Physics Solution Manual for 1100 and 2101

# 0 kgs so the average force is f m v v 0 300 kgs

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Unformatted text preview: all. 7. REASONING The impulse-momentum theorem (Equation 7.4) states that the impulse of an applied force is equal to the change in the momentum of the object to which the force is applied. We will use this theorem to determine the final momentum from the given value of the initial momentum. The impulse is the average force times the time interval during which the force acts, according to Equation 7.1. The force and the time interval during which it acts are given, so we can calculate the impulse. SOLUTION According to the impulse-momentum theorem, the impulse applied by the retrorocket is J = mvf − mv0 (7.4) The impulse is J = F ∆t (Equation 7.1), which can be substituted into Equation 7.4 to give F ∆t = mvf − mv0 or mvf = F ∆t + mv0 where mvf is the final momentum. Taking the direction in which the probe is traveling as the positive direction, we have that the initial momentum is mv0 = +7.5 × 107 kg⋅m/s and the force is F = −2.0 × 106 N . The force is negative, because it points opposite to the direction of the motion. With these data, we find that the final momentum after the retrorocket ceases to fire is ( ) mvf = F ∆t + mv0 = −2.0 × 106 N (12 s ) + 7.5 × 107 kg ⋅ m/s = +5.1× 107 kg ⋅ m/s 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 impulsemomentum theorem that ΣF = mvf − mv 0 ∆t = ( 75 kg ) ( 0 m/s ) − ( 75 kg ) ( −6.4 m/s ) 2.0 × 10 −3 s = +2.4 × 105 N Chapter 7 Problems 347 b. Again using the impulse-momentum theorem, we find that ΣF = mvf − mv 0 = ∆t ( 75 kg ) ( 0 m/s ) − ( 75 kg ) ( −6.4 m/s ) 0.10 s = +4.8 × 103 N c. The net average force acting on the man is ΣF = FGround + W , where FGround 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 FGround = ΣF − W . Since the weight is W = −mg, we have Stiff − legged FGround = ΣF − W ( ) ( ) = +2.4 × 105 N − − ( 75 kg ) 9.80 m/s 2 = +2.4 × 105 N Knees − bent FGround = ΣF − W = +4.8 × 103 N − − ( 75 kg ) 9.80 m/s 2 = +5.5 × 103 N 9. REASONING AND SOLUTION a. According to Equation 7.4, the impulse-momentum theorem, ( ΣF ) ∆t = mvf − mv0 . Since the only horizontal force exerted on the puck is the force F exerted by the goalie, ΣF = F . Since the goalie catches the puck, vf = 0 m/s . Solving for the average force exerted on the puck, we have F= m( v f − v 0 ) ∆t = ( 0.17 kg ) ( 0 m/s ) − ( +65 m/s ) 5.0 ×10 −3 s = –2.2 ×103 N By Newton’s third law, the force exerted on the goalie by the puck is equal in magnitude and opposite in direction to the force exerted on the puck by the goalie. Thus, the average force exerted on the goalie is +2.2 × 103 N . b. If, instead of catching the puck, the goalie slaps it with his stick and returns the puck straight back to the player with a velocity of –65 m/s, then the average force exerted on the puck by the goalie is F= m( v f − v 0 ) ∆t = 0.17 b kg g(–65 m / s) – (+65 m / s) 5.0 × 10 –3 s = – 4.4 × 10 3 N The average force exerted on the goalie by the puck is thus +4.4 × 103 N . 348 IMPULSE AND MOMENTUM The answer in part (b) is twice that in part (a). This is consistent with the conclusion of Conceptual Example 3. The change in the momentum of the puck is greater when the puck rebounds from the stick. Thus, the puck exerts a greater impulse, and hence a greater force, on the goalie. 10. REASONING During the time interval ∆t, a mass m of water strikes the turbine blade. The incoming water has a momentum mv0 and that of the outgoing water is mvf. In order to change the momentum of the water, an impulse ( ΣF ) ∆t is applied to it by the stationary turbine blade. Now ( Σ F ) ∆t = F ∆t , since only the force of the blade is assumed to act on the water in the horizontal direction. These variables are related by the impulse-momentum theorem, F ∆t = mvf − mv0, which can be solved to find the average force F exerted on the water by the blade. SOLUTION Solving the impulse-momentum theorem for the average force gives F= mv f − mv 0 ∆t = m ( v − v0 ) ∆t f The ratio m / ( ∆t ) is the mass of water per second that strikes the blade, or 30.0 kg/s, so the average force is F= m ( v − v 0 ) = ( 30.0 kg/s ) ( −16.0 m/s ) − ( +16.0 m/s ) = −960 N ∆t f The magnitude of the average force is 960 N . 11. REASONING We will divide this problem into two parts, because the forces acting on the student change abruptly at the instant of impact. In the first part, the student falls freely from rest, under the sole influence of the conservative gravitational force. Thus, the studen...
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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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