Solution_Problem_of_the_Week_–_10

Solution_Problem_of_the_Week_–_10 - ...

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Unformatted text preview: Solution Problem of the Week – 10 A block of steel with mass 2.4 kg sits on a low-friction surface. A bullet of mass 0.04 kg traveling horizontally bounces off the block as shown in the diagram, looking down from above. We take the plane of the motion to be the xy plane (with z pointing outward, toward the viewer). Y X The incident velocity vector of the bullet is <280, 120, 0> m/s and the final velocity vector of the bullet is <-200, 65, 0> m/s. (a) What is the vector velocity of the block of steel just after the collision? Start from a fundamental principle, and show all your work. System: Bullet, Block. Surroundings: Earth, low-friction surface. Fundamental Principle: Momentum Principle. Forces acting on the block in the z-direction are equal and opposite (gravity, normal force of ice on block). We will make the assumption that the force due to the Earth’s gravity acting on the bullet can be ignored in the process described above. In addition, we will assume that the duration of the scattering event is so short that heat transfer between the system and surroundings can be ignored. Note also that the speeds are all much less than the speed of light, so we can use the nonrelativistic relationships for energy and momentum. The Momentum Principle then becomes: ! ! !psystem = Fnet !t and because there are no forces in the surroundings acting in x or y direction, !psystem , x = 0, !psystem , y = 0 . psystem , x = pbullet , x + pblock , x , psystem , y = pbullet , y + pblock , y . Evaluating the above for the initial and final states gives: 0.04 280,120, 0 kg ! m / s + 2.4 0, 0, 0 kg ! m / s = 0.04 "200, 65, 0 kg ! m / s + 2.4 vblock , x , vblock , y , 0 kg ! m / s vblock , x = 8 m / s, vblock , y = 0.92 m / s . (b) What is the increase ΔEthermal in the thermal energy of the block and bullet? Start from a fundamental principle, and show all your work. Energy Principle. Because no work is done on the system by the surroundings, the energy of the system is the same in its initial and final state. !Esystem = !Ek + !Ethermal + Q = Wext = 0 . In the initial state, only the bullet has kinetic energy. In the final state, both the bullet and block have kinetic energy. Q = !( EK f ! EKi ) = !(9.6 " 10 2 ! 1.86 " 10 3 ) J = 900 J ...
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This note was uploaded on 12/07/2011 for the course PHYS 172 taught by Professor ? during the Fall '08 term at Purdue.

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