3261 = [tex]\frac{1.72v^{2} }{2}[/tex]
v = 1.6446 m/s
In the collision is conserved the lineal momentum of the system then:
[tex]m_{1}Vo_{1} + m_{2}Vo_{2} = m_{1}Vf_{1} + m_{2}Vf_{2}[/tex]
(1.72)(0) + (0.0085)(431) = (1.72)(1.6446) + (0.085)[tex]Vf_{2}[/tex]
3.6635 = 2.8287 + 0.0085[tex]Vf_{2}[/tex]
[tex]Vf_{2}[/tex] = 98.21 m/s
