A 55-g bullet traveling at 250 m/s penetrates a block of ice at $0 ^\circ \textrm{C}$ and comes to rest within the ice. Assuming that the temperature of the bullet doesn’t change appreciably, how much ice is melted as a result of the collision?

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. The heat energy that's gained by the ice cube is going to equal the kinetic energy lost by the bullet. And the kinetic energy is 1/2 times mass of the bullet times its initial speed squared. And we're only concerned with the magnitude of the energy. So, never mind, you know, final minus initial will give us a negative here but I didn't bother writing it, we're just taking the absolute value of its change kinetic energy. This is how much energy turns into heat. And this is all going into the phase change of the ice and so this heat equation is going to be mass of the ice cube times the latent heat of fusion for ice. And well this isn't the mass of the ice cube, it's the mass of ice which ends up melting. And then divide both sides by latent heat of fusion here. And so the amount of ice that will melt is going to be 0.055 kilograms, mass of the bullet, times its initial speed of 250 meters per second squared, divided by 2 times 333 times 10 to the 3 joules per kilogram, latent heat of fusion for water. And that's about 5.2 grams of ice will end up melting.