Physics Solution Manual for 1100 and 2101

According to the impulsemomentum theorem equation 74

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Unformatted text preview: ION Applying Equation 7.1, we write the impulse of each average force as follows: J1 = F1∆t1 and J 2 = F 2 ∆t 2 But the impulses J1 and J2 are the same, so we have that F1∆t1 = F 2 ∆t2 . Writing this result in terms of the magnitudes of the forces gives F 1∆t1 = F 2 ∆t2 or F ∆t2 = 1 ∆t1 F2 The ratio of the force magnitudes is given as F 1 / F 2 = 3 , so we find that F ∆t2 = 1 ∆t1 = 3 ( 3.2 ms ) = 9.6 ms F2 2. REASONING The impulse-momentum theorem, as expressed in Equation 7.4, states that the impulse acting on each car is equal to the final momentum of the car minus its initial momentum: ΣF) ∆ (424t = 13 Impulse m vf − m v0 { { Final momentum Initial momentum or ΣF = m vf − m v0 ∆t where ΣF is the net average force that acts on the car, and ∆t is the time interval during which the force acts. SOLUTION We assume that the velocity of each car points in the +x direction. The net average force acting on each car is: 344 IMPULSE AND MOMENTUM Car A m v f − m v 0 (1400 kg ) ( +27 m/s ) − (1400 kg ) ( 0 m/s ) = = +4200 N 9.0 s ∆t Car B 3. ΣF = ΣF = m v f − m v 0 (1900 kg )( +27 m/s ) − (1900 kg )( 0 m/s ) = = +5700 N ∆t 9.0 s 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 ( ΣF ) is equal to F . SOLUTION From Equation 7.4, the impulse that the roof applies to the hailstones is: F2 mvf mv 0 1 ∆t = { − { = m ( v f − v 0 ) 3 Impulse Final momentum Initial momentum Solving for F (with up taken to be the positive direction) gives m F = ( vf − v 0 ) = (0.060 kg/s) [ (+15 m/s) − (−15 m/s)] = +1.8 N ∆t 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 . 4. REASONING According to the impulse-momentum theorem, the rocket’s final momentum m v f differs from its initial momentum by an amount equal to the impulse ( ΣF ) ∆t of the net force exerted on it: ( ΣF ) ∆t = mvf − mv 0 (Equation 7.4). We are ignoring gravitational and frictional forces, so this impulse is due entirely to the force generated by the motor. The magnitude of the motor’s impulse is given as ( ΣF ) ∆t = 29.0 N ⋅ s , so we will obtain the rocket’s final speed by using Equation 7.4. Chapter 7 Problems SOLUTION 345 The rocket starts from rest, so v0 = 0 m/s, and the impulse-momentum theorem becomes ( ΣF ) ∆t = mvf . Therefore, the rocket’s final speed vf is the magnitude of the motor impulse divided by the rocket’s mass: vf = 5. ( ΣF ) ∆t = 29.0 N ⋅ s = 166 m/s m 0.175 kg SSM REASONING The impulse that the volleyball player applies to the ball can be found from the impulse-momentum theorem, Equation 7.4. Two forces act on the volleyball while it’s being spiked: an average force F exerted by the player, and the weight of the ball. As in Example 1, we will assume that F is much greater than the weight of the ball, so the weight can be neglected. Thus, the net average force ( ΣF ) is equal to F . SOLUTION From Equation 7.4, the impulse that the player applies to the volleyball is F2 mv f mv 0 1 ∆t = { − { 3 Impulse Final momentum Initial momentum = m( v f − v 0 ) = (0.35 kg) [ (–21 m/s) – (+4.0 m/s)] = –8.7 kg ⋅ m/s The minus sign indicates that the direction of the impulse is the same as that of the final velocity of the ball. 6. REASONING During the collision, the bat exerts an impulse on the ball. The impulse is the product of the average force that the bat exerts and the time of contact. According to the impulse-momentum theorem, the impulse is also equal to the change in the momentum of the ball. We will use these two relations to determine the average force exerted by the bat on the ball. SOLUTION The impulse J is given by Equation 7.1 as J = F∆t , where F is the average force that the bat exerts on the ball and ∆t is the time of contact. According to the impulsemomentum theorem, Equation 7.4, the net average impulse ( ΣF ) ∆t is equal to the change in the ball’s momentum; ( ΣF ) ∆t = mv f − mv 0 . Since we are ignoring the weight of the ball, the bat’s force is the net force, so ΣF = F . Substituting this value for the net average force into the impulse-momentum equation and solving for the average force gives F= mv f − mv 0 ∆t = ( 0.149 kg ) ( −45.6 m/s ) − ( 0.149 kg ) ( +40.2 m/s ) = 1.10 ×10−3 s −11 600 N 346 IMPULSE AND MOMENTUM where the positive direction for the velocity has been chosen as the direction of the incoming b...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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