- #1

Rijad Hadzic

- 321

- 20

## Homework Statement

A bullet flying horizontally hits a wooden block that is initially at rest on a frictionless, horizontal surface. The bullet gets stuck in the block, and the bullet-block system has a final speed of [itex] V_f[/itex]. Find the final speed of the bullet-block system in terms of the mass of the bullet [itex] m_b [/itex], the speed of the bullet before the collision, [itex] v_b[/itex], the mass of the block [itex] m_wb [/itex] and the amount of thermal energy generated during the collision [itex] E_t[/itex]

## Homework Equations

[itex] K_{final} = (1/2)mv^2 [/itex]

[itex] K_{inital} + P_{inital} + W_{total} = K_{final} + P_{final} + E_t [/itex]

## The Attempt at a Solution

So it is asking for final speed so I use my first equation and rewrite it:

[itex] v = \sqrt {(2K_f)/(m)} [/itex]

in this case, m is going to equal the mass of the block + the bullet, so I write:

[itex] v = \sqrt {(2K_f)/(m_{wb} + m_b ) } [/itex]

Now all I have to do is find [itex] K_f [/itex]

Using equation 2 I can rewrite to:

[itex] K_f = K_i + W_t - E_t [/itex] I drop the potential energies because there is no potential energy in the system

K_i = initial kinetic energy, right? So (1/2) (mass of bullet)(velocity of bullet)^2 = (1/2)(m_b)(v_b)^2

I just left W_t as W_t and looked to E_t.

Solving [itex] K_f = K_i + W_t - E_t [/itex] for E_t I get:

[itex]E_t = K_i - K_f + W_t [/itex]

[itex]E_t = -\Delta K + W_t [/itex] if [itex] K_f - K_ i = \Delta K [/itex] that's why I have [itex] -\Delta K [/itex].

Now I plug in E_t and it cancels a few things...

[itex] 2K_f = m_bv_b^2 + 2W_tot -2(-\Delta K + W) [/itex]

[itex] 2k_f = m_bv_b^2 + \Delta K [/itex]

for my final answer of

[itex] v = \sqrt {(m_bv_b^2 + 2\Delta K) / (m_b + m_{wb})} [/itex]

but my books answer is...:

[itex] v = \sqrt {(m_bv_b^2 - 2\Delta E_t ) / (m_b + m_{wb})} [/itex]

does anyone know where I went wrong??