Physics Solution Manual for 1100 and 2101

0kg msg 32 b m cos s 140 5 b5 kg g3kgmm s g 32 b

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Unformatted text preview: t’s total mechanical energy E is conserved up to the instant of impact. We will use the energy conservation principle to determine the student’s initial height H in terms of the student’s velocity vimpact at that instant. The second part of the student’s motion begins at impact, when the force Fground due to the ground overwhelms the gravitational force and brings the student to rest. The force of the ground is nonconservative, so instead of the energy conservation principle, we will apply the impulse-momentum theorem Fground ∆t = mvf − mv0 (Equation 7.4) to analyze the collision. Because this time interval begins at impact, v0 is the student’s impact velocity: v0 = vimpact. Chapter 7 Problems SOLUTION We begin with the energy conservation principle 1 mv 2 f 2 1 mv 2 0 2 349 + mgh0 = + mghf (Equation 6.9b) applied to the student’s fall to the ground. Falling from rest implies v0 = 0 m/s, and the student’s final velocity is the impact velocity: vf = vimpact. Thus, we have 0 + m gh0 = 1 2 2 m vimpact + m ghf or g ( h0 − hf ) = 1 24 43 1 v2 2 impact or H= 2 vimpact 2g (1) H For the student’s collision with the ground, the impulse-momentum theorem gives Fground ∆t = mvf − mv0 (Equation 7.4). The collision brings the student to rest, so we know that vf = 0 m/s, and Equation 7.4 becomes Fground ∆t = −mv0 . Solving for the impact speed v0, we obtain v0 = vimpact = − Fground ∆t (2) m Substituting Equation (2) into Equation (1) yields 2 H= 2 vimpact 2g Fground ∆t − m = Fground ∆t = 2g 2 gm2 ( ) 2 ( +18 000 N )( 0.040 s ) = 6.7 m = 2 2 2 9.80 m/s ( 63 kg ) 2 ( ) 12. REASONING AND SOLUTION According to the impulse-momentum theorem (Equation 7.4) ( Σ F ) ∆t = m ( v f − v 0 ) (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 2 1 mgh0 = 2 mv B where h0 is the height of the ball when it is dropped and vB is the speed of the ball just before it strikes the ground. Solving for vB gives vB = 2 gh0 Similarly, 2 mghf = 1 mvA 2 (2) 350 IMPULSE AND MOMENTUM where hf is the maximum height of the ball when it rebounds and vA is the speed of the ball just after it rebounds from the ground. Solving for vA gives vA = 2 ghf (3) Substituting equations (2) and (3) into equation (1), where v0 = −vB, and vf = vA gives (taking "upward" as the positive direction) ( ΣF ) ∆t = m 2g + hf − ( − h0 ) ( ) = (0.500 kg) 2 9.80 m/s 2 + 0.700 m − ( − 1.20 m ) = +4.28 N ⋅ s Since the impulse is positive, it is directed upward . 13. SSM WWW REASONING The impulse applied to the golf ball by the floor can be found from Equation 7.4, the impulse-momentum theorem: ( ΣF ) ∆t = mv f − mv 0 . Two forces act on the golf ball, the average force F exerted by the floor, and the weight of the golf ball. Since F is much greater than the weight of the golf ball, the net average force ( ΣF ) is equal to F . Only the vertical component of the ball's momentum changes during impact with the floor. In order to use Equation 7.4 directly, we must first find the vertical components of the initial and final velocities. We begin, then, by finding these velocity components. SOLUTION The figures below show the initial and final velocities of the golf ball. Before impact After impact vf sin 30.0° v0 cos 30.0° v0 30.0° vf cos 30.0° 30.0° vf v0 sin 30.0° If we take up as the positive direction, then the vertical components of the initial and final velocities are, respectively, v 0y = −v0 cos 30.0° and vf y = +vf cos 30.0° . Then, from Equation 7.4 the impulse is F ∆t = m( v fy − v 0y ) = m (+ vf cos 30.0°) – (– v0 cos 30.0°) Chapter 7 Problems 351 Since v 0 = v f = 45 m / s , the impulse applied to the golf ball by the floor is F∆t = 2mv0 cos 30.0° = 2(0.047 kg)(45 m/s)(cos 30.0°) = 3.7 N ⋅ s 14. REASONING This is a problem in vector addition, and we will use the component method for vector addition. Using this method, we will add the components of the individual momenta in the direction due north to obtain the component of the vector sum in the direction due north. We will obtain the component of the vector sum in the direction due east in a similar fashion from the individual components in that direction. For each jogger the momentum is the mass times the velocity. SOLUTION Assuming that the directions north and east are positive, the components of the joggers’ momenta are as shown in the following table: Direction due east Direction due north 85 kg jogger 8 b5 kg 170.0kgm⋅ ms/g g2 / s b = 0 kg⋅m/s 55 kg jogger 55 b kg g3.0kg ⋅ /msg 32 ° b m / cos s = 140 5 b5 kg g3.kgmm/ s/ g 32 ° b 0 ssin = 87 ⋅ Total 310 kg⋅m /s 87 kg⋅m/s Using the Pythagorean theorem, we find that the magnitude of the total momentum is 3 87 b10 kg ⋅ m / sg+ b kg ⋅ m / sg = 2 2 322 kg ⋅ m / s The total momentum vector points north of east by an angle θ, which is given by θ = tan −1 F87...
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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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