Physics Solution Manual for 1100 and 2101

Thus the work done by nonconservative forces is zero

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Unformatted text preview: mOd sin 60.0o + mO d sin 60.0o + mS ( 0 ) = = 0 nm mO + mO + mS This result makes sense, because the sulfur atom’s x coordinate is x3 = 0 nm, and the oxygen atoms, which have equal masses, have x coordinates of equal magnitude but opposite signs: x1 = −x2. b. The y coordinate of the sulfur atom is zero (y3 = 0 nm), but the y coordinates of the two oxygen atoms are both positive, so the center of mass of the sulfur dioxide molecule has a positive y coordinate: ycm = = + mO d cos 60.0o + mO d cos 60.0o + mS ( 0 nm ) mO + mO + mS +2 mO d cos 60.0o 4 mO = = +2mO d cos 60.0o 2mO + 2mO o + d cos 60.0o + ( 0.143 nm ) cos 60.0 = = +0.0358 nm 2 2 53. SSM WWW REASONING The system consists of the lumberjack and the log. For this system, the sum of the external forces is zero. This is because the weight of the system is balanced by the corresponding normal force (provided by the buoyant force of the water) 382 IMPULSE AND MOMENTUM and the water is assumed to be frictionless. The lumberjack and the log, then, constitute an isolated system, and the principle of conservation of linear momentum holds. SOLUTION a. The total linear momentum of the system before the lumberjack begins to move is zero, since all parts of the system are at rest. Momentum conservation requires that the total momentum remains zero during the motion of the lumberjack. m1vf1 + m2vf2 = 1 24 40 3 14 244 4 3 Total momentum just before the jump Initial momentum Here the subscripts "1" and "2" refer to the first log and lumberjack, respectively. Let the direction of motion of the lumberjack be the positive direction. Then, solving for v1f gives v f1 = – m2 v f2 m1 =– (98 kg)(+3.6 m / s) = –1.5 m / s 230 kg The minus sign indicates that the first log recoils as the lumberjack jumps off. b. Now the system is composed of the lumberjack, just before he lands on the second log, and the second log. Gravity acts on the system, but for the short time under consideration while the lumberjack lands, the effects of gravity in changing the linear momentum of the system are negligible. Therefore, to a very good approximation, we can say that the linear momentum of the system is very nearly conserved. In this case, the initial momentum is not zero as it was in part (a); rather the initial momentum of the system is the momentum of the lumberjack just before he lands on the second log. Therefore, m v f1 + m2 v f2 = m1 v 01 + m2 v 02 11 4 44 4 2 3 14 244 4 3 Total momentum just after lumberjack lands Initial momentum In this expression, the subscripts "1" and "2" now represent the second log and lumberjack, respectively. Since the second log is initially at rest, v 01 = 0 . Furthermore, since the lumberjack and the second log move with a common velocity, v f1 = v f2 = v f . The statement of momentum conservation then becomes m1 v f + m2 v f = m2 v 02 Solving for vf, we have vf = m2 v 02 m1 + m2 = (98 kg)(+3.6 m / s) = + 1.1 m / s 230 kg + 98 kg The positive sign indicates that the system moves in the same direction as the original direction of the lumberjack's motion. Chapter 7 Problems 383 54. REASONING Since all of the collisions are elastic, the total mechanical energy of the ball is conserved. However, since gravity affects its vertical motion, its linear momentum is not conserved. If hf is the maximum height of the ball on its final bounce, conservation of energy gives 1 mvf2 + mghf 24 14 244 3 1 2 = mv0 + mgh0 24 14 244 3 Ef E0 Solving this equation for hf gives hf = 2 v0 − vf2 2g + h0 SOLUTION In order to use this expression, we must obtain the values for the velocities v0 and vf. The initial velocity has only a horizontal component, v0 = v0 x .The final velocity also has only a horizontal component since the ball is at the top of its trajectory, vf = vf x . No forces act in the horizontal direction so the momentum of the ball in this direction is conserved, hence v0 = vf . Therefore, hf = h0 = 3.00 m 55. REASONING a. During the collision between the bullet and the wooden block, linear momentum is conserved, since no net external force acts on the bullet and the block. The weight of each is balanced by the tension in the suspension wire, and the forces that the bullet and block exert on each other are internal forces. This conservation law will allow us to find the speed of the bullet/block system immediately after the collision. b. Just after the collision, the bullet/block rise up, ultimately reaching a final height hf before coming to a momentary rest. During this phase, the tension in the wire (a nonconservative force) does no work, since it acts perpendicular to the motion. Thus, the work done by nonconservative forces is zero, and the total mechanical energy of the system is conserved. An application of this conservation law will enable us to determine the height hf. SOLUTION a. The principle of conservation of linear momentum states that the total momentum after the collision is equal to that before the...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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