Chapter 9 Hm 6 Solutions

Chapter 9 Hm 6 Solutions - Solutions to Homework #6 Chapter...

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Unformatted text preview: Solutions to Homework #6 Chapter 9 6. Picture the Problem : The ball falls vertically downward, landing with a speed of 2.5 m/s and rebounding upward with a speed of 2.0 m/s. Strategy: Use equation 9-1 to find the change in momentum of the ball when it rebounds. Solution: 1. (a) Use equation 9-1 to find p r : 2. (b) Subtract the magnitudes of the momenta: 3. (c) The quantity in part (a) is more directly related to the net force acting on the ball during its collision with the floor, first of all because t = F p r r (equation 9-3) and as we can see from above that f i p p - p r . Secondly, we expect the floor to exert an upward force on the ball but we calculated a downward (negative) value in part (b). Insight: If the ball were to rebound at 2.5 m/s upward we would find 2 1.1 kg m/s mv = = p r and f i p p- = . Such a collision with the floor would be called elastic as discussed in section 9-6. 14. Picture the Problem : The ball rebounds from the bat in the manner indicated by the figure at right. Strategy: The impulse is equal to the vector change in the momentum. Analyze the x and y components of p r separately, then use the components to find the direction and magnitude of I r . Solution: 1. (a) Find x p : ( 29 ( 29 ( 29 f i 0.14 kg 36 m/s 5.0 kg m/s x x x p m v v =- =- - = 2. Find y p : ( 29 ( 29 ( 29 f i 0.14 kg 18 0 m/s 2.5 kg m/s y y y p m v v =- =- = 3. Use equation 9-6 to find I r : ( 29 ( 29 5.0 kg m/s 2.5 kg m/s = = + I p x y r r 4. Find the direction of I r : 1 1 2.5 tan tan 27 5.0 y x I I -- = = = above the horizontal 5. Find the magnitude of I r : ( 29 ( 29 2 2 2 2 5.0 kg m/s 2.5 kg m/s 5.6 kg m/s x y I I I = + = + = 6. (b) If the mass of the ball were doubled the impulse would double in magnitude. There would be no change in the direction. 7. (c) If p r of the ball is unchanged, the impulse delivered to the ball would not change, irregardless of the mass of the bat. Insight: The impulse brings the ball to rest horizontally but gives it an initial horizontal speed. Verify for yourself that this ball will travel straight upward 16.5 m (54 feet) before falling back to i v r f v r Earth. An easy popup! 20. Picture the Problem : The astronaut and the satellite move in opposite directions after the astronaut pushes off. The astronaut travels at constant speed a distance d before coming in contact with the space shuttle. Strategy: As long as there is no friction the total momentum of the astronaut and the satellite must remain zero, as it was before the astronaut pushed off. Use the conservation of momentum to determine the speed of the astronaut, and then multiply the speed by the time to find the distance....
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This note was uploaded on 10/17/2011 for the course PHY 101 taught by Professor Ashkenkai during the Fall '08 term at FIU.

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Chapter 9 Hm 6 Solutions - Solutions to Homework #6 Chapter...

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