Chapter 9: Linear Momentum and CollisionsAnswers to Even-Numbered Conceptual Questions2. Doubling an object’s speed increases its kinetic energy by a factor of four, and its momentum by a factor oftwo.4. No. Consider, for example, a system of two particles. The total momentum of this system will be zero if theparticles move in opposite directions with equal momentum. The kinetic energy of each particle is positive,however, and hence the total kinetic energy is also positive.6. Yes. Just point the fan to the rear of the boat. The resulting thrust will move the boat forward.8. (a) The force due to braking—which ultimately comes from friction with the road—reduces the momentumof the car. The momentum lost by the car does not simply disappear, however. Instead, it shows up as anincrease in the momentum of the Earth. (b) As with braking, the ultimate source of the force accelerating thecar is the force of static friction between the tires and the road.10. Yes. For example, we know that in a one-dimensional elastic collision of two objects of equal mass the objects“swap” speeds. Therefore, if one object is at rest before the collision, it is possible for one object to be at restafter the collision as well. See Figure 9-7(a).12. No. Any collision between cars will be at least partially inelastic, due to denting, sound production, heating,and other effects.14. The center of mass of the hourglass starts at rest in the upper half of the glass and ends up at rest in thelower half. Therefore, the center of mass accelerates downward when the sand begins to fall—to get itmoving downward—and then accelerates upward when most of the sand has fallen—to bring it to restagain. It follows from equation 9-18 that the weight read by the scale is less than Mg when the sand beginsfalling, but is greater than Mg when most of the sand has fallen.16. (a) Assuming a very thin base, we conclude that the center of mass of the glass is at its geometric center ofthe glass. (b) In the early stages of filling, the center of mass is below the center of the glass. When the glassis practically full, the center of mass is again at the geometric center of the glass. Thus, as water is added, thecenter of mass first moves downward, then turns around and moves back upward to its initial position.18. As this jumper clears the bar, a significant portion of his body extends below the bar due to the extremearching of his back. Just as the center of mass of a donut can lie outside the donut, the center of mass of thejumper can be outside his body. In extreme cases, the center of mass can even be below the bar at all timesduring the jump.Solutions to Problems and Conceptual Exercises1. Picture the Problem: The car and the baseball each travel in a straight line at constant speed. The data given in the Exercise 9-1 include pcar=15,800 kg⋅m/s and mball=0.142 kg.Strategy: Use equation 9-1 to set the magnitude of the momentum of the baseball equal to the magnitude of the momentum of the car, and solve for mball.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.9 – 1Chapter 9: Linear Momentum and Collisions James S. Walker, Physics, 4th EditionSolution: Set pball=pcarand solve for vball: pball=mballvball=pcarvball=pcarmball=15,800 kg⋅m/s0.142 kg=1.11×106 ms×3600 sh×1 mi1609 m= 2.49×105 mi/hInsight: The huge required speed for the baseball (3,240 times the speed of sound!) is due to the huge mass difference between the two objects.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.9 – 2Chapter 9: Linear Momentum and Collisions James S. Walker, Physics, 4th Edition2. Picture the Problem: The ducks approach each other and the goose recedes as indicated in the figure at right.Strategy: Sum the velocity vectors from the three birds using the component method to find the total momentum.Solution: 1. Find the momen-tum of the left duck: rpd1=mdvdˆx = 4.00 kg( )1.10 m/s( )ˆx= 4.40 kg⋅m/s( )ˆx2. Find the momentum of the top duck: rpd2=mdvdˆy= 4.00 kg( )−1.10 m/s( )ˆy= −4.40 kg⋅m/s( )ˆy3. Find the momentum of the goose: rpg=mgvgˆy= 9.00 kg( )−1.30 m/s( )ˆy= −11.7 kg⋅m/s( )ˆy4. Add the momenta: rptotal=rpd1+rpd2+rpg=4.40 kg⋅m/sˆx−4.40 kg⋅m/sˆy−11.7 kg⋅m/sˆyrptotal= 4.40 kg⋅m/s( )ˆx+ −16.1 kg⋅m/s( )ˆyInsight: The total momentum vector now has a magnitude of 16.7 kg·m/s and points 74.7° below the positive x axis.3. Picture the Problem: The owner walks slowly toward the northeast while the cat runs eastward and the dog runs northward.Strategy: Sum the momenta of the dog and cat using the component method. Use the known components of the total momentum to find its magnitude and direction. Let north be in the y direction, east in the x direction. Use the momentum together with the owner’s mass to find the velocity of the owner.Solution: 1. Use the component method of vector addition to find the owner’s momentum: rptotal=rpcat+rpdog=mcatrvcat+mdogrvdog= 5.30 kg( )3.04 m/s ˆx( )+ 26.2 kg( )2.70 m/s ˆy( )rptotal= 16.1 kg⋅m/s( )ˆx+ 70.7 kg⋅m/s( )ˆy2. Divide the owner’s momentum by his mass to get the components of the owner’s velocity: rpowner=mownerrvowner=rptotalvowner=rptotalm0=16.1 kg⋅m/s( )ˆx+ 70.7 kg⋅m/s( )ˆy74.0 kg= 0.218 m/s( )ˆx+ 0.955 m/s( )ˆy3. Use the known components to find the direction and magnitude of the owner’s velocity: vowner= 0.218 m/s( )2+ 0.955 m/s( )2= 0.980 m/sθ =tan−10.9550.218⎛⎝⎜⎞⎠⎟= 77.1° north of eastInsight: We bent the rules of significant figures a bit in step 3 in order to avoid rounding error. The owner is moving much slower than either the cat or the dog because of his larger mass.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.9 – 3Chapter 9: Linear Momentum and Collisions James S. Walker, Physics, 4th Edition4. Picture the
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