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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

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.

Page1 / 8

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online