Download ANSWERS TO QUESTIONS 1. A conservation law states that the

ANSWERS TO QUESTIONS
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A conservation law states that the total amount of some quantity is always constant. One can use
such a law by equating the total amount of the conserved quantity before an interaction or process
to the total amount after. (See Section 3.1.)
Newton’s second law using momentum is more general because it applies to situations in which an
object’s mass is not constant. The earlier form of the second law that uses mass and acceleration
does not. (See pages 88-89.)
Yes. If the horse is not moving its linear momentum is zero. The linear momentum of a moving
turtle would be greater.
The astronaut should face directly away from the spacecraft and throw the tools in that direction.
That will give the astronaut momentum toward the spacecraft.
Collisions.
I walked up and down the attic stairs several times, drove my car around town… in many instances I
or my car did work against the force of gravity. In nearly any movement of my body some of my
muscles are doing work—even when I’m sitting still my heart is pumping blood and doing work.
There are many more possible answers to this question, though “working” hard to think of them
does not count as work in the physics sense.
Without information about the direction of the force and the direction of movement all we can say
is that the work done is between +10 J (if the motion is in exactly the same direction as the force)
and –10 J (if the motion is opposite the direction of the force). Any value in between is possible.
The air bag compresses as the person’s body presses against it, so a smaller force acts over a longer
distance compared to the person hitting the steering wheel or dash. Also the force is spread over a
larger area and the pressure is smaller.
You do not do work on the stairs because the stairs do not move. You do work on yourself. In
particular, with each step your leg raises the rest of your body, thereby doing work on it. When you
lift this leg to the next step, the upper part of your body does work on it.
In physics it is forces that do work. We could speak of “work done on me by the force due to the
earth’s gravity,” or "the work done on me by the earth,” as long as we understand the second
statement means the same thing. Similarly, we can associate the work done by other forces with
the agent responsible for exerting those forces.
You can do work on the basketball and give it kinetic energy by pushing on it with your hand, as in
throwing or dribbling. You could heat it up, too, giving it more internal energy.
Electrical energy flowing in wires and electronic devices, light, sound, gravitational potential energy
of yourself if you are seated in a chair, chemical energy stored inside you from the food you’ve
eaten, internal energy of your warm body and so on.
No. Twice as much work will give the ball twice as much kinetic energy. But since KE is proportional
to the speed squared, the speed will be the square root of 2 (= 1.41 ) times larger.
It must be spinning.
The gravitational potential energy of an object is negative when the object is below the reference
level. A ball in a hole with the level ground above it as the reference level is one example.
Elastic potential energy arises when something “springy” is stretched or compressed. The chair I’m
sitting on has a compressed suspension system that rebounds when I get up, and my bookshelves
are sagging in the middle due to the weight of the books—if they’re not simply permanently bent
out of shape they have elastic potential energy.
17. a) Chemical energy in the camper’s body converted to internal energy both in the camper’s body
and the hot sticks. b) Elastic potential energy in the bow converted to kinetic energy of the arrow,
which is converted to gravitational potential energy of the arrow as it rises to the high point—plus
some internal energy due to heating by air resistance. c) Chemical energy in the carpenter’s body
converted into kinetic energy of the hammer which is converted into internal energy of the atoms
in the nail. d) Kinetic energy and gravitational potential energy of the meteoroid converted to
internal energy via heating due to air resistance.
18. Nuclear energy in the sun is converted to light energy, then converted into electrical energy in the
solar cells, then converted into chemical energy in the battery as it is charged, then converted into
heat energy and light energy in the spotlights, and the light energy is converted into internal energy
when it is absorbed by the surroundings. If some of the light is absorbed by plants or solar cells,
some of the energy is converted to chemical energy (in the plants) or electrical energy (in the solar
cells).
19. By building up speed the truck gains kinetic energy which can be converted into gravitational
potential energy as it climbs the hill. High jumpers and pole vaulters do something similar.
20. As the ball falls, its potential energy is continuously converted into kinetic energy, plus a little bit of
heat because of air resistance. When the ball strikes the floor, its kinetic energy is converted to
elastic potential energy as it is flattened. Most of this is immediately converted back into kinetic
energy as the ball rebounds, but some of the elastic potential energy is converted into internal
energy. On its way up, kinetic energy is converted into potential energy until it reaches its highest
point and stops. This point is below the original starting point because the ball has less energy—
some of its original energy has been converted into internal energy.
21. The maximum height would be greater on the moon. Since the same amount of KE is converted
into PE—which is given by mgd —d will be about six times larger on the moon since g is about one
sixth as large as it is on earth.
22. In an elastic collision, the total kinetic energy of the colliding objects is conserved. Two billiard balls
colliding and bouncing apart is (nearly) an elastic collision. In an inelastic collision, the total kinetic
energy after the collision is not the same as it was before. Two ice-skaters colliding and clinging
together is one example of an inelastic collision. (See Section 3.6.)
23. Football tackles where the tackler “wraps up” and hangs onto the opponent would be inelastic, as
would similar collisions between participants in other sports like hockey or soccer or figure skating.
Things like the impact of the head of a golf club with the ball, a volleyball being hit by a hand, or a
bowling ball hitting a pin are probably closer to elastic collisions.
24. To be stopped after the collision the total linear momentum must be zero. Consequently the total
linear momentum before the collision must also be zero. Since B’s mass is one half that of A, B’s
speed before the collision must be exactly twice that of A.
25. Yes. A spacecraft can gain energy from a close pass of a planet in a “slingshot” maneuver, for
instance. Any “action at a distance” force between two objects can transfer energy between them
if it pulls or pushes at least partly in the direction of movement.
26. The crane with double the power must be lifting its beam twice as fast.
27. It takes essentially the same amount of work to run up stairs as it does to walk up. It takes more
power to run up, so one’s energy reserves are tapped more quickly in this case and resting is
required.
28. Since the satellite’s angular momentum is given by mvr, v can decrease while r increases to keep
the angular momentum constant. A satellite orbiting the earth has its highest speed at perigee
(closest approach) and lowest speed at apogee (highest point), but at every point in its orbit its
angular momentum is the same.
29. Pulling in the legs reduces the radius of the legs’ motion, reducing the angular momentum of the
legs. But since the total angular momentum of the diver stays the same (it is conserved), the diver’s
rotation rate increases—the diver spins faster and can complete more rotations before hitting the
water.
30. If the body’s orbital (“moving in a circle”) and spin angular momentum components are equal and
opposite, its total angular momentum will be zero. This will be true when the orbital and spin
angular momenta have the same magnitude and the orbital and spinning motions are in opposite
senses—one clockwise, the other counterclockwise.
ANSWERS TO EVEN NUMBERED PROBLEMS
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The speedboat. (The houseboat has a momentum of 10,000 kgm/s, the speedboat has momentum
12,000 kgm/s.)
F = 240 N
v = 12m/s
Ratio of speeds = 1.5. Speed of 60 kg skater is 1.5 times the speed of the 90 kg skater.
v = –5.33 m/s
a) Work = 100 ft-lb
b) F = 33 lb
PE = 39,200 J
a) KE = 3.4 × 1011 J See Example 3.6.
b) PE = 6.0 × 1010 J See Example 3.7.
a) KE = 0.5 J
b) v = 7.1 m/s
v = 15.3m/s
v = 22.9 m/s = 51.2 mph
a) d = 7.3m = 24ft
b) The world record has changed very often in recent years. As of 1990 it was about 20 ft.
a) Assuming he was thrown straight upward, his kinetic energy at launch would all become
gravitational potential energy at his maximum height of 10 m. This gives v = 14 m/s.
b) he would land at the same speed he was launched, 14 m/s.
KE lost = 54,000 J
a) t = 29.4 s
b) t = 14.7 s
a) Work = 392,000 J
b) P = 26,100 W
a) Work = 113,400 ft-lb
b) P = 163ft-lb/s = 0.296 hp
ANSWERS TO CHALLENGES
1.
Assuming all of the collisions are inelastic, a) would result in the least amount of damage and b) and
c) would result in about the same amount of damage. In a), the car would not be decelerated to a
complete stop—assuming the two cars become interlocked—so the forces on the car would be
smaller and cause less damage. In both b) and c) the car would come to a complete stop—assuming
that in c) the two cars again interlock—in about the same amount of time. The forces acting on the
car in each case would be about the same resulting in about the same amount of damage.
2.
The total mass of one cart and the mass of the cart on which the object to be “weighed” is placed
would have to be known. The ratio of the speeds of the carts after the plunger is released would
have to be measured. If the carts’ wheels have negligible friction, these could be measured by
placing a backstop in the path of each and, by trial and error, moving them until the two carts hit
the two backstops at the same time. The ratio of the speeds would be the same as the ratio of the
two distances that the carts traveled. Or the individual speeds could be measured using ultrasonic
ranging modules or photogates. The mass is ultimately computed using the equation
3.
For the collision:
(mv)before = (mv)after
0.01 kg = vbefore = 2.01 kg vafter
vbefore = (2.01 kg ÷ 0.01 kg) x vafter
vbefore = 201 vafter
For the swing upward:
vafter =
2gd  2  9.8m / s2  0.1m = 1.4 m/s
vbefore = 201 vafter = 201 x 1.4 m/s
vbefore = 281.4 m/s
4.
a) (mv)before = (mv)after
0.01 kg x vbefore = 2.01 kg x vafter
(mv)before = 4,000 kg x 10 m/s + 1,000 kg x (–20 m/s)
(mv)before = 20,000 kg-m/s (east)
(mv)after = (4,000 kg + 1,000 kg) x v
(mv)after = (5,000 kg) x v
(mv)before = (mv)after
20,000 kg-m/s = (5,000 kg) x v
v = 4 m/s (east)
v1
m
 2 .
v2
m1
b) KEbefore = (1/2) x (4,000 kg) x (10 m/s)2 + (1/2) x (1,000 kg) x (–20 m/s)2
KEbefore = 400,000 J
KEafter = (1/2) x (5,000 kg) x (4 m/s)2
KEafter = 40,000 J
KElost = 360,000 J
5.
If the length of the chain is represented by r, the person drops a vertical distance r and at the low
point has speed:
v  2gd  2gr
The person moves along a circular path with radius r. So:
a = (v2) ÷ r = (2gr) ÷ r
a = 2g
6.
As the car decelerates, it does work as its kinetic energy is converted into internal energy, heating
the brakes. The distance that it takes to bring the car to a stop is determined by equating the work
and the original kinetic energy.
Work = KE
Fd  (1 / 2)mv 2
mv2
d
2F
Constant deceleration means the force is constant. Therefore the stopping distance is proportional
to the square of the speed. If the speed is doubled, the stopping distance is quadrupled.
7.
a) KE = (1/2) x (7.3 kg) x (14 m/s)2 = 715.4 J
b) Work = KE
Fd = KE
F = (KE) ÷ d = (715.4 J) ÷ (3 m)
F = 238.5 N
c) P = Work ÷ t = (715.4 J) ÷ (0.5 s)
P = 1,431 W = 1.92 hp
8.
Since angular momentum is conserved, mvr is constant. If r becomes 60 times as large, v becomes
1/60th as large.
v = (1/60) = 54,500 m/s
v = 908 m/s