Physics Solution Manual for 1100 and 2101

We have that xa v0 x ta and xb v0 xtb or substituting

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Unformatted text preview: v0y t 0 m/s tB With these data, Equation 3.5b gives y = v0 y t + a y t 2 = ( 0 m/s ) t + a y t 2 = a y t 2 1 2 1 2 1 2 (1) SOLUTION Applying Equation (1) to both shots, we find that yB yA = −HB −H A = 1 a t2 2 yB 1 a t2 2 yA or HB HA = 2 tB (2) 2 tA To use this result to calculate the ratio HB/HA, it is necessary to determine the times tA and tB. To do this, we consider the horizontal part of the motion and note that there is no acceleration in the horizontal direction. Therefore, the horizontal component v0x of the bullet’s initial velocity remains unchanged throughout the motion, and the horizontal component x of the displacement is simply v0x times the time t during which the motion occurs. We have that xA = v0 x tA and xB = v0 xtB or Substituting these results into Equation (2) gives tA = xA v0 x and tB = xB v0 x (3) Chapter 3 Problems 2 x H B tB ( xB / v0 x ) =2= = B H A tA ( x / v )2 xA A 0x 2 121 2 It is given that xB = 2xA, so that we find HB HA 37. x = B x A 2 2 xA = xA 2 =4 SSM REASONING a. The drawing shows the initial velocity v0 of the package when it is released. The initial speed of the package is 97.5 m/s. The component of its displacement along the ground is labeled as x. The data for the x direction are indicated in the data table below. +y v0 +x 50.0° y x x-Direction Data x ax ? 0 m/s2 vx v0x t +(97.5 m/s) cos 50.0° = +62.7 m/s Since only two variables are known, it is not possible to determine x from the data in this table. A value for a third variable is needed. We know that the time of flight t is the same for both the x and y motions, so let’s now look at the data in the y direction. y-Direction Data y ay −732 m −9.80 m/s vy 2 v0y t +(97.5 m/s) sin 50.0° = +74.7 m/s ? Note that the displacement y of the package points from its initial position toward the ground, so its value is negative, i.e., y = −732 m. The data in this table, along with the appropriate equation of kinematics, can be used to find the time of flight t. This value for t can, in turn, be used in conjunction with the x-direction data to determine x. 122 KINEMATICS IN TWO DIMENSIONS +y b. +x The drawing at the right shows the velocity of the package just before impact. The angle that the velocity makes with respect to the ground can be found from the inverse ( ) tangent function as θ = tan −1 v y / vx . Once the time has been found in part (a), the values of vy and vx can be determined from the data in the tables and the appropriate equations of kinematics. vx θ vy v SOLUTION a. To determine the time that the package is in the air, we will use Equation 3.5b (y =v 0y ) t + 1 a y t 2 and the data in the y-direction data table. Solving this quadratic equation 2 for the time yields t= t= ( a )(− y) 2( a ) 2 −v0 y ± v0 y − 4 1 2 − ( 74.7 m/s ) ± 1 2 y y ( 74.7 m/s ) − 4 ( 1 ) ( −9.80 m/s 2 ) ( 732 m ) 2 = 1 2 ( 2 ) ( −9.80 m/s 2 ) 2 − 6.78 s and 22.0 s We discard the first solution, since it is a negative value and, hence, unrealistic. The displacement x can be found using t = 22.0 s, the data in the x-direction data table, and Equation 3.5a: x = v0 x t + 1 ax t 2 = ( +62.7 m/s )( 22.0 s ) + 2 ( 0 m/s2444) = +1380 m ) ( 22.0 s3 1444 1 2 2 2 =0 b. The angle θ that the velocity of the package makes with respect to the ground is given by ( ) θ = tan −1 v y / vx . Since there is no acceleration in the x direction (ax = 0 m/s2), vx is the same as v0x, so that vx = v0x = +62.7 m/s. Equation 3.3b can be employed with the y-direction data to find vy : v y = v0 y + a y t = +74.7 m/s + ( −9.80 m/s 2 ) ( 22.0 s ) = −141 m/s Therefore, vy −1 −141 m/s = tan = −66.0° vx +62.7 m/s θ = tan −1 where the minus sign indicates that the angle is 66.0° below the horizontal . ______________________________________________________________________________ Chapter 3 Problems 123 38. REASONING We can obtain an expression for the car’s initial velocity v0 by starting with the relation x = v0 x t + 1 a x t 2 (Equation 3.5a). Here x is the horizontal component of the car’s 2 displacement, v0x is the horizontal component of the car’s initial velocity (v0x = v0 for a horizontal launch), and ax is its acceleration in the horizontal direction (ax = 0 m/s2 for projectile motion). Solving for the initial velocity, we find that ( ) x = v0t + 1 0 m/s2 t 2 = v0t 2 or v0 = x t (1) The time t in Equation (1) is known, but x is not. To find x, we note that at t = 1.1 s, the car’s displacement has a magnitude of ∆r = 7.0 m. The displacement ∆r of the car has a horizontal component x and a vertical component y. The magnitude of the displacement is related to x and y by the Pythagorean theorem: (2) ( ∆r )2 = x 2 + y 2 Solving Equation (2) for x and substituting the result into Equation (1) gives v0 = x = t ( ∆r )2 − y 2 (3) t To determine y, we turn to the relation y = v0 yt + 1 a yt 2 (Equation 3.5b). Setting 2 v0y = 0 m/s (again, for a horizontal launch), we find that Equation (3) becomes y = ( 0...
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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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