h2_s11 - Phys 2101 Homework 2 Solution Spring `11 These...

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Phys 2101 Homework 2 Solution Spring `11 These solutions use the parameter values from the problems printed in the book, not those that appear in your personal homework assignment. The logic used to get to the answer is the same, however. 1. P.4-38. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The coordinate origin is at the release point (the initial position for the ball as it begins projectile motion in the sense of §4-5), and we let 0 be the angle of throw (shown in the figure). Since the horizontal component of the velocity of the ball is v x = v 0 cos 40.0°, the time it takes for the ball to hit the wall is 22.0 1.15 s. 25.0 cos 40.0 x x t v (a) The vertical distance is 22 00 11 ( sin ) (25.0sin 40.0 )(1.15) (9.80)(1.15) 12.0 m. y v t gt   (b) The horizontal component of the velocity when it strikes the wall does not change from its initial value: v x = v 0 cos 40.0° = 19.2 m/s. (c) The vertical component becomes (using Eq. 4-23) sin 25.0 sin 40.0 4.80 m/s. y v v gt   (d) Since v y > 0 when the ball hits the wall, it has not reached the highest point yet. 2. We denote the two forces F F 1 2 and . According to Newton’s second law, F F ma F ma F 1 2 2 1 = so , . (a) In unit vector notation i N F ˆ ) 20 ( 1 and         2 2 2 2 ˆ ˆ ˆ ˆ 12.0 sin 30.0 m/s i 12.0 cos 30.0 m/s 6.00 m/s i 10.4m/s j. j a    Therefore,               2 ˆ ˆ ˆ 2.00kg 6.00 m/s i 2.00 kg 10.4 m/s j 20.0 N i ˆˆ 32.0 N i 20.8 N j. F  
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(b) The magnitude of F 2 is 2 2 2 2 2 2 2 | | ( 32.0 N) ( 20.8 N) 38.2 N. xy F F F   (c) The angle that F 2 makes with the positive x axis is found from tan = ( F 2 y / F 2 x ) = [( 20.8 N)/( 32.0 N)] = 0.656. Consequently, the angle is either 33.0° or 33.0° + 180° = 213°. Since both the x and y components are negative, the correct result is 213°. An alternative answer is 213 360 147     . 3. Since the tire remains stationary, by Newton’s second law, the net force must be zero: net 0. A B C F F F F ma From the free-body diagram shown on the right, we have net, net, 0 cos cos 0 sin sin x C A y A C B F F F F F F F   To solve for B F , we first compute . With 220 N A F , 170 N C F and 47 ,  we get cos (220 N)cos47.0 cos 0.883 28.0 170 N A C F F  Substituting the value into the second force equation, we find sin sin (220 N)sin 47.0 (170 N)sin 28.0 241 N. B A C F F F 
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4. The solutions to parts (a) and (b) have been combined here. The free-body diagram is shown below, with the tension of the string T , the force of gravity mg , and the force of the air F . Our coordinate system is shown. Since the sphere is motionless the net force on it is zero, and the x and the y components of the equations are: T sin F = 0 T cos mg = 0, where = 37°. We answer the questions in the reverse order.
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h2_s11 - Phys 2101 Homework 2 Solution Spring `11 These...

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