Billy the kid and the kinetic energy

Hello there!
Once again, I really need some help with a physics problem I have not been able to figure out. Unfortunately, I do not even have a clue how to solve this one and have spent quite a lot time on it. Here is the question:

Billy the Kid is throwing an apple with mass 200 grams vertical in to the air. He is shooting with his gun at it and the bullett is hitting the apple at the apple's highest point of its throw. The bullet has a mass of 10 grams and its velocity right before the impact is 300 m/s.
Due to the full penetration, the bullet's kinetic energy decreases to only 1/4 of its original value.
How great is the energy of deformation during the penetration ?

Apparently, my trys ended unsuccessfully and I never got the result of 331,875 J

I am not expecting a whole solution, a hint would be great! I really want to figure that out and would be very happy about every help I can get!
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 Quote by fara0815 Hello there! Once again, I really need some help with a physics problem I have not been able to figure out. Unfortunately, I do not even have a clue how to solve this one and have spent quite a lot time on it. Here is the question: Billy the Kid is throwing an apple with mass 200 grams vertical in to the air. He is shooting with his gun at it and the bullett is hitting the apple at the apple's highest point of its throw. The bullet has a mass of 10 grams and its velocity right before the impact is 300 m/s. Due to the full penetration, the bullet's kinetic energy decreases to only 1/4 of its original value. How great is the energy of deformation during the penetration ? Apparently, my trys ended unsuccessfully and I never got the result of 331,875 J I am not expecting a whole solution, a hint would be great! I really want to figure that out and would be very happy about every help I can get!
I think you are supposed to assume that the bullet does not remove any of the apple mass - it just deforms it (not realistic).

What is the energy of the system (bullet and apple) before and after the impact? The rest of the energy is lost in the process of penetration.

Hint: you have to use conservation of momentum

AM
 Thank you for your fast reply! I have tried conversation of momentum and conversation of energy but it does not make any sense to me. There is a hint that belongs to this problem: The decrement of the bullet's kinetic energy is not the energy of deformation. Here is what I tried: $$\Delta E_{kin} = \frac{1}{2}mv^2_{before} - \frac{1}{2}mv_{after}^2$$ this one did not bring the answer either: $$m_b * V_b + m_a * V_a = m_b * V_b + m_a * V_a$$ I assume that the apple's velocity is zero since it is at its highest point of the throw.

Billy the kid and the kinetic energy

The bullet has a certain amount of Kinetic Energy before it hits the apple. And afterwards it has a quarter of this, so three quarters were expended in the collision.

That doesn't work out as 331,875 Joules, or anywhere near. Note that 10g is is 0.01Kg

If the bullet retained a quarter of its Kinetic Energy, and the latter is related to velocity squared, it means the bullet is now travelling at half its former velocity. And it didn't have the energy to propel a 200g apple up to 150ms. Its ounds like the bullet goes straight straight through, which suggests that the mass of the apple is a spurious number.

Hmmn. Apologies if I've missed something obvious, but I don't think I can help.

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 Quote by fara0815 I have tried conversation of momentum and conversation of energy but it does not make any sense to me.
As AM suggested, apply conservation of momentum.
 There is a hint that belongs to this problem: The decrement of the bullet's kinetic energy is not the energy of deformation.
Right: Some of the bullet's KE becomes the apple's KE.

 Here is what I tried: $$\Delta E_{kin} = \frac{1}{2}mv^2_{before} - \frac{1}{2}mv_{after}^2$$
This will certainly be useful to find the final speed of the bullet.

 this one did not bring the answer either: $$m_b * V_b + m_a * V_a = m_b * V_b + m_a * V_a$$
You'd better distinguish initial and final velocities!

 I assume that the apple's velocity is zero since it is at its highest point of the throw.
Right, the apple's initial velocity is zero.
 mh, if the bullet's final velocity is 150 m/s ( $$V_{b final}$$ ) and I use this in the conversation of momentum $$m_bV_{b intial}= m_{apple}V_{apple} + m_bV_{b final}$$ I would get the apple's velocity with $$V_{apple}= \frac{m_bV_{b intial}-m_bV_{b final}}{m_{apple}}= 7.5 m/s$$ Is that right? If so, how can I get to the energy of deformation? It is not the apple's kinetic energy, which is $$E_{kin apple}= \frac{1}{2}m_{apple}V_{apple}^2$$, is it? If so, the KE would be 5,625 J and that does not bring me to the answer :(

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 Quote by fara0815 if the bullet's final velocity is 150 m/s ( $$V_{b final}$$ ) and I use this in the conversation of momentum $$m_bV_{b intial}= m_{apple}V_{apple} + m_bV_{b final}$$ I would get the apple's velocity with $$V_{apple}= \frac{m_bV_{b intial}-m_bV_{b final}}{m_{apple}}= 7.5 m/s$$ Is that right?
Looks good.

 If so, how can I get to the energy of deformation? It is not the apple's kinetic energy, which is $$E_{kin apple}= \frac{1}{2}m_{apple}V_{apple}^2$$, is it?
Nope.

The bullet starts out with a given amount of KE. After blasting through that apple, both apple and bullet have some final total KE. (Figure that out.) That final KE will be less than the original; the difference is the deformation energy.
 Great! I got it! It is $$E_{kin.bullet.initial}=E_{kin.apple}+E_{kin.bullet.final}+E_{kin.deform ation}$$ With that I get $$E_{kin.deformation}=E_{kin.bullet.initial}-E_{kin.bullet.final}-E_{kin-apple}$$ Thank you very much for leading me there, I really appreciate it !