Physics Solution Manual for 1100 and 2101

The distances above and below the crossing point must

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Unformatted text preview: Upward moving stone In these expressions t is the time it takes for either stone to reach the crossing point, and a is the acceleration due to ydown gravity. Note that yup is the displacement of the upward H moving stone above the base of the cliff, ydown is the displacement of the downward moving stone below the yup top of the cliff, and H is the displacement of the cliff-top above the base of the cliff, as the drawing shows. The distances above and below the crossing point must add to equal the height of the cliff, so we have yup − ydown = H where the minus sign appears because the displacement ydown points in the negative direction. Substituting the two expressions for yup and ydown into this equation gives ( ) up down v0 t + 1 at 2 − v0 t + 1 at 2 = H 2 2 This equation can be solved for t to show that the travel time to the crossing point is t= H u v0 p down − v0 Substituting this result into the expression from Equation 2.8 for yup gives 78 KINEMATICS IN ONE DIMENSION up yup = v0 t + 1 at 2 = v up 0 2 H up down v −v 0 0 H + 1 a up down 2 v0 − v0 2 1 6.00 m 2 = ( 9.00 m/s ) + 2 −9.80 m/s 9.00 m/s − ( −9.00 m/s ) ( ) 6.00 m 9.00 m/s − ( −9.00 m/s ) 2 = 2.46 m Thus, the crossing is located a distance of 2.46 m above the base of the cliff, which is below the halfway point of 3.00 m, as expected. 61. SSM REASONING Once the man sees the block, the man must get out of the way in the time it takes for the block to fall through an additional 12.0 m. The velocity of the block at the instant that the man looks up can be determined from Equation 2.9. Once the velocity is known at that instant, Equation 2.8 can be used to find the time required for the block to fall through the additional distance. SOLUTION When the man first notices the block, it is 14.0 m above the ground and its displacement from the starting point is y = 14.0 m – 53.0 m . Its velocity is given by ( ) 2 Equation 2.9 v 2 = v0 + 2ay . Since the block is moving down, its velocity has a negative value, v = − v0 + 2ay = – ( 0 m/s )2 + 2(–9.80 m/s 2 )(14.0 m – 53.0 m) = –27.7 m/s The block then falls the additional 12.0 m to the level of the man's head in a time t which satisfies Equation 2.8: y = v0t + 1 at 2 2 where y = –12.0 m and v0 = –27.7 m/s . Thus, t is the solution to the quadratic equation 4.90t 2 + 27.7t –12.0 = 0 where the units have been suppressed for brevity. From the quadratic formula, we obtain –27.7 ± (27.7) 2 − 4(4.90)(–12.0) t= = 0.40 s or –6.1 s 2(4.90) The negative solution can be rejected as nonphysical, and the time it takes for the block to reach the level of the man is 0.40 s . ______________________________________________________________________________ Chapter 2 Problems 79 62. REASONING Once its fuel is gone, the rocket is in free fall, so its motion consists of two intervals of constant but different acceleration. We will take upward as the positive direction. From launch to engine burn-out, the acceleration is a1 = +86.0 m/s2, and the rocket’s displacement is y1. Its velocity at the end of the burn, v1, is also the initial velocity for the second portion of its flight: engine burn-out to maximum altitude. During this second portion, the rocket slows down with the acceleration of gravity a2 = −9.80 m/s2 and undergoes an additional displacement of y2 in reaching its maximum height. Its maximum altitude is the sum of these two vertical displacements: h = y1 + y2. SOLUTION First we consider the time period t1 = 1.70 s from the ignition of the engine until the fuel is gone. The rocket accelerates from v0 = 0 m/s to v = v1, rising a displacement ( y1, as given by Equation 2.8 y = v0t + 1 at 2 2 ): 2 2 2 y1 = v0t1 + 1 a1t1 = ( 0 m/s ) t1 + 1 a1t1 = 1 a1t1 2 2 2 (1) Equation 2.4 ( v = v0 + at ) gives its velocity v1 at the instant the fuel runs out: v1 = v0 + a1t1 = 0 m/s + a1t1 = a1t1 (2) From that moment onward, the second part of the rocket’s motion is free fall (a2 = −9.80 m/s2). It takes a time t2 for the rocket’s velocity to decrease from v0 = v1 to ( 2 2 ) v2 = 0 m/s at its maximum altitude. We solve Equation 2.9 v = v0 + 2ay to find its upward displacement y2 during this time: ( 0 m/s ) = v12 + 2a2 y2 2 y2 = or 2 −v1 2a2 Substituting for v1 from Equation (2), we find for y2 that y2 = − ( a1t1 ) 2a2 2 (3) Using Equations (1) and (3), we find that the rocket’s maximum altitude, relative to the ground, is h = y1 + y2 = 1 a t2 2 11 ( a1t1 )2 = 1 a t 2 1 − a1 − 2a2 2 11 a2 80 KINEMATICS IN ONE DIMENSION Using the values given, we find that h= 63. REASONING 1 2 ( 86.0 m/s 2 ) (1.70 s )2 1 − 86.0 m/s 2 = 1210 m −9.80 m/s 2 To find the initial velocity v0,2 of the second stone, we will employ Equation 2.8, y = v0,2t2 + 1 2 at2 2 . In this expression t2 is the time that the second stone is in the air, and it is equal to the time t1 that the first stone is in the air minus the time t3.20 it takes for the first stone to fall 3.20 m: t2 = t1 − t3.20 We can find t1 and t3.20 by applying Equation 2.8 to the first stone. SOLU...
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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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