How Do You Calculate the Final Speed of a Bullet-Block System?

[Rijad Hadzic]

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 V_f. Find the final speed of the bullet-block system in terms of the mass of the bullet m_b, the speed of the bullet before the collision, v_b, the mass of the block m_wb and the amount of thermal energy generated during the collision E_t

Homework Equations

K_{final} = (1/2)mv^2
K_{inital} + P_{inital} + W_{total} = K_{final} + P_{final} + E_t

The Attempt at a Solution

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

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

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

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

Now all I have to do is find K_f

Using equation 2 I can rewrite to:

K_f = K_i + W_t - E_t 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 K_f = K_i + W_t - E_t for E_t I get:

E_t = K_i - K_f + W_t
E_t = -\Delta K + W_t if K_f - K_ i = \Delta K that's why I have -\Delta K.

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

2K_f = m_bv_b^2 + 2W_tot -2(-\Delta K + W)

2k_f = m_bv_b^2 + \Delta K

for my final answer of

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

but my books answer is...:

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

does anyone know where I went wrong??

 

[kuruman]

does anyone know where I went wrong??

Energy is not conserved, momentum is.

 

[Rijad Hadzic]

Energy is not conserved, momentum is.

Hmm I don't think I understand.. Do you know where in my logic I went wrong? I was using only the formulas I was given so I don't see where I went wrong :/

 

[kuruman]

OK, if set aside momentum conservation for the time being, what does $W_t$ represent in your equations?

 

[Rijad Hadzic]

OK, if set aside momentum conservation for the time being, what does $W_t$ represent in your equations?

Energy going from one system (K_e of bullet) to the next (Bullet block).Correct??

 

[kuruman]

Consider writing expressions for the energy of your system of bullet + block before the bullet is embedded and after it is embedded. You have
$E_{before} = K_{bullet}$ and $E_{after} = K_{bullet + block} + E_t$
If you start from this and do the algebra correctly, you should get the book's answer. However, I think this problem is over-determined because the answer does not conserve momentum. Have you studied momentum conservation?

 

[Rijad Hadzic]

Consider writing expressions for the energy of your system of bullet + block before the bullet is embedded and after it is embedded. You have
$E_{before} = K_{bullet}$ and $E_{after} = K_{bullet + block} + E_t$
If you start from this and do the algebra correctly, you should get the book's answer. However, I think this problem is over-determined because the answer does not conserve momentum. Have you studied momentum conservation?

Not yet, we are not on momentum conservation. I will do the problem and report back!

 

[Comeback City]

@kuruman is correct. Momentum conservation should be the way to go. This is a perfectly inelastic collision.

 

[gneill]

The OP wants to match the book's solution which specifically involves the use of the energy lost to heat over the collision. Provided that you account for this energy you can write the energy balance equation.

 

[Rijad Hadzic]

The OP wants to match the book's solution which specifically involves the use of the energy lost to heat over the collision. Provided that you account for this energy you can write the energy balance equation.

So I CAN use the equations I listen in the OP? If so do you know where I went wrong in my method?

I have a hunch that:

Solving K_f = K_i + W_t - E_t for E_t I get:

E_t = K_i - K_f + W_t
E_t = -\Delta K + W_t if K_f - K_ i = \Delta K that's why I have -\Delta K.

was the wrong step, but I'm not too sure..

 

[gneill]

I don't know why you're distinguishing two energies W_t and E_t. There's only one value that is the difference between the initial and final kinetic energies, so use one variable for it.

 

[kuruman]

When you conserve energy you have to stay with one system. Here your system is bullet + block. There is no $W_t$
Look at post #6. Set E_before = E_after and proceed.

BTW, I stand corrected. Momentum conservation is not an issue here as gneill noted.

 

[Rijad Hadzic]

I don't know why you're distinguishing two energies W_t and E_t. There's only one value that is the difference between the initial and final kinetic energies, so use one variable for it.

Wow.. I think I understand now. So W_t and E_t are the same thing? Since the definition of work is energy transferred and E_t is the only energy being transferred in the system?

 

[gneill]

Wow.. I think I understand now. So W_t and E_t are the same thing?

Well, it's up to you to define whatever quantities you introduce. Originally you wrote your equation with these two varables but didn't spell out explicitly what they stood for.

You could, for example, account for the heat energy in the block and the bullet and the air and the horizontal surface all as separate quantities. They'd still all have to sum to one value that is the difference between the initial and final kinetic energies.