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

In summary: If you solve for E_t using the equationE_t = K_i - K_f + W_t E_t = -\Delta K + W_t you get the same result as if you used the equationE_t = K_i - K_f - W_t E_t = -\Delta K - W_t
  • #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??
 
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  • #2
Rijad Hadzic said:
does anyone know where I went wrong??
Energy is not conserved, momentum is.
 
  • #3
kuruman said:
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 :/
 
  • #4
OK, if set aside momentum conservation for the time being, what does ##W_t## represent in your equations?
 
  • #5
kuruman said:
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??
 
  • #6
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?
 
  • #7
kuruman said:
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!
 
  • #8
@kuruman is correct. Momentum conservation should be the way to go. This is a perfectly inelastic collision.
 
  • #9
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.
 
  • #10
gneill said:
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 [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].

was the wrong step, but I'm not too sure..
 
  • #11
I don't know why you're distinguishing two energies Wt and Et. There's only one value that is the difference between the initial and final kinetic energies, so use one variable for it.
 
  • #12
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 Ebefore = Eafter and proceed.

BTW, I stand corrected. Momentum conservation is not an issue here as gneill noted.
 
  • #13
gneill said:
I don't know why you're distinguishing two energies Wt and Et. 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?
 
  • #14
Rijad Hadzic said:
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.
 

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

1. How does the speed of a bullet affect its trajectory when flying horizontally?

The speed of a bullet does not significantly affect its trajectory when flying horizontally. This is because horizontal motion is not affected by gravity and is therefore constant regardless of the bullet's speed.

2. Can a bullet fired horizontally still cause damage?

Yes, a bullet fired horizontally can still cause damage depending on its size, velocity, and the type of material it hits. However, the damage may be less severe compared to a bullet fired at an angle.

3. What factors can affect the accuracy of a bullet fired horizontally?

The accuracy of a bullet fired horizontally can be affected by factors such as wind speed and direction, air density, and the bullet's shape and weight.

4. Is it possible for a bullet fired horizontally to change its direction?

No, once a bullet is fired horizontally, its direction will remain constant due to the absence of external forces acting on it. However, factors such as wind or hitting an object can cause it to deviate from its original path.

5. How far can a bullet fly horizontally before losing its velocity?

The distance a bullet can fly horizontally before losing its velocity depends on its initial speed and the air resistance it encounters. In a vacuum, a bullet can theoretically travel forever. However, in reality, a bullet will eventually slow down and fall to the ground due to air resistance and gravity.

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