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hw9 - karna(pk4534 HW 9 coker(58245 This print-out should...

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karna (pk4534) – HW 9 – coker – (58245) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. I could find very few problems in the ”bank” that were relevant to Ch. 10 in Ohanian and Markert. 001 (part 1 of 2) 10.0 points An object is moving so that its kinetic energy is 95 . 4 J and the magnitude of its momentum is 47 . 8 kg · m / s. Determine the speed of the object. Correct answer: 3 . 99163 m / s. Explanation: Let : K = 95 . 4 J and p = 47 . 8 kg · m / s . The kinetic energy is given by K = 1 2 m v 2 , and the magnitude of the momentum is p = m v. Dividing one equation by the other yields K p = v 2 . v = 2 K p = 2 (95 . 4 J) 47 . 8 kg · m / s = 3 . 99163 m / s . 002 (part 2 of 2) 10.0 points Determine the mass of the object. Correct answer: 11 . 9751 kg. Explanation: m = p v = 47 . 8 kg · m / s 3 . 99163 m / s = 11 . 9751 kg . 003 10.0 points A balloon of mass M is floating motionless in the air. A person of mass less than M is on a rope ladder hanging from the balloon. The person begins to climb the ladder at a uniform speed v relative to the ground. How does the balloon move relative to the ground? 1. Down with a speed less than v correct 2. Up with speed v 3. Down with speed v 4. Up with a speed less than v 5. The balloon does not move. Explanation: Let the mass of the person be m . Total momentum is conserved (because the exterior forces on the system are balanced), especially the component in the vertical di- rection. When the person begins to move, we have m v + M v M = 0 , = v M = m M v < v = | v M | = m M v < v, since m < M = m M < 1 . Thus the balloon moves in the opposite direction. 004 10.0 points Two blocks of masses m and M [ M = 2 . 45 m ] are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them. A cord holding them together is burned, after which the block of mass M moves to the right with a speed of 2 . 01 m / s.
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karna (pk4534) – HW 9 – coker – (58245) 2 m M m M Before After (a) (b) v What is the speed of the block of mass m ? Correct answer: 4 . 9245 m / s. Explanation: From conservation of momentum Δ p = 0, in our case we obtain 0 = M v M m v m . Therefore v m = M m v M = (2 . 45) v M = (2 . 45) (2 . 01 m / s) = 4 . 9245 m / s . 005 10.0 points A baseball player uses a pitching machine to help him improve his batting average. He places the 77 kg machine on a frozen pond. The machine fires a 0 . 153 kg baseball horizon- tally at a speed of 36 . 8 m / s. What is the magnitude of the recoil velocity of the machine? Correct answer: 0 . 0731221 m / s. Explanation: Let : m 1 = 0 . 153 kg , m 2 = 77 kg , and v 1 f = 36 . 8 m / s . We take the system to consist of the base- ball and the pitching machine. Because of the force of gravity and the normal force, the system is not really isolated. However, both of these forces are directed perpendicularly to the motion of the system. Therefore, momen- tum is constant in the x direction because there are no external forces in this direction (assuming the surface is frictionless). The to- tal momentum of the system before firing is zero. Therefore, the total momentum after firing must be zero, that is, m 1 v 1 f + m 2
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