HW 07 solution

HW 07 solution - gutierrez (ig3472) HW7 ditmire (58216) 1...

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Unformatted text preview: gutierrez (ig3472) HW7 ditmire (58216) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points What is the momentum of a 0.159 kg baseball thrown with a velocity of 38 m / s toward home plate? Correct answer: 6 . 042 kg m / s. Explanation: Let : m = 0 . 159 kg and v = 38 m / s . vectorp = mvectorv p = mv = (0 . 159 kg) (38 m / s) = 6 . 042 kg m / s toward home plate. 002 10.0 points A 4 . 25 kg steel ball strikes a massive wall at 14 . 6 m / s at an angle of 46 . 7 with the plane of the wall. It bounces off with the same speed and angle. x y 1 4 . 6 m / s 1 4 . 6 m / s 46 . 7 46 . 7 If the ball is in contact with the wall for . 208 s , what is the magnitude of average force exerted on the ball by the wall? Correct answer: 409 . 183 N. Explanation: Let : m = 4 . 25 kg , v = 14 . 6 m / s , = 46 . 7 , and t = 0 . 208 s . x y vectorv f m vectorv i m F t = p. Only the component of the balls velocity perpendicular to the wall will change. This velocity component before hitting the wall is v = v cos = (14 . 6 m / s) cos 46 . 7 = 10 . 0129 m / s . After hitting the wall, this component is- 10 . 0129 m / s, because the rebound angle is also 46 . 7 . The change in momentum during contact with the wall is therefore p = mv f- mv = m (- v )- mv =- 2 mv =- 2 (4 . 25 kg) (10 . 0129 m / s) =- 85 . 1101 kg m / s , so the average force on ball is F = vextendsingle vextendsingle vextendsingle vextendsingle p t vextendsingle vextendsingle vextendsingle vextendsingle = 85 . 1101 kg m / s . 208 s = 409 . 183 N . 003 10.0 points A machine gun fires 54 . 3 g bullets at a speed of 732 m / s. The gun fires 141 bullets / min. What is the average force the shooter must exert to keep the gun from moving? gutierrez (ig3472) HW7 ditmire (58216) 2 Correct answer: 93 . 4069 N. Explanation: Let : m = 54 . 3 g = 0 . 0543 kg , v = 732 m / s , and n = 141 bullets / min . To keep the gun from moving, the force you exert should balance the force the machine gun exerts on you. Letting n be the number of bullets fired per minute, we have F = P t = mv n = (0 . 0543 kg) (732 m / s) (141 bullets / min) 1 min 60 s = 93 . 4069 N . 004 10.0 points The separation between the hydrogen and chlorine atoms of the HCl molecule is about 2 . 56 10 10 m. Determine the location of the center of mass of the molecule as measured from the hydro- gen atom. (Chlorine is 35 . 3 times more mas- sive than hydrogen.) Correct answer: 2 . 48948 10 10 m. Explanation: From x cm = m i x i m i , we obtain x cm = m H 0 + n m H d m H + n m H = (35 . 3) (2 . 56 10 10 m) 1 + 35 . 3 = 2 . 48948 10 10 m ....
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This note was uploaded on 04/13/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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HW 07 solution - gutierrez (ig3472) HW7 ditmire (58216) 1...

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