RCW5 solutions - Recitation ClassWork 5 solutions 1 Chapter 7 problem 3 Physics 203 REASONING The impulse that the roof of the car applies to the

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Recitation ClassWork 5 solutions Physics 203 1. Chapter 7 problem 3 REASONING The impulse that the roof of the car applies to the hailstones can be found from the impulse- momentum theorem, Equation 7.4. Two forces act on the hailstones, the average force F exerted by the roof, and the weight of the hailstones. Since it is assumed that F is much greater than the weight of the hailstones, the net average force ( ) Σ F is equal to F . SOLUTION From Equation 7.4, the impulse that the roof applies to the hailstones is: ( ) Impulse Final Initial momentum momentum tm m m ∆= = f 0 f0 F v v vv  Solving for F (with up taken to be the positive direction) gives [ ] ( ) (0.060 kg/s) ( 15 m/s) ( 15 m/s) = +1.8 N m t  = = + −−   F This is the average force exerted on the hailstones by the roof of the car. The positive sign indicates that this force points upward. From Newton's third law, the average force exerted by the hailstones on the roof is equal in magnitude and opposite in direction to this force. Therefore, Force on roof = 1.8 N The negative sign indicates that this force points downward . 2. Problem 8 REASONING We will apply the impulse momentum theorem as given in Equation 7.4 to solve this problem. From this theorem we know that, for a given change in momentum, greater forces are associated with shorter time intervals. Therefore, we expect that the force in the stiff-legged case will be greater than in the knees-bent case. SOLUTION a. Assuming that upward is the positive direction, we find from the impulse-momentum theorem that ( )( ) ( )( ) 5 3 75 kg 0 m/s 75 kg 6.4 m/s 2.4 10 N 2.0 10 s mm t Σ= = =+ × × F b. Again using the impulse-momentum theorem, we find that ( )( ) ( )( ) 3 75 kg 0 m/s 75 kg 6.4 m/s 4.8 10 N 0.10 s t = F c. The net average force acting on the man is Ground FF W , where F Ground is the average upward force exerted on the man by the ground and W is the downward-acting weight of the man. It follows, then, that = Σ− Ground F FW . Since the weight is W = mg , we have
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( ) ( ) ( ) ( ) 5 25 3 23 2.4 10 N 75 kg 9.80 m/s 4.8 10 N 75 kg 9.80 m/s 5.5 10 N = Σ−  = =  = = = Ground Ground Stiff legged F F W Knees bent F F W 3. Problem 12 REASONING AND SOLUTION According to the impulse-momentum theorem (Equation 7.4) ( ) () tm Σ ∆= f0 F vv (1) Conservation of mechanical energy can be used to relate the velocities to the heights. If the floor is used to define the zero level for the heights, we have mgh mv 0 1 2 2 = B where h 0 is the height of the ball when it is dropped and v B
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This note was uploaded on 06/22/2009 for the course PHYS 1101 taught by Professor Richardson, b during the Fall '08 term at Cornell University (Engineering School).

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RCW5 solutions - Recitation ClassWork 5 solutions 1 Chapter 7 problem 3 Physics 203 REASONING The impulse that the roof of the car applies to the

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