# Kinetic enengy needed to break a board

1. Jan 29, 2013

### Howlin

1. The problem statement, all variables and given/known data

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.

2. Relevant 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
m8$V_{i}$ = (m+n)*$V_{f}$

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

3. 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?

2. Jan 29, 2013

### Staff: Mentor

$E_{b} = \frac{1}{2}mV^{2}_{i} - \frac{1}{2}\frac{(mV_{i})^{2}}{(m+n)}$ - 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.