How much ice melts when a projectile is fired at it?

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A projectile weighing 0.025 kg is fired at 240 m/s into ice at 0 degrees C, and the kinetic energy is calculated as 720 J. The heat of fusion for ice is noted as 3.35x10^5 J/kg. The mass of ice that melts is determined by dividing the kinetic energy by the heat of fusion, resulting in approximately 0.002 kg of melted ice. The discussion confirms that the calculations are correct, emphasizing the omission of rotational kinetic energy. The heat lost by the projectile while cooling from 30 to 0 degrees C is ignored as per the problem's instructions.
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Homework Statement


Newt needs to melt some ice, so he fires a test projectile (0.025 kg) at 30.0 degrees C at a speed of 240 m/s into a large block of ice at 0 degrees C, in which it becomes embedded (comes to rest). Approximately what mass of ice melts? (i.e. ignore the heat lost by the projectile in going from 30-0 degrees C).


Homework Equations


e=(mv^2)/2


The Attempt at a Solution


The heat of fusion of water is 3.35x10^5J.
e=(0.025 * 240^2)/2=720J
720J/3.35x10^5J=0.002, so 0.002kg of ice melts.
I get the feeling that's incorrect, especially since I didn't do anything with the temperature (although it mentions that I should ignore the heat lost by the projectile).

Thanks for any help you can provide.
 
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Energy release of the projectile is its kinetic energy. So assuming the ice is at 0 C, the amount that melts equals the projectile KE divided by the latent heat of the ice. The rotational kinetic energy of the projectile is omitted. This is what you have done and it is correct. I assume the heat of fusion of ice is 3.35X10^5 J/kg. You left off the kg.
 
LawrenceC said:
Energy release of the projectile is its kinetic energy. So assuming the ice is at 0 C, the amount that melts equals the projectile KE divided by the latent heat of the ice. The rotational kinetic energy of the projectile is omitted. This is what you have done and it is correct. I assume the heat of fusion of ice is 3.35X10^5 J/kg. You left off the kg.

Thanks a million!
 
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