Kinetic enengy needed to break a board

[Howlin]

Homework Statement

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

where E_{b} 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 V_{i} is the velocity of the object before hitting the board.

Homework Equations

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

K_{f}= E_{b} + .5*(m+ n)*V^{2}_{f} [Eq2]
E_{b} 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 V_{i} is the velocity of the object before hitting the board.

momentum before = momentum after
m8V_{i} = (m+n)*V_{f}

V_{f}=(m*V_{i})/(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

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

Now i can sub in eq3 into eq 4

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

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