Recit%209%20ch%207 - Recit. 9, week of Mar. 21-25: Ch. 7 #...

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Recit. 9, week of Mar. 21-25: Ch. 7 # 14, 32, 40, 45 14. An 85-kg jogger is heading due east at a speed of 2.0 m/s. A 55-kg jogger is heading 32 o north of east at a speed of 3.0 m/s. Find the magnitude and direction of the sum of the momenta of the two joggers. REASONING This is a problem in vector addition, and we will use the component method for vector addition. Using this method, we will add the components of the individual momenta in the direction due north to obtain the component of the vector sum in the direction due north. We will obtain the component of the vector sum in the direction due east in a similar fashion from the individual components in that direction. For each jogger the momentum is the mass times the velocity. SOLUTION Assuming that the directions north and east are positive, the components of the joggers’ momenta are as shown in the following table: Direction due east Direction due north 85 kg jogger 85 2 0 170 kg m / s kg m / s b gb g . = 0 kg m/s 55 kg jogger 55 3 0 32 140 kg m / s kg m / s b gb g . cos ° = 55 3 0 32 87 kg m / s kg m / s b gb g . sin ° = Total 310 kg m /s 87 kg m/s Using the Pythagorean theorem, we find that the magnitude of the total momentum is 310 87 322 2 2 kg m / s kg m / s kg m / s + = b g b g The total momentum vector points north of east by an angle θ , which is given by = F H G I K J = ° - tan 1 87 310 16 kg m / s kg m / s
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32. A car (mass = 1100 kg) is traveling at 32 m/s when it collides head-on with a sport utility vehicle (mass = 2500 kg) traveling in the opposite direction. In the collision, the two vehicles come to a halt. At what speed was the sport utility vehicle traveling? REASONING
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Recit%209%20ch%207 - Recit. 9, week of Mar. 21-25: Ch. 7 #...

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