hw 21 sol - Patel (pnp223) homework 21 Turner (92160) 1...

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Unformatted text preview: Patel (pnp223) homework 21 Turner (92160) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points An attack helicopter is equipped with a 20- mm cannon that fires 135 g shells in the forward direction with a muzzle speed of 324 m / s. The fully loaded helicopter has a mass of 1770 kg. A burst of 89 . 8 shells is fired in a 5 . 33 s interval. What is the resulting average force on the helicopter? Correct answer: 736 . 933 N. Explanation: The impulse imparted to the shells equals the change in momentum: F av t = mv The mass change is m = n m = (89 . 8 shells) (135 g) = 12 . 123 kg , so the average force is F = v m t = (324 m / s) (12 . 123 kg) 5 . 33 s = 736 . 933 N . Since the velocity of the shells is much greater than the velocity of the helicopter, there is no need to use relative velocity. 002 (part 2 of 2) 10.0 points By what amount is its forward speed re- duced? Correct answer: 2 . 21913 m / s. Explanation: From conservation of momentum p = 0, so v = mv M = (12 . 123 kg) (324 m / s) 1770 kg = 2 . 21913 m / s , 003 10.0 points A(n) 2 . 9 kg object moving at a speed of 4 . 3 m / s strikes a(n) 0 . 98 kg object initially at rest. Immediately after the collision, the 2 . 9 kg object has a velocity of 1 . 2 m / s directed 29 from its initial line of motion. What is the speed of the 0 . 98 kg object immediately after the collision? Correct answer: 9 . 77155 m / s. Explanation: Let : m 1 = 2 . 9 kg , v 1 i = 4 . 3 m / s , v 1 f = 1 . 2 m / s , m 2 = 0 . 98 kg , and = 29 . p 1 i p 1 f p 2 f 2 9 Momentum is conserved, therefore vectorp 1 i = vectorp 1 f + vectorp 2 f p 2 2 f = p 2 1 i + p 2 1 f 2 p 1 i p 1 f cos = (12 . 47 kg m / s) 2 + (3 . 48 kg m / s) 2 2 (12 . 47 kg m / s) (3 . 48 kg m / s) cos29 = 91 . 702 kg 2 m 2 / s 2 , and p 2 f = radicalBig (91 . 702 kg 2 m 2 / s 2 ) = 9 . 57612 kg m / s , so p 2 f = m 2 v 2 f . Solving for v 2 f , we have v 2 f = p 2 f m 2 = (9 . 57612 kg m / s) (0 . 98 kg) = 9 . 77155 m / s . Patel (pnp223) homework 21 Turner (92160) 2 004 (part 1 of 2) 10.0 points A 1480 kg car skidding due north on a level frictionless icy road at 265 . 95 km / h collides with a 2234 . 8 kg car skidding due east at 197 km / h in such a way that the two cars stick together....
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This note was uploaded on 07/11/2009 for the course PHY 92160 taught by Professor Turner during the Spring '09 term at University of Texas at Austin.

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hw 21 sol - Patel (pnp223) homework 21 Turner (92160) 1...

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