Kinetic enengy needed to break a board

[Howlin]

Homework Statement

I need to get [itex]E_{b}[/itex] = .5(m*n*[itex]V^{2}_{i}[/itex])/(m+n) [eq 6]

where [itex]E_{b}[/itex] is the energy required to break the board, m is the mass of the object hitting the boards, n is the mass of the board and [itex]V_{i}[/itex] is the velocity of the object before hitting the board.

Homework Equations

[itex]K_{i}[/itex]=.5*m*[itex]V^{2}_{i}[/itex] [eq 1)
where [itex]K_{i}[/itex] is the initial kinetic energy, m is the mass of the object, [itex]V_{i}[/itex] is the inital velocity

[itex]K_{f}[/itex]= [itex]E_{b}[/itex] + .5*(m+ n)*[itex]V^{2}_{f}[/itex] [Eq2]
[itex]E_{b}[/itex] is the energy required to break the board, m is the mass of the object hitting the boards, n is the mass of the board and [itex]V_{i}[/itex] is the velocity of the object before hitting the board.

momentum before = momentum after
m8[itex]V_{i}[/itex] = (m+n)*[itex]V_{f}[/itex]

[itex]V_{f}[/itex]=(m*[itex]V_{i}[/itex])/(m+n) [eq3]

The Attempt at a Solution

I know make eq 1 and 2 equal (assuming no energy is lost) rearrange the equation to get E on its own

[itex]E_{b}[/itex] = .5*m*[itex]V^{2}_{i}[/itex] - .5*(m+ n)*[itex]V^{2}_{f}[/itex] [eq4]

Now i can sub in eq3 into eq 4

[itex]E_{b}[/itex] = .5*m*[itex]V^{2}_{i}[/itex] - .5*(m+ n)*[(m*[itex]V_{i}[/itex])/(m+n)[itex])^{2}[/itex]

[itex]E_{b}[/itex] = .5*m*[itex]V^{2}_{i}[/itex] - .5*[(m*[itex]V_{i}[/itex][itex])^{2}[/itex]/(m+n)]

This is where I am stuck, I cannot get to equation 6.
Can anyone help me?

 

[mfb]

##E_{b} = \frac{1}{2}mV^{2}_{i} - \frac{1}{2}\frac{(mV_{i})^{2}}{(m+n)}## - this is your last equation

##E_{b} = \frac{1}{2}mV^{2}_{i}\frac{m+n}{m+n} - \frac{1}{2}\frac{mmV_{i}^{2}}{(m+n)}##
##E_{b} = \frac{1}{2}mV^{2}_{i}\frac{m+n-m}{m+n}##
And the next step is the final result.
The idea was just to combine both terms.