Physics Solution Manual for 1100 and 2101

00 ms cos 750 2329 ms we find v0 xb 600 m 2329

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Unformatted text preview: _______________________________________________ 2 2 Chapter 3 Problems 157 79. REASONING We can use information about the motion of clown A and the collision to determine the initial velocity components for clown B. Once the initial velocity components are known, the launch speed v0 B and the launch angle θ B for clown B can be determined. e j 1 2 SOLUTION From Equation 3.5b y = v 0 y t + 2 a y t we can find the time of flight until the collision. Taking upward as positive and noting for clown A that v0y = (9.00 m/s) sin 75.0° = 8.693 m/s, we have b gc h 1 1.00 m = 8.693 m / s t + 2 –9.80 m / s 2 t 2 Rearranging this result and suppressing the units gives 4.90t 2 – 8.693t + 1.00 = 0 The quadratic equation reveals that t= 8.693 ± –8. 4 b 693g– 4 b.90g1.00g 1.650 s b= 2 b.90 g 4 2 or 0.1237 s Using these values for t with the magnitudes v0xA and v0xB of the horizontal velocity components for each clown, we note that the horizontal distances traveled by each clown must add up to 6.00 m. Thus, v0 xA t + v0 xB t = 6.00 m or v0 xB = 6.00 m – v0 xA t Using v0xA = (9.00 m/s) cos 75.0° = 2.329 m/s, we find v0 xB = 6.00 m – 2.329 m/s = 1.307 m/s 1.650 s or v0 xB = 6.00 m – 2.329 m/s = 46.175 m/s 0.1237 s The vertical component of clown B’s velocity is v0yB and must be the same as that for clown A, since each clown has the same vertical displacement of 1.00 m at the same time. Hence, v0yB = 8.693 m/s (see above). The launch speed of clown B, finally, is 2 2 v 0 B = v 0 xB + v 0 yB . Thus, we find v0 B = or 1 8 b.307 m / sg+ b.693 m / sg = 8.79 m / s 2 2 158 KINEMATICS IN TWO DIMENSIONS v0 B = 46.175 8 b m / sg+ b.693 m / sg = 47.0 m / s 2 2 For these two possible launch speeds, we find the corresponding launch angles using the following drawings, neither of which is to scale: +y +y v0B 8.693 m/s θB 1.307 m/s θB +x 8.693 m/s = 81.5° 1.307 m/s θ = tan –1 v0B 8.693 m/s 46.175 m/s +x 8.693 m/s = 10.7° 46.175 m/s θ = tan –1 Since the problem states that θB > 45°, the solution is v0 B = 8.79 m/s and θ B = 81.5° . CHAPTER 4 FORCES AND NEWTON'S LAWS OF MOTION ANSWERS TO FOCUS ON CONCEPTS QUESTIONS 1. (b) If only one force acts on the object, it is the net force; thus, the net force must be nonzero. Consequently, the velocity would change, according to Newton’s first law, and could not be constant. 2. (d) This situation violates the first law, which predicts that the rabbit’s foot tends to remain in place where it was when the car begins accelerating. The car would leave the rabbit’s foot behind. That is, the rabbit’s foot would swing away from, not toward, the windshield. 3. (e) Newton’s first law states that an object continues in a state of rest or in a state of motion at a constant speed along a straight line, unless compelled to change that state by a net force. All three statements are consistent with the first law. 4. (a) Newton’s second law with a net force of 7560 N – 7340 N = 220 N due north gives the answer directly. 5. (c) Newton’s second law gives the answer directly, provided the net force is calculated by vector addition of the two given forces. The direction of the net force gives the direction of the acceleration. 6. (e) Newton’s second law gives the answer directly. One method is to determine the total acceleration by vector addition of the two given components. The net force has the same direction as the acceleration. 7. (e) Answers a and b are false, according to the third law, which states that whenever one body exerts a force on a second body, the second body exerts an oppositely directed force of equal magnitude on the first body. It does not matter whether one of the bodies is stationary or whether it collapses. Answer c is false, because according to the third law, Sam and his sister experience forces of equal magnitudes during the push-off. Since Sam has the greater mass, he flies off with the smaller acceleration, according to the second law. Answer d is false, because in catching and throwing the ball each astronaut applies a force to it, and, according to the third law, the ball applies an oppositely directed force of equal magnitude to each astronaut. These reaction forces accelerate the astronauts away from each other, so that the distance between them increases. 8. (b) Newton’s third law indicates that Paul and Tom apply forces of equal magnitude to each other. According to Newton’s second law, the magnitude of each of these forces is the mass times the magnitude of the acceleration. Thus, we have mPaulaPaul = mTomaTom, or mPaul/mTom = aTom/aPaul. 160 FORCES AND NEWTON'S LAWS OF MOTION 9. (e) Newton’s law of gravitation gives the answer directly. According to this law the weight is directly proportional to the mass of the planet, so twice the mass means twice the weight. However, this law also indicates that the weight is inversely proportional to the square of the planet’s radius, so three times the radius means one ninth the weight. Together, th...
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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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