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A battleship that is $6.00 \times 10^{7} \mathrm{kg}$ and is originally at rest fires a $1100-\mathrm{kg}$ artillery shell horizontally with a velocity of 575 $\mathrm{m} / \mathrm{s} .$ (a) If the shell is fired straight aft (toward the rear of the ship), there will be negligible friction opposing the ship's recoil. Calculate its recoil velocity. (b) Calculate the increase in internal kinetic energy (that is, for the ship and the shell). This energy is less than the energy released by the gun powder-significant heat transfer occurs.

(a) $v _ { \mathrm { bf } } = - 0.01 \mathrm { ms } ^ { - 1 }$(b) $\Delta K E = 1.50 \mathrm { kJ }$

Physics 101 Mechanics

Chapter 8

Linear Momentum and Collisions

Moment, Impulse, and Collisions

Cornell University

Rutgers, The State University of New Jersey

Simon Fraser University

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the first months has given as six times 10 to the seven kilograms. The second mouse has given us 1100 kilograms with a velocity initial off 575 meters per second. And we want to know the recoil velocity. So initially they're both at rest this ship and the shell inside of it. And then afterwards the ship is going to recoil. Wealthy show is shot forward. So our initial woman to zero equals my mo mentum off the ship. Plus a moment, um, of the show. So solving for the initial velocity or the final velocity of the ship, we get a recoil speed of about negative 0.1 meters per seconds. What is very small for part B, whereas Thio calculate the increase in internal kinetic energy. So initially we have zero kinetic energy and that the end, we're gonna compare it to the comedic energy off the ship and the commitment energy of the show together. So we have six times 10 to 7. You're the one who dropped that negative since it is being squared and we get a Toto energy of 1.8 times 10 to the eighth jewels of increase energy

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