Physics Solution Manual for 1100 and 2101

The two paths differ by the extra distance covered by

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Unformatted text preview: = 35 m/s. The final velocity is 2 v = v0 + at = 35 m/s + ( 9.8 m/s )(5.0 s) = − 14 m/s (2.4) Instantaneous speed is the magnitude of the instantaneous velocity, so we drop the minus sign and find that Speed = 14 m/s 45. REASONING AND SOLUTION The figure at the right shows the paths taken by the pellets fired from gun A and gun B. The two paths differ by the extra distance covered by the pellet from gun A as it rises and falls back to the edge of the cliff. When it falls back to the edge of the cliff, the pellet from gun A will have the same speed as the pellet fired from gun B, as Conceptual Example 15 discusses. Therefore, the flight time of pellet A will be greater than that of B by the amount of time that it takes for pellet A to cover the extra distance. A ____ B The time required for pellet A to return to the cliff edge after being fired can be found from Equation 2.4: v = v0 + a t. If "up" is taken as the positive direction then v0 = +30.0 m/s and v = –30.0 m/s. Solving Equation 2.4 for t gives t= v − v0 a = (−30.0 m/s) − (+30.0 m/s) = −9.80 m/s2 6.12 s Notice that this result is independent of the height of the cliff. ______________________________________________________________________________ 46. REASONING Assuming that air resistance can be neglected, the acceleration is the same for both the upward and downward parts, namely −9.80 m/s2 (upward is the positive direction). Moreover, the displacement is y = 0 m, since the final and initial positions of the ball are the 70 KINEMATICS IN ONE DIMENSION same. The time is given as t = 8.0 s. ( y = v0 t + 1 a t2 2 Therefore, we may use Equation 2.8 ) to find the initial velocity v0 of the ball. ( ) SOLUTION Solving Equation 2.8 y = v0 t + 1 a t 2 for the initial velocity v0 gives 2 v0 = y − 1 a t2 2 ( ) 0 m − 1 −9.80 m/s2 ( 8.0 s ) 2 2 = = +39 m/s t 8.0 s ______________________________________________________________________________ 47. REASONING AND SOLUTION In a time t the card will undergo a vertical displacement y given by y = 1 at2 2 where a = –9.80 m/s2. When t = 60.0 ms = 6.0 × 10–2 s, the displacement of the card is 0.018 m, and the distance is the magnitude of this value or d1 = 0.018 m . Similarly, when t = 120 ms, d 2 = 0.071 m , and when t = 180 ms, d3 = 0.16 m . ______________________________________________________________________________ 48. REASONING The minimum time that a player must wait before touching the basketball is the time required for the ball to reach its maximum height. The initial and final velocities are known, as well as the acceleration due to gravity, so Equation 2.4 ( v = v0 + a t ) can be used to find the time. SOLUTION Solving Equation 2.4 for the time yields v − v0 0 m / s − 4.6 m / s = 0.47 s a −9.8 m / s 2 ______________________________________________________________________________ t= 49. = SSM REASONING AND SOLUTION Equation 2.8 can be used to determine the displacement that the ball covers as it falls halfway to the ground. Since the ball falls from rest, its initial velocity is zero. Taking down to be the negative direction, we have y = v0t + 1 at 2 = 1 at 2 = 1 (–9.80 m/s 2 )(1.2 s) 2 = –7.1 m 2 2 2 Chapter 2 Problems 71 In falling all the way to the ground, the ball has a displacement of y = –14.2 m . Solving Equation 2.8 with this displacement then yields the time 2y 2(–14.2 m) = = 1.7 s a –9.80 m/s 2 ______________________________________________________________________________ t= 50. REASONING The initial velocity of the compass is +2.50 m/s. The initial position of the compass is 3.00 m and its final position is 0 m when it strikes the ground. The displacement of the compass is the final position minus the initial position, or y = –3.00 m. As the compass falls to the ground, its acceleration is the acceleration due to gravity, a = –9.80 m/s2. Equation 2.8 y = v0t + 1 a t 2 can be used to find how much time elapses ( 2 ) before the compass hits the ground. SOLUTION Starting with Equation 2.8, we use the quadratic equation to find the elapsed time. t= ( 1 a ) ( − y ) = − ( 2.50 m/s ) ± ( 2.50 m/s )2 − 4 ( −4.90 m/s2 ) − ( −3.00 m ) 2 1a 2 2( 2 ) 2 ( −4.90 m/s ) 2 −v0 ± v0 − 4 There are two solutions to this quadratic equation, t1 = 1.08 s and t2 = –0.568 s. The second solution, being a negative time, is discarded. ______________________________________________________________________________ 51. REASONING AND SOLUTION a. 2 v 2 = v0 + 2ay ( ) v = ± (1.8 m/s ) + 2 –9.80 m/s 2 ( –3.0 m ) = ±7.9 m/s 2 The minus is chosen, since the diver is now moving down. Hence, v = −7.9 m/s . b. The diver's velocity is zero at his highest point. The position of the diver relative to the board is y=– 2 v0 2a =– (1.8 m/s )2 ( 2 –9.80 m/s 2 ) = 0.17 m The position above the water is 3.0 m + 0.17 m = 3.2 m . ______________________________________________________________________________ 72 KINEMATICS IN ONE DIMENSION ( ) 2 Equation 2.9 v 2 = v0 + 2ay can be used to find out how far above the 52. REASONING cliff's edge...
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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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