Physics Solution Manual for 1100 and 2101

Relative to its initial position the vertical

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Unformatted text preview: the bullet after it enters the building is, taking down as the negative direction, equal to –0.50 m. Therefore, the vertical component of the velocity of the bullet as it passes through the window is, from Equation 3.5b, v0 y (window) = 1 2 y − a yt 2 t = y1 –0.50 m 1 − a yt = − (−9.80 m/s 2 )(0.0203 s) = –24.5 m/s 2 t 0.0203 s 2 The vertical displacement of the bullet as it travels between the buildings is (according to Equation 3.6b with v0 y = 0 m/s) y= v2 y 2a y = (−24.5 m/s)2 = −30.6 m 2(–9.80 m/s 2 ) Therefore, the distance H is H = 30.6 m + 0.50 m = 31 m The time for the bullet to reach the window, according to Equation 3.4b, is t1 = 2y 2 y 2(–30.6 m) = = = 2.50 s v0 y + v y v y (–24.5 m/s) 134 KINEMATICS IN TWO DIMENSIONS Hence, the distance D is given by D = v0 xt1 = (340 m/s)(2.50 s) = 850 m 50. REASONING AND SOLUTION Let H be the initial height of the can above the muzzle of the rifle. Relative to its initial position the vertical coordinate of the bullet at time, t, is y = v0 y t − 1 gt 2 2 Relative to the can’s initial position the vertical coordinate of the can at the same time is 2 y' = (1/2)gt . NOTE: y = H – y' if the bullet hits the can. Then y = v0yt – y'. The horizontal distance traveled by the bullet in time t is x = v0xt. Solving for t and substituting gives y = (v0y/v0x)x – y' Now v0y/v0x = tan θ and it is seen from the figure in the text that H = x tan θ if the bullet is to hit the can so y = H – y'. Hence, both objects are at the same place at the same time, and the bullet will always strike the can. 51. REASONING The drawing shows the trajectory of the ball, along with its initial speed v0, horizontal displacement x, and vertical displacement y. The angle that the initial velocity of the ball makes with the horizontal is θ. The known data are shown in the tables below: +y +x v0 y θ x x-Direction Data x ax +26.9 m 0 m/s vx 2 v0x +(19.8 m/s) cos θ t Chapter 3 Problems 135 y-Direction Data y ay +2.74 m −9.80 m/s2 vy t v0y +(19.8 m/s) sin θ There are only two known variables in each table, so we cannot directly use the equations of kinematics to find the angle θ. Our approach will be to first use the x direction data and obtain an expression for the time of flight t in terms of x and v0x. We will then enter this expression for t into the y-direction data table. The four variables in this table, y, ay, v0y, and t, can be related by using the appropriate equation of kinematics. This equation can then be solved for the angle θ. SOLUTION Using the x-direction data, Equation 3.5a can be employed to find the time t that the ball is in the air: ( since a x = v0 x t + 1 ax t 2 = v0 x t 2 Solving for t gives t= x = 0 m/s 2 ) x +26.9 m = v0 x + (19.8 m/s ) cos θ Using the expression above for the time t and the data in the y-direction data table, the displacement in the y direction can be written with the aid of Equation 3.5b: y = v0 y t + 1 a y t 2 2 +26.9 m + +2.74 m = 19.8 m/s sin θ 19.8 m/s cos θ 144 2444 4 3 ( ) ( ) =t 2 + 6.9 1 ( −9.80 m/s ) (19.82m/s )m θ 2 cos 144 2444 4 3 2 = t2 Evaluating the numerical factors and using the fact that sin θ /cos θ = tan θ , the equation above becomes ( −9.04 m ) +2.74 m = ( +26.9 m ) tan θ + cos 2 θ 1 Using = 1 + tan 2 θ , this equation can be rearranged and placed into a quadratic form: 2 cos θ ( −9.04 m ) tan 2 θ + ( 26.9 m ) tan θ − 11.8 m = 0 The solutions to this quadratic equation are 136 KINEMATICS IN TWO DIMENSIONS tan θ = −26.9 m ± ( 26.9 m ) − 4 ( −9.04 m )( −11.8 m ) = 2 ( −9.04 m ) 2 0.535 and 2.44 The two angles are θ1 = tan −1 0.535 = 28.1° and θ 2 = tan −1 2.44 = 67.7° ______________________________________________________________________________ 52. REASONING Since car A is moving faster, it will eventually catch up with car B. Each car is traveling at a constant velocity, so the time t it takes for A to catch up with B is equal to the displacement between the two cars (x = +186 m) divided by the velocity vAB of A relative to B. (If the relative velocity were zero, A would never catch up with B). We can find the velocity of A relative to B by using the subscripting technique developed in Section 3.4 of the text. vAB = velocity of car A relative to car B vAG = velocity of car A relative to the Ground = +24.4 m/s vBG = velocity of car B relative to the Ground = +18.6 m/s We have chosen the positive direction for the displacement and velocities to be the direction in which the cars are moving. The velocities are related by vAB = vAG + vGB SOLUTION The velocity of car A relative to car B is vAB = vAG + vGB = +24.4 m/s + (−18.6 m/s) = +5.8 m/s, where we have used the fact that vGB = − vBG = −18.6 m/s. The time it takes for car A to catch car B is x +186 m t= = = 32.1 s v AB +5.8 m/s ______________________________________________________________________________ Chapter 3 Problems 53. SSM REASONING The velocity vSG of the swimmer relative to the ground is the ve...
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